5
Multiple Integrals and their
Applications
aaaaa
5.1 INTRODUCTION TO DEFINITE INTEGRALS AND DOUBLE INTEGRALS
Definite Integrals
Y
The concept of definite integral
∫a f ( x ) dx
b
…(1)
y = f( x)
is physically the area under a curve y = f(x), (say), the
A
x-axis and the two ordinates x = a and x = b. It is
defined as the limit of the sum
x=a
x=b
f(x1)δx1 + f(x2)δx2 + … + f(xn)δxn
X
O
when n → ∞ and each of the lengths δx1, δx2, …, δxn
Fig. 5.1
tends to zero.
Here δx1, δx2, …, δxn are n subdivisions into which the range of integration has been
divided and x1, x2, …, xn are the values of x lying respectively in the Ist, 2nd, …, nth
subintervals.
Double Integrals
A double integral is the counter part of the above
definition in two dimensions.
Let f(x, y) be a single valued and bounded function of
two independent variables x and y defined in a closed
region A in xy plane. Let A be divided into n elementary
areas δA1, δA2, …, δAn.
Let (xr, yr) be any point inside the rth elementary area
δA r.
Consider the sum
Y
A
δA r
X
O
Fig. 5.2
f ( x1 , y1 ) δ A1 + f ( x2 , y2 ) δ A2 + … + f ( xn , yn ) δ An = ∑ f ( xr , yr ) δ Ar
n
…(2)
r =1
Then the limit of the sum (2), if exists, as n → ∞ and each sub-elementary area approaches
to zero, is termed as ‘double integral’ of f(x, y) over the region A and expressed as ∫ ∫ f ( x, y ) dA .
A
355
356
Engineering Mathematics through Applications
Thus
Lt
∫ ∫ f ( x, y ) dA = n →∞
A
δ Ar → 0
n
∑ f ( xr , yr ) δAr
…(3)
r =1
Observations: Double integrals are of limited use if they are evaluated as the limit of the sum. However, they
are very useful for physical problems when they are evaluated by treating as successive single integrals.
Further just as the definite integral (1) can be interpreted as an area, similarly the double integrals (3) can be
interpreted as a volume (see Figs. 5.1 and 5.2).
5.2 EVALUATION OF DOUBLE INTEGRAL
Evaluation of double integral ∫ ∫ f ( x, y ) dx dy
R
y-axis
is discussed under following three possible cases:
Case I: When the region R is bounded by two continuous
curves y = ψ (x) and y = φ (x) and the two lines (ordinates)
x = a and x = b.
In such a case, integration is first performed with
respect to y keeping x as a constant and then the
resulting integral is integrated within the limits x = a
and x = b.
Mathematically expressed as:
x=b
(
y = Ψ (x )
y = φ( x )
∫ ∫ f ( x, y ) dx dy = ∫ ∫
R
x=a
R
y = a  x =θ(y)
R
A
x=a
O
y = φ(x)
B
P
x=b
x = axis
Fig. 5.3
)
Case 2: When the region R is bounded by two continuous
curves x = φ (y) and x = Ψ (y) and the two lines (abscissa)
y = a and y = b.
In such a case, integration is first performed with
respect to x. keeping y as a constant and then the
resulting integral is integrated between the two limits
y = a and y = b.
Mathematically expressed as:
∫ ∫ f ( x, y ) dx dy = ∫  ∫
y = Ψ(x)
D
f ( x , y ) dy dx
Geometrically the process is shown in Fig. 5.3,
where integration is carried out from inner rectangle
(i.e., along the one edge of the ‘vertical strip PQ’ from
P to Q) to the outer rectangle.
y = b  x =Ψ( y)
Q
C

f ( x , y ) dx  dy

Geometrically the process is shown in Fig. 5.4,
where integration is carried out from inner rectangle
(i.e., along the one edge of the horizontal strip PQ
from P to Q) to the outer rectangle.
Case 3: When both pairs of limits are constants, the region
of integration is the rectangle ABCD (say).
y = axis
x = Ψ (y)
x = φ( y )
y=b
B
D
P
Q
R
y=a
A
C
O
x = axis
Fig. 5.4
y-axis
y=a
Q
D
R
y=b
O
C
S
A
P
x=a
Fig. 5.5
B
x=b
x-axis
Multiple Integrals and their Applications
357
In this case, it is immaterial whether f(x, y) is integrated first with respect to x or y, the
result is unaltered in both the cases (Fig. 5.5).
Observations: While calculating double integral, in either case, we proceed outwards from the innermost
integration and this concept can be generalized to repeated integrals with three or more variable also.
Example 1: Evaluate
1+ x 2
1
∫∫
0
0
(
1
dydx
1 + x2 + y2
)
[Madras 2000; Rajasthan 2005].
Solution: Clearly, here y = f(x) varies from 0 to 1 + x2
and finally x (as an independent variable) goes between 0
to 1.
I=
=
1
∫  ∫
1 + x2
1
1 + x2
0
0
∫ ∫

0
y
1
=  tan−1 
0a
a 0
∫
∫
=
∫
=
π
log (1 + 2 )
4
1
1 + x2
0
1 + x2
C
(1, 1. 414)
(2, 0)
O
(0, 0)
(10) (1. 732, 0)
dx
Fig. 5.6
2


−1 1 + x
− tan−1 0 dx
 tan
2

1+ x
=
1
B
A
(0, 1)

1
dy dx , a2 = (1 + x2)
2
+
a y

1
(0, 2)
(1. 732, 2)

1
dy dx
2
2
(1 + x ) + y 
2
0
D (2, 2. 36)
{
}
1
 π − 0 dx = π log x + 1 + x2 1
0

4
1 + x2  4
1
0
Example 2: Evaluate ∫ ∫ e2x +3y dxdy over the triangle bounded by the lines x = 0, y = 0 and
x + y = 1.
Solution: Here the region of integration is the triangle OABO as the line x + y = 1 intersects
the axes at points (1, 0) and (0, 1). Thus, precisely the region R (say) can be expressed as:
0 ≤ x ≤ 1, 0 ≤ y ≤ 1 – x (Fig 5.7).
∴
2x + 3y
I = ∫∫e
dxdy
B
R
1  1− x

= ∫  ∫ e2 x + 3 y dy  dx

0 0
1− x
1
= ∫  e2 x +3 y  dx

0
0 3
1
Y
(0, 1)
Q
x=1
x=0
(1, 0)
O
(0, 0)
P
y=0
A
Fig. 5.7
X
358
Engineering Mathematics through Applications
∫ (e
1
− e2x ) dx
=
1
3
=
1  e3 − x e2 x 
−
3  −1
2 0
=
−1  2 e2   3 1  
 e +  −  e +  
3 
2
2 
=
1
1
2
2e3 − 3e2 + 1 = (2e + 1)( e − 1)  .



6
6
3−x
0
1
Example 3: Evaluate the integral ∫ ∫ xy ( x + y )dxdy over the area between the curves y = x2
R
and y = x.
Solution: We have y = x2 and y = x which implies
x2 – x = 0 i.e. either x = 0 or x = 1
Further, if x = 0 then y = 0; if x = 1 then y = 1. Means the
two curves intersect at points (0, 0), (1, 1).
∴ The region R of integration is doted and can be
expressed as: 0 ≤ x ≤ 1, x2 ≤ y ≤ x.
∴


0
∫ ∫
1
∫ ∫ xy ( x + y ) dxdy =
R
=
x
x2
Y
y = x2
Q

xy ( x + y ) dy  dx

P
O (0, 0)
x
y3  
 2 y2
+
x
x

 dx
0 
2
3  x2 


∫
y=x
A(1, 1)
X
1 
Fig. 5.8
1
 x4 x4   x6 x7  
+  −  +   dx

0 2
3  2
3 
=
∫
=
∫  6 x
1
0
5
4
−
1 6 1 7
x − x  dx
2
3 
1
1 1
1
3
 5 x5 1 x7 1 x8 
= ×
−
−
= −
−
=

 6 5 2 7 3 8  0 6 14 24 56
x2 y2
2
Example 4: Evaluate ∫ ∫ ( x + y ) dxdy over the area bounded by the ellipse 2 + 2 = 1.
a
b
[UP Tech. 2004, 05; KUK, 2009]
Solution: For the given ellipse
x2 y2
+
= 1 , the region of integration can be considered as
a2 b2
Multiple Integrals and their Applications
bounded by the curves y = −b 1 −
x2
x2
,
y
b
1
=
−
and finally x goes from – a to a
a2
a2
I = ∫ ∫ ( x + y ) dxdy =
2
∴
I=
[Here


−a
∫ ∫
a
b 1− x2 / a2
−b
359
1− x2 / a2


–a 
∫ ∫
a
b 1 − x 2 / a2
–b 1− x2 / a2

( x2 + y2 + 2xy) dy dx


( x2 + y2 ) dy dx

∫ 2xy dy = 0 as it has the same integral value for both limits i.e., the term xy, which is
an odd function of y, on integration gives a zero value.]
a

I=4 

0
∫∫
b 1− x2 / a2
Y

( x2 + y2 ) dy dx
Q

0

y 
I = 4  x2 y + 
3 0

0
∫
X
O
2
2
3 b 1− x / a
a
dx
x=–a
P
x=a
Fig. 5.9
1
3
2
2
 
 2 b3 
 2
x
x
2
I = 4  x b  1 − 2  +  1 − 2   dx

3
a 
a  

0
a
∫
⇒
On putting x = a sinθ, dx = a cosθ dθ; we get
∫
I = 4b
π /2
0
∫
= 4ab
b3
 2
2
3 
 ( a sin θ cos θ) + cos θ a cos θdθ
3
π /2
0
 2
b3
2
2
4 
 a sin θ cos θ + cos θ dθ
3
|
|
1  p + 1  q + 1

 

π /2
2
2   2 
sinp x cosq x dx =
Now using formula
|
0
 p + q + 2


2
∫
|
and
∫
∫∫
π /2
0
cosn x dx =
 n + 1
 2 
π
,
2
 n + 2
 2 
|
¬
¬
 3¬ 3¬
5 1

2
( x + y )2 dxdy = 4ab a2 2 ¬2 + b3 2 ¬2
23
23

(in particular when p = 0 , q = n)




360
Engineering Mathematics through Applications


3 π
π π
π
 2 2 2
b2 2 2
= 4ab  a
+

2.2.1
3 2.2.1 



¬
1
Q  = π
 2
2
2
 πa2 πb2  πab ( a + b )
= 4ab 
+
=
16 
4
 16
ASSIGNMENT 1
∫∫
dx dy
1 1
1. Evaluate
(1 − x2 )(1 − y2 )
0 0
2. Evaluate ∫ ∫ xy dx dy, where A is the domain bounded by the x-axis, ordinate x = 2a and
R
the curve x2 = 4ay.
[M.D.U., 2000]
3. Evaluate ∫ ∫ eax + by dy dx , where R is the area of the triangle x = 0, y = 0, ax + by = 1 (a > 0,
b > 0). [Hint: See example 2]
21
12
13
31
y
y
4. Prove that ∫ ∫ ( xy + e ) dy dx = ∫ ∫ ( xy + e ) dx dy .
1
5. Show that
1
x−y
∫ ∫ ( x + y)
dx
0
∞∞
6. Evaluate ∫ ∫ e
3
1
0
− x2 (1+ y2 )
1
x dx dy
00
x−y
∫ ∫ ( x + y)
dy ≠ dy
0
3
dx .
0
[Hint: Put x2(1 + y2) = t, taking y as const.]
5.3 CHANGE OF ORDER OF INTEGRATION IN DOUBLE INTEGRALS
The concept of change of order of integration evolved to help in handling typical integrals
occurring in evaluation of double integrals.
When the limits of given integral ∫a ⋅ ∫y = φ(x) f ( x, y ) dy dx are clearly drawn and the region
b
y = Ψ( x )
of integration is demarcated, then we can well change the order of integration be performing
integration first with respect to x as a function of y (along the horizontal strip PQ from P to
Q) and then with respect to y from c to d.
Mathematically expressed as:
I=
d
∫∫
c
x = Ψ ( y)
x = φ( y )
f ( x, y ) dx dy.
Sometimes the demarcated region may have to be split into two-to-three parts (as the case
may be) for defining new limits for each region in the changed order.
Multiple Integrals and their Applications
361
1 − x2
1
Example 5: Evaluate the integral ∫
∫ y2 dydx by changing the order of integration.
0
0
[KUK, 2000; NIT Kurukshetra, 2010]
Solution: In the above integral, y on vertical strip (say PQ) varies as a function of x and then
the strip slides between x = 0 to x = 1.
Here y = 0 is the x-axis and y = 1 − x 2 i.e., x2 + y2 = 1 is the circle.
In the changed order, the strip becomes P’Q’, P’ resting on the curve x = 0, Q’ on the circle
x = 1 − y 2 and finally the strip P’Q’ sliding between y = 0 to y = 1.
Y
 1− y2 
2
=
I ∫ y  ∫ dx dy
0
 0

1
∴
1
I = ∫ y [ x ]0
2
0
1
1− y2
Q
Q´
P´
dy
y=0
X
P
I = ∫ y2 (1 −
1
y2 2
0
)
dx
x=1
π
Substitute y = sin θ, so that dy = cos θ d θ and θ varies from 0 to 2 .
x=0
Fig. 5.10
π
2
I = ∫ sin2 θ cos2 θ dθ
0
I=
( 2 − 1) ⋅ ( 2 − 1) π =
4⋅2
2
π
16
 π
Q 2 sinp θ cos θ dθ = (p − 1)(p − 3)…(q − 1)(q − 3) × π , only if both p and q are + ve even integers]
 ∫0
(p + q)(p + q − 2)……
2

4a
2 ax
dydx
by changing the order of integration.
Example 6: Evaluate ∫0 x∫2
4a
Y
[M.D.U. 2000; PTU, 2009]
Solution: In the given integral, over the vertical
strip PQ (say), if y changes as a function of x such
x2
and Q lies on the
that P lies on the curve y =
4a
curve y = 2 ax and finally the strip slides between
x = 0 to x = 4a.
x2
i.e. x2 = 4ay is a parabola
Here the curve y =
4a
with

y=0
implying x = 0


y = 4a
implying x = ± 4a 
(4a, 4a)
y2 = 4ax
Q
P´
A
Q´
x2 = 4ay
P
X
O
x = 4a
Fig. 5.11
362
Engineering Mathematics through Applications
i.e., it passes through (0, 0) (4a, 4a), (– 4a, 4a).
Likewise, the curve y = 2 ax or y2 = 4ax is also a parabola with
x = 0 ⇒ y = 0 and x = 4a ⇒ y = ± 4a
i.e., it passes through (0, 0), (4a, 4a), (4a, – 4a).
Clearly the two curves are bounded at (0, 0) and (4a, 4a).
∴ On changing the order of integration over the strip P’Q’, x changes as a function of y
such that P’ lies on the curve y2 = 4ax and Q’ lies on the curve x2 = 4ay and finally P’Q’ slides
between y = 0 to y = 4a.
I=
whence
=
=
4a 
∫  ∫
y2
x=
4a
0
∫
x = 2 ay
4a
[x] y
2 ay
2
0

dx  dy

dy
4a
4a 
y2 
 2 ay − 4a  dy
∫0
4a
3


3
y2
y3 

4 a
1
= 2 a
−
=
4 a) 2 −
(
( 4a)3

3 12a
3
12a


2

0
=
32a2 16a2 16a2
−
=
.
3
3
3
a
x
a
2
2
Example 7: Evaluate ∫ ∫x ( x + y ) dxdy by changing the order of integration.
0
a
Solution: In the given integral ∫0 ∫x / a ( x2 + a2 ) dx dy , y varies along vertical strip PQ as a
function of x and finally x as an independent variable varies from x = 0 to x = a.
a
x /a
Here y = x/a i.e. x = ay is a straight line and y = x / a , i.e.
x = ay2 is a parabola.
For x = ay; x = 0 ⇒ y = 0 and x = a ⇒ y = 1.
Means the straight line passes through (0, 0), (a, 1).
For x = ay2; x = 0 ⇒ y = 0 and x = a ⇒ y = ± 1.
Means the parabola passes through (0, 0), (a, 1), (a, – 1),.
Further, the two curves x = ay and x = ay2 intersect at common
points (0, 0) and (a, 1).
On changing the order of integration,
a
∫∫
0
x/ a
x/ a
Y
y = x /a
y=1
Q
O
P´
Q´
P
x = ay
2
(at P’)
X
x=a
(x2 + y2 ) dxdy = ∫y =0  ∫x = ay (x2 + y2 ) dxdy
y =1
y=0
(0, 0)
Fig. 5.12
Multiple Integrals and their Applications
I=
363
ay
 x3

+ xy2  dy

0  3
 ay2
∫
1
3


ay )
3
(
1

=
+ ay.y2  −  ( ay2 ) + ay2 ⋅ y2   dy


  3
0  3


∫
=
1 
 a3
 3 a3 6
4
 + a y − y − ay  dy

0  3
3

∫
1
1
 a3
 y4 a3 y7 ay5 
=  + a
−
−

 4
3 7
5 0
 3
 a3
a3   a a  
= 
−
 +  − 
 3 × 4 3 × 7   4 5  
=
a3
a
a
+
=
(5a2 + 7 ) .
28 20 140
∫∫
a
Example 8: Evaluate
0
y2
dy dx.
y 4 – a2 x 2
a
ax
[SVTU, 2006]
Solution: In the above integral, y on the vertical strip (say PQ) varies as a function of x and
then the strip slides between x = 0 to x = a.
Here the curve y = ax i.e., y2 = ax is the parabola and the curve y = a is the straight line.
On the parabola, x = 0 ⇒ y = 0; x = a ⇒ y = ± a i.e., the parabola passes through points
(0, 0), (a, a) and (a, – a).
On changing the order of integration,
∫ ∫
0
a
0
=
∫
0
y2
a
x =0
( at P´ )
∫  ∫
y2
a
0


a
x=
y2
a

y2
dx dy
y4 − a2 x2 


dx
 dy
2

 y2 
2

 a  − x

1
y2
y2  −1 x  a
sin
dy
 y2  
a 

 a  

0
y-axis
(a, a)
Q
P´
y=a
Q´
P
y=0
O
(0, 0)
x=a
=




x=0
I=
a
(a, –a)
Fig. 5.13
x-axis
364
Engineering Mathematics through Applications
=
∫
a
=
∫
a
0
0
y2
sin−1 1 − sin−1 0 dy
a 
y2 π
π y3
dy =
2a 3
a 2
a
=
0
πa2
.
6
1 2-x
Example 9: Change the order of integration of ∫ ∫ xy dy dx and hence evaluate the same.
0 x2
[KUK, 2002; Cochin, 2005; PTU, 2005; UP Tech, 2005; SVTU, 2007]
1
 2−x

Solution: In the given integral ∫  ∫ xydy dx, on the vertical strip PQ(say), y varies as a
2

0 x
Y
function of x and finally x as an independent variable,
varies from 0 to 1.
y=2
B(0, 2)
Here the curve y = x2 is a parabola with
y=0
implying x = 0 
Q

y = 1 implying x = ±1 

i.e., it passes through (0, 0), (1, 1), (– 1, 1).
Likewise, the curve y = 2 – x is straight line
with
y = 0 ⇒ x = 2

y = 1 ⇒ x = 1
y = 2 ⇒ x = 0
Q´´
P´´
y = x2
A
C
P´
Q´
y=1
y=2–x
y=0
P
O
x=0
(1, 1)
(2, 0)
X
x=1
Fig. 5.14
i.e. it passes though (1, 1), (2, 0) and (0, 2)
On changing the order integration, the area OABO is divided into two parts OACO and
ABCA. In the area OACO, on the strip P’Q’, x changes as a function of y from x = 0 to x = y .
Finally y goes from y = 0 to y = 1.
Likewise in the area ABCA, over the strip p”Q”, x changes as a function of y from x = 0 to
x = 2 – y and finally the strip P”Q” slides between y = 1 to y = 2.
∴
1 y

2  2−y

0 0

1 0

∫  ∫ xy dx dy + ∫  ∫ xy dx dy
 x2
= y
2
0
1
∫
2

 x2
 dy +  y
 2

1
y
∫
0
y2
dy +
2
y (2 − y)
dy
2
2
∫
=
1 1  2 4y3 y4 
+
2y −
+ 
6 2 
3
4 1
0
∫
0
=
1
2
1
2
I=
1 5
3
+
= .
6 24 8

 dy

2− y
Multiple Integrals and their Applications
2– x2
∫∫
1
Example 10: Evaluate
0
x
x
dydx by changing order of integration.
x + y2
[KUK, 2000; MDU, 2003; JNTU, 2005; NIT Kurukshetra, 2008]
2
Y
Soluton: Clearly over the strip PQ, y varies as a
function of x such that P lies on the curve y = x and Q
lies on the curve y = 2 − x 2 and PQ slides between
ordinates x = 0 and x = 1.
y=x
M
between y = 0 to y = 1 and y = 1 to y =
vely.
whence
Now the exp.
∴
∫∫
1
y
0 0
O
∫ ∫
2
1
2 − y2
0
Q´´
y =
N (1, 1)
2
y=1
Q´
P
y=0
X
x2 + y2 = 2
x=1
x=0
2 respecti-
x
dx dy +
x2 + y2
Q
P´´
L
P´
The curves are y = x, a straight line and y = 2 − x2 ,
i.e. x2 + y2 = 2, a circle.
The common points of intersection of the two are
(0, 0) and (1, 1).
On changing the order of integration, the same
region ONMO is divided into two parts ONLO and
LNML with horizontal strips P’Q’ and P”Q” sliding
I=
365
Fig. 5.15
x
dx dy
x2 + y2
1
x
d 2
2 2
x
y
=
+
(
)
x2 + y2 dx
I=
I=
=
1
∫ ( x
2
0
1
∫ ( x
2
0
)
1
a
y =a 
+
1 y

y2 2 
0
)
dy +
∫
2
dy +
∫
2
1
1
1
 2
( x + y2 )2  0
2 − y2
1
 2
( x + y2 )2  0
2 − y2
dy
dy
2
y2
y2 

1
2 − 1)
+  2 y −  = ( 2 − 1)
2 0 
2 0
2
(
Example 11: Evaluate
+
1 y

y2 2 
0
∫∫
0
a+ a2 – y2
a– a2 – y2
dy dx by changing the order of integration.

dx  dy
y =o
x =a−

Clearly in the region under consideration, strip PQ is horizontal with point P lying on the
Solution: Given
∫  ∫
x =a + a 2 − y 2
a 2 − y2
curve x = a − a2 − y 2 and point Q lying on the curve x = a + a2 − y2 and finally this strip
slides between two abscissa y = 0 and y = a as shown in Fig 5.16.
366
Engineering Mathematics through Applications
Y
x=0
Now, for changing the order of integration, the
region of integration under consideration is same but
this time the strip is P’Q’ (vertical) which is a function
of x with extremities P’ and Q’ at y = 0 and
y = 2ax − x2 respectively and slides between x = 0
and x = 2a.
2a  2ax − x2
I= ∫
0 
Thus
∫
0
2ax − x
2a

dx
dy dx = ∫ y 

0
0
2
2a
2a
0
0
C
Q' x = 2a
Q
P
A
(0, 0)
O
(a, 0)
= ∫ 2ax − x2 dx = ∫ x 2a − x dx
2a sin θ so that dx = 4a sinθ cos θ dθ,
π
For x = 0, θ = 0 and for x = 2a, θ =
2
Also,
X
x2 + y2 = 2ax
x=
Take
B (2a, 0)
P'
Fig. 5.16
π
2
I = ∫ 2a sin θ ⋅ 2a − 2a sin2 θ ⋅ 4a sin θ ⋅ cos θ dθ
Therefore,
0
π
2
= 8a2 ∫ sin 2 θ cos2 θ dθ = 8a2 ⋅
0
(2 − 1)(2 − 1) π = πa2
4 ( 4 − 2) 2
2

 using ∫ sinp θ cosq θ dθ = (p − 1)(p − 3)…(q − 1)(q − 3)… π ,


(p + q)(p + q − 2)……………… 2
0



p and q both positive even integers 
π
2
3 4− y
Example 12: Changing the order of integration, evaluate ∫ ∫
0
1
( x + y ) dx dy.
[MDU, 2001; Delhi, 2002; Anna, 2003; VTU, 2005]
Solution: Clearly in the given form of integral, x
changes as a function of y (viz. x = f(y) and y as an
independent variable changes from 0 to 3.
Thus, the two curves are the straight line x = 1 and
the parabola, x = 4 − y and the common area under
consideration is ABQCA.
For changing the order of integration, we need to
convert the horizontal strip PQ to a vertical strip P’Q’
over which y changes as a function of x and it slides for
values of x = 1 to x = 2 as shown in Fig. 5.17.
∴
I=
2
∫  ∫
1
( 4 − x2 )
0

( x + y ) dy dx =

∫
2
1
C (1, 3)
dx
y=3
Q'
Q
P
O
A
P´
Fig. 5.17
4 − x2
y2 
+
xy

2  0

Y
B y=0
(2, 0)
X
Multiple Integrals and their Applications
=
=
∫
2
 x ( 4 − x2 ) +


1
∫
367
( 4 − x2 )2  dx


2
2
x4


2
4
8
x
x
−
+
+
− 4x2   dx
(
)




2

1
2

x4
x5 4 3 
=  2x2 −
+ 8x +
− x
4
10 3  1

= 2 ( 22 − 12 ) −
=6−
1 4
( 2 − 14 ) + 8 (2 − 1) + 101 ( 25 − 15 ) − 34 (23 − 13 )
4
15
31 28 241
.
+8+
−
=
4
10 3
60
a
2
Example 13: Evaluate ∫
∫
0
0
a2 − y 2
(
)
log x 2 + y 2 dx dy ( a > 0 ) changing the order of integration.
[MDU, 2001]
Solution: Over the strip PQ (say), x changes as a function of y such that P lies on the curve
Y
x2 + y2 = a2
=
x = y and Q lies on the curve x = a2 − y2 and
a .
the strip PQ slides between y = 0 to y =
2
Here the curves, x = y is a straight line
B
x = 0 ⇒ y = 0

a
a 
x=
⇒y =
2
2 
and
x
y
Q´
P
GH
a
a
,
2,
2
JK
y=
Q´´
Q
y=0
O
(0, 0)
 a
a 
,

i.e. it passes through (0, 0) and 
2
2
P´ P´´ A
a
2
X
x=a
Also x = a2 − y2 , i.e. x2 + y2 = a2 is a circle
with centre (0, 0) and radius a.
x=0
 a
a 
,
Thus, the two curves intersect at 
.
2
2
x =
a
2
Fig. 5.18
On changing the order of integration, the same region OABO is divided into two parts
with vertical strips P’Q’ and P”Q” sliding between x = 0 to x =
a
a
and x =
to x = a
2
2
respectively.
Whence,
I=
∫
a/ 2
0


∫
x
0
log ( x2 + y2 ) ⋅ dy  dx +

∫
a
a/

2

∫
0
a2 − x2

log ( x2 + y2 ) ⋅ 1 dy  dx

…(1)
368
Engineering Mathematics through Applications
Now,
∫ log ( x
2
Ist
Function

+ y2 ) 1 dy =  log ( x2 + y2 ) ⋅ y −

∫x
2

1
2y ⋅ y dy 
2
+y

IInd
Function

y2 + x2 − x2 
dy 
=  y log ( x2 + y2 ) − 2
x2 + y2


∫

=  y log ( x2 + y2 ) − 2 y + 2x2

∫

1
dy

( x2 + y2 ) 
y 

1
=  y log ( x2 + y2 ) − 2y + 2x2  tan−1  

x
x 

…(2)
On using (2),
I1 =
∫
a/ 2
=
∫
a/ 2
=
∫
a/ 2
=
∫
a/ 2
0
0
0
0
x


2
2
−1 y  
 y log ( x + y ) − 2y + 2x  tan x   dx

0
 x log 2x2 − 2x + 2x tan−1 1 dx
 x log 2x2 − 2x + 2x π  dx

4 
π
x log 2x2 dx + 2  − 1
4

∫
a/ 2
0
x dx
For first part, let 2x2 = t so that 4x dx = dt and limits are t = 0 and t = a2.
∴
∫
I1 =
a2
0
a/ 2
0
a2
1
 π − 1 a2
t ( log t − 1) +
0
4
 2 , (By parts with logt = logt · 1)
4
=
=
2
π
log t ⋅ dt + 2  − 1 x
4
 2
4
a2
π a2 a2
log a2 − 1) +
−
(
4
8
2
…(3)
Agian, using (2),
I2 =
⇒
∫
=
a
a/ 2
∫
a
a/ 2


2
2
−1 y  
 y log ( x + y ) − 2y + 2x  tan x  

0
a2 − x2
dx
2
2
 2
−1 a − x
2
2
2
2
 a − x log a − 2 a − x + 2x tan
x

…(4)

 dx

Multiple Integrals and their Applications
π
π
to
4
2
Let x = a sinθ so that dx = a cos θ dθ and limits,
∴
∫
I2 =
=
π/ 2
π/ 4
∫
π /2
π/4
2
2
2 

2
2
2
2
− 1 a − a sin θ
(log a − 2) a − a sin θ + 2a sin θ tan
 a cos θ dθ
a sin θ


a2 (log a2 − 2) cos2 θ dθ + a2
= a2 ( log a2 − 2 )
=
369
∫
π /2
π /4
∫
π /2
π/4
2 sin θ cos θ tan−1 (cot θ) dθ
(1 + cos 2θ) dθ + a2
2
π /2
sin 2θ 
a2
log a2 − 2 ) θ +
+ a2
(
2
4 π/4

∫
∫
π /2
π /4
π /2
π /4

π

sin 2θ tan− 1  tan  − θ  dθ
2


 π − θ sin 2θ dθ
2

Ist
Fun.
IInd
Fun.
π /2

 π π  1 
a2
− cos 2θ 


2
2 π
a
a
log
2
= (
− )
−
−
+ 
−θ
−

2
2  π/4
  2 4  2 
  2
a2
a2
π 1
log a2 − 2)  −  −
(
 4 2 2
2
I2 =
∫
π /2
π /4
∫
π /2
π /4

(−1)  − cos 2θ  dθ 
2

π
− cos 2θ 
cos 2θ dθ ,  − θ 
is zero for both
2

2 
the limits)
π
 π a2
π a2 a2 a2
 a2
log a2 −
=
+
− log a2  − (sin 2θ)π 2
 8
 4
4
4
2
4
 π a2
π a2 a2 a2
 a2
log a2 −
=
+
− log a2  +
 8
 4
4
2
4
…(5)
On using results (3) and (5), we get
I = I1 + I2
 a2
a2 πa2 a2   πa2
πa2 a2 a2
a2 
log a2 −
=  log a2 − +
− +
+
− log a2 + 
4
4
8
2  8
4
2
4
4
=
πa2
πa2 πa2
log a2 −
=
(log a2 − 1)
8
8
8
=
1
πa2
πa2 
2 log a − 1) =
log a −  .
(
8
4 
2
∞x
Example 14: Evaluate by changing the order of integration. ∫ ∫ xe
– x 2 /y
dx dy
00
[VTU, 2004; UP Tech., 2005; SVTU, 2006; KUK, 2007; NIT Kurukshetra, 2007]
370
Engineering Mathematics through Applications
∞
∫∫
Solution: We write
0
x
0
xe−x
2 /y
dxdy =
x = ∞( = b )
y = f2 ( x ) = x
x = 0(= a
y = f1 (x ) = 0
∫
) ∫
xe − x
2/
y
dxdy
Here first integration is performed along the vertical strip with y as a function of x and
then x is bounded between x = 0 to x = ∞.
We need to change, x as a function of y and finally the limits of y. Thus the desired
geometry is as follows:
In this case, the strip PQ changes to P’Q’ with x as function of y, x1 = y and x2 = ∞ and
finally y varies from 0 to ∞.
Therefore Integtral
∞∞
I = ∫ ∫ xe − x
2/y
dxdy
0 y
Put x2 = t so that 2x dx = dt Further, for
I =
∞
∞
0
y
∫∫
1
=
2
∫
x = y , t = y2 ,

x = ∞, t = ∞ 
Q
dt
dy,
2
∞
e −t / y
 dy
− 1/ y y2 
e − t/ y
2
∞
O
1   e− y  
−
y

2   − 1   0
=
∞
1
 −ye−y − e−y  0
2
=
1
1
(0) − (0 − 1) = .
2
2
1
∞
e− y 
dy
− 1  0
∞
e –y
dy dx.
x –y
0
[NIT Jalandhar, 2004, 2005; VTU, 2007]
Example 15: Evaluate the integral
∫ ∫
Soluton: In the given integral, integration is performed first
with respect to y (as a function of x along the vertical strip say
PQ, from P to Q) and then with respect to x from 0 to ∞.
On changing the order, of integration integration is
performed first along the horizontal strip P'Q' (x as a function
of y) from P' to Q' and finally this strip P'Q' slides between
the limits y = 0 to y = ∞.
Y
Q
y
=
x
x=∞
Q´
P´
x=0
∞
0
Fig. 5.19
∞
=
∫
X
P
∫
∫
∞
Q'
P'


∞ y
=
−  0 − e − y  dy
0
2
∞ ye − y
=
dy (By parts)
0
2
0
Y
P
y=0
O (0, 0)
Fig. 5.20
X
Multiple Integrals and their Applications
I=
∴
∫
∞ −y
e 

y 
0
∞
=
∫
=
e− y
−1
0
∫
y
0

dx  dy

e −y
( y ) dy =
y
∞
0
371
∫
∞
0
e −y dy
1
1
= −1  ∞ − 0 
e
e 
= – 1(0 – 1) = 1
Example 16: Change the order of integration in the double integral
∫ ∫
2a
0
[Rajasthan, 2006; KUK, 2004-05]
0
x 2 − 2ax
f ( x , y) dydx =
a
a − a 2 − y2
0
y 2 /2 a
∫∫
y = 2a
+
2a
0
a + a2 − y2
∫∫
Part II
f ( x, y) dydx +
Q ´´´
x=0
P´´´
ΙΙΙ
Ι
P´
P
P´´
Q´
ΙΙ
y=a
Q ´´
(a , 0)
y=0
O
(0, 0)
X
f ( x, y) dydx
( x – a) 2 + y2 = a2
2a
2a
a
y 2 / 2a
∫ ∫
Fig. 5.21
f ( x, y) dydx
Y
Part III
Example 17: Find the area bounded by the lines y
= sin x, y = cos x and x = 0.
Q
Solution: See Fig 5.22.
Clearly the desired area is the doted portion
where along the strip PQ, P lies on the curve
y = sin x and Q lies on the curve y = cos x and finally
the strip slides between the ordinates x = 0 and
P
π 1 
 ,

4 2
y = sinx
O
x=0
π
x= .
4
Q
y2 = 2 ax
Part I
a
f ( x, y ) dx dy.
x = 2a
∫ ∫
2ax
2ax - x2
Y
Solution: Clearly from the expressions given above,
the region of integration is described by a line which
starts from x = 0 and moving parallel to itself goes
over to x = 2a, and the extremities of the moving line
lie on the parts of the circle x2 + y2 – 2ax = 0 the parabola
y2 = 2ax in the first quadrant.
For change and of order of integration, we need to
consider the same region as describe by a line moving
parallel to x-axis instead of Y-axis.
In this way, the domain of integration is divided
into three sub-regions I, II, III to each of which
corresponds a double integral.
Thus, we get
2a
2ax
π
2
π
y = cosx
Fig. 5.22
3π
2
2π
X
372
Engineering Mathematics through Applications
π
4  cos x

∫ ∫ dx dy = ∫  ∫ dy dx


∴
R
0
sin x
π
4
= ∫ (cos x − sin x) dx
0
= (sin x + cos x )0
π /4
 1
  1

=
− 0 + 
− 1 .
 2
  2

=
(
)
2 −1
ASSIGNMENT 2
∫∫
a
1. Change the order of integration
0
a
y
x
dxdy
x + y2
2
a
a2 − y2
2. Change the order integration in the integral ∫ ∫
−a 0
a ⋅cos α
3. Change the order of integration in
∫
4. Change the order of integration in
∫∫
∫
a2 − x 2
x tan α
0
a
lx
0
mx
f ( x, y ) dx dy
f ( x , y ) dy dx
f (x, y)dxdy
[PTU, 2008]
5.4 EVALUATION OF DOUBLE INTEGRAL IN POLAR COORDINATES
θ=β r = Ψ(θ)
To evaluate
∫
∫
θ=α r = φ(θ )
f ( r, θ) dr dθ, we first integrate with respect to r between the limits
r = φ(θ) to r = ψ(θ) keeping θ as a constant and then the
resulting expression is integrated with respect to θ from θ =
α to θ = β.
θ=
r =Ψ(θ)
r = Ψ(θ)
D
r = φ(θ)
Geometrical Illustration: Let AB and CD be the two
continuous curves r = φ(θ) and r = Ψ(θ) bounded between
the lines θ = α and θ = β so that ABDC is the required
region of integration.
Let PQ be a radial strip of angular thickness δθ when OP
makes an angle θ with the initial line.
Here ∫r =φ(θ) f ( r, θ) dr refers to the integration with
π
2
Q
C
δθ
P
θ=
β
B
θ
θ=
α
θ=0
O
respect to r along the radial strip PQ and then integration
with respect to θ means rotation of this strip PQ from AC to CD.
A
Fig. 5.23
Multiple Integrals and their Applications
373
θ) above the initial line.
Example 18: Evaluate ∫∫ r sinθ dr dθ over the cardiod r = a (1 – cosθ
Solution: The region of integration under consideration is the cardiod r = a(1 – cos θ) above
the initial line.
In the cardiod
r = a(1 – cos θ); for
θ = 0, r = 0, 

π
θ = , r = a, 
2

θ = π, r = 2a 
As clear from the geometry along the radial strip OP, r (as a function of θ) varies from
r = 0 to r = a(1 – cos θ) and then this strip slides from θ = 0 to θ = π for covering the area above
the initial line.
Hence
θ = π/2
π  r = a(1− cos θ)

I = ∫
r dr sin θ dθ
∫

0
0
A(a , π/2)
P
π  2 a(1− cos θ )
r
= ∫
 sin θ dθ
0 2 0

θ=π
=
C
Fig. 5.24

f n +1 ( x ) 
n
Q f ( x ) f ´ ( x ) dx =

n+1 

∫
∫∫
r2sinθ dr dθ =
R
2a3
θ above
, where R is the semi circle r = 2a cosθ
3
θ = π/2
Solution: The region R of integration is the semi-circle
r = 2a cos θ above the initial line.
For the circle
r = 2a cosθ, θ = 0 ⇒ r = 2a 
r = 2a cosθ


π
θ=
⇒ r=0 
2

Otherwise also,
θ=0
4a2
a2 
a2
3
1 − cos π ) − (1 − cos 0) = [ 8 − 0 ] =
.
(
 6
6 
3
Example 19: Show that
the initial line.
π
 (1 − cos θ)3 

 ,
3

 0
O
(0, 0)
(2a, π)
a2 π
2
=
∫ (1 − cos θ ) sin θ dθ
2 0
a2
=
2
θ
B
θ=0
θ
(0, 0)
O
(a, 0)
r = 2a cosθ ⇒ r2 = 2ar cosθ
x 2 + y2 = 2ax
(x2 – 2ax + a2) + y2 = a2
(x – a)2 + (y – 0)2 = a2
Fig. 5.25
(2a, 0)
374
Engineering Mathematics through Applications
i.e., it is the circle with centre (a, 0) and radius r = a
π
2 2 a cos θ
2
r sin θdr dθ
Hence the desired area ∫ ∫
0 0
π
2
 2a cos θ

= ∫  ∫ r2 dr  sin θdθ

0
0
=
∫
π /2 
r3
 3

0
π /2
2 a cos θ
0

 sin θdθ

=
−1
3
=
−8a 3  cos 4 θ 
3  4  0 , using
=
2a 3
.
3
Example 20: Evaluate
∫
0
(2a)3 cos3 θ sin θ dθ
π /2
∫∫
n +1
f (x)
∫ f (x) f '(x)dx = n + 1
n
r dr dθ
θ.
over one loop of the lemniscate r2 = a2 cos2θ
a2 + r 2
[KUK, 2000; MDU, 2006]
π
and maximum value of r is a.
4
See Fig. 5.26, in one complete loop, r varies from 0 to r = a cos 2θ and the radial strip
π
π.
slides between θ = −
to
4
4
Hence the desired area
Solution: The lemniscate is bounded for r = 0 implying θ = ±
π/ 4
A =
∫ ∫
∫
=
π /4
∫
2
a cos 2θ
0
π /4
∫− π/4 ( a2 + r2 )
∫
π /4
r
(a
− π /4 0

=

/4
−π 
=
a cos2θ
1
+ r2 )
1
2
dr dθ
θ
d ( a2 + r2 )
1
2
a cos2θ
2

dr  dθ

dθ
π /4
(
∫ (
− π /4
4
π/
P(r, θ)
θ=0
O
θ
=
0
)
 a 2 + a 2 cos 2θ 1 2 − a  d θ

− π /4 

=a
=
)
2 cos θ − 1 dθ
Fig. 5.26
–
π/
4
Multiple Integrals and their Applications
π /4
∫ (
= 2a
)
2 cos θ − 1 dθ
0
= 2a 

(
375
2 sin θ − θ
)0π/4 
1

π
π
= 2a  2
−
= 2a  1 −  .

2 4 
4

θ and
Example 21: Evaluate ∫ ∫ r 3 dr dθ , over the area included between the circles r = 2a cosθ
θ (b < a).
r = 2b cosθ
[KUK, 2004]
Solution: Given r = 2a cosθ or r2 = 2a r cosθ
x2 + y2 = 2ax
(x + a)2 + (y – 0)2 = a2
i.e this curve represents the circle with centre (a, 0) and radius a.
Likewise, r = 2b cosθ represents the circle with centre (b, 0) and radius b.
We need to calculate the area bounded between the two circles, where over the radial
strip PQ, r varies from circle r = 2b cosθ to r = 2a cosθ and finally θ varies from −
π
2
2 a cos θ
3
Thus, the given integral ∫ ∫ r3 dr dθ = ∫ ∫
R
=
=
−
r dr dθ
π 2b cos θ
2
θ=
π
2
1
4
∫
r = 2 b cosθ
Q
4 2a cos θ
r 
dθ
− π /2 
 4 2b cos θ
∫
π /2
π
π
to .
2
2
π /2
( 2a cos θ)4 − ( 2b cos θ)4  dθ

− π /2 
P
θ
O
(0, 0)
2b
2a
π
2
= 4 ( a4 − b4 ) ∫ cos4 θ dθ
−
π
2
π
2
r = 2a cos θ
Fig 5. 27
= 8 ( a4 − b4 ) ∫ cos4 θ dθ
= 8 (a − b
4
=
4
0
) 43 ⋅⋅ 12 π2
3
π ( a4 − b4 ) .
2
Particular Case: When r = 2 cosθ and r = 4 cosθ i.e., a = 2 and b = 1, then
3
3
45π
units .
I = π ( a4 − b4 ) = π ( 24 − 14 ) =
2
2
2
376
Engineering Mathematics through Applications
ASSIGNMENT 3
1. Evaluate ∫ ∫ r sin θ dr dθ over the area of the caridod r = a(1 + cosθ) above the initial line.

Hint : I =

π
∫∫
0
a (1+ cos θ )
0

r sin θ dr dθ

2. Evaluate ∫ ∫ r3 dr dθ , over the area included between the circles r = 2a cosθ and r = 2b cosθ
(b > a).
[Madras, 2006]
π


2 r = 2b cos θ
 Hint : I = ∫  ∫ r 3 dr  dθ

  (See Fig. 5.27 with a and b interchanged)

π  r = 2a cos θ
−
2


3. Find by double integration, the area lying inside the cardiod r = a(1 + cosθ) and outside
the parabola r(1 + cosθ) = a.
[NIT Kurukshetra, 2008]

Hint : 2


∫
π /2 
0


∫
a(1+ cos θ)
a
1+ cos θ
 
rdr  dθ
 

5.5 CHANGE OF ORDER OF INTERGRATION IN DOUBLE INTEGRAL IN POLAR
COORDINATES
θ=β r =Ψ(θ )
In the integral ∫θ=α ∫r =φ (θ) f ( r, θ) dr dθ , interation is first performed with respect to r along a
‘radial strip’ and then this trip slides between two values of θ = α to θ = β.
In the changed order, integration is first performed with respect to θ (as a function of r
along a ‘circular arc’) keeping r constant and then integrate the resulting integral with respect
to r between two values r = a to r = b (say)
Mathematically expressed as
( )
()
∫θ=α ∫r=φ(θ) f ( r, θ) dr dθ = I = ∫r= a ∫θ= f (r) f (r, θ ) dθ dr
θ=β r =Ψ θ
r = b θ=η r
π /2 2a cosθ
Example 22: Change the order of integration in the integral ∫0 ∫0
Solution: Here, integration is first performed with
respect to r (as a function of θ) along a radial strip
OP (say) from r = 0 to r = 2a cos θ and finally this
radial strip slides between θ = 0 to θ =
y2
a)2
y2
r = 2a cosθ or θ = cos −1
a2
O
(0, 0)
r
2a
P
R
⇒
+ = 2ax ⇒ (x –
+ =
i.e., it is circle with centre (a, 0) and radius a.
On changing the order of integration, we have to
first integrate with respect to θ (as a function of r) along
x2
π
2
π
.
2
r = 2a cosθ ⇒ r2 = 2a rcosθ
Curve
θ=
f ( r , θ ) dr dθ
θ
Q (a , 0)
(2 a, 0)
x = 2a
Fig. 5.28
θ=0
Multiple Integrals and their Applications
377
−1 r
the ‘circular strip’ QR (say) with pt. Q on the curve θ = 0 and pt. R on the curve θ = cos
2a
and finally r varies from 0 to 2a.
π
2
I=∫
∴
 cos−1 2r

a

f ( r, θ) dr dθ = ∫  ∫ f (r, θ) dθ dr

0
0
2a cos θ
∫
0
0
2a
π
Example 23: Sketch the region of integration
aeπ / 4
∫ ∫
π /2
2 log r
a
0
π /2
a
2log
∫ ∫
of integration.
Solution: Double integral
ae 4
r
a
f ( r , θ ) r dr dθ and change the order
r =β
f ( r, θ) r dr dθ is identical to ∫
r =α
θ= f2(r)
f (r, θ)rdrdθ, whence
∫
θ= f1(r)
integration is first performed with respect to θ as a function of r i.e., θ = f(r) along the
r
‘circular strip’ PQ (say) with point P on the curve θ = 2 log and point Q on the curve
a
π
θ=
and finally this strip slides between between r = a to r = aeπ/4. (See Fig. 5.29).
2
r
θ
r
The curve θ = 2 log implies = log
a
a
2
r
eθ/2 =
or r = a eθ/2
a
Now on changing the order, the integration is first performed with respect to r as a
function of θ viz. r = f(θ) along the ‘radial strip’ PQ (say) and finally this strip slides between
π
θ = 0 to θ = . (Fig. 5.30).
2
θ = π /2
C(aeπ/4, π/2)
M
r = aeθ/2
θ = 2log r/a
or
r = aeθ/2
Q
L
(a , π/2) B
P
θ
P
r=a
O
O
Fig. 5.29
I=
π /2 
Fig. 5.30
∫  ∫
θ= 0
r=
aeθ / 2
r =a

f (r, θ ) r dr  dθ

(a , 0)
A
r=a
θ=0
∴
Q
δθ
r = aeπ/4
θ=0
378
Engineering Mathematics through Applications
5.6. AREA ENCLOSED BY PLANE CURVES
Y
1. Cartesian Coordinates: Consider the area bounded
by
the
two
continuous
curves
y = φ(x) and y = Ψ(x) and the two ordinates x = a, x =
b (Fig. 5.31).
Now divide this area into vertical strips each of
width δx.
Let R(x, y) and S(x + δx, y + δy) be the two
neigbouring points, then the area of the elementary
shaded portion (i.e., small rectangle) = δxδy
But all the such small rectangles on this strip PQ
are of the same width δx and y changes as a function
of x from y = φ(x) to y = Ψ(x)
Q
C
D
S δy
δx
R
B
A
P
O
(0, 0)
x=a
x=b
X
Fig. 5.31
Ψ( x )
∴ The area of the strip PQ = Lt ∑ δxδy = δx Lt ∑ dy = δ x∫φ( x( ) ) dy
δ y →0
Now on adding such strips from x = a, we
get the desired area ABCD,
Ψ(x)
Ψ( x)
dy
φ( x)
Lt ∑ δx∫
δy →0 φ(x)
b
Ψ( x)
dx φ( x) dy
a
=∫
∫
b Ψ(x)
Y
P
y=a
x = Ψ(y)
δy
a φ(x)
Q
δx
x = φ(y)
A
B
X
O
Fig. 5.32
Q
Q´
δθ
P´
r
2 Polar Coordinates: Let R be the region
enclosed by a polar curve with P(r, θ) and Q(r +
δr, θ + δθ) as two neighbouring points in it.
Let PP’QQ’ be the circular area with radii OP
and OQ equal to r and r + δr respectively.
Here the area of the curvilinear rectangle is
approximately
= PP’ · PQ’ = δr · r sin δθ = δr ⋅ r δθ = r δr δθ.
If the whole region R is divided into such small
curvilinear rectangles then the limit of the sum
Σrδrδθ taken over R is the area A enclosed by the
curve.
i.e.,
A = Lt ∑ rδr δθ = ∫ ∫rdr dθ
δ r →0
δθ→0
D
C
y=b
= ∫ ∫ dxdy
Likewise taking horizontal strip P’Q’ (say)
as shown, the area ABCD is given by
y = b x =Ψ( y )
∫y = a ∫x =φ( y) dx dy
Ψ x
δy →0 φ( x )
P
δr
δθ
θ
θ=0
O
X
Fig. 5.33
R
Example 24: Find by double integration, the area lying between the curves y = 2 – x2 and
y = x.
Solution: The given curve y = 2 – x2 is a parabola.
Multiple Integrals and their Applications
⇒ y = 0
y = 1
⇒

⇒ y = −2
where in
y = 1
⇒
⇒ y = −2
i.e., it passes through points (0, 2), (1, 1), (2, – 2),
(– 1, 1), (– 2, – 2).
Likewise, the curve y = x is a straight line
379
Y
x=0
x =1
x=2
x = −1
x = −2
where
y = 2 – x2
B(0, 2)
Q
y
=
x
A(1, 1)
C
X
O (0, 0)
y=0 ⇒ x=0 

y=1 ⇒ x=1 
y = −2 ⇒ x = − 2
x =1
P
i.e., it passes through (0, 0), (1, 1), (– 2, – 2)
Now for the two curves y = x and y = 2 – x2 to
intersect, x = 2 – x 2 or x 2 + x – 2 = 0 i.e.,
x = 1, –2 which in turn implies y = 1, –2
respectively.
y=–2
D(– 2, –2)
Fig. 5.34
Thus, the two curves intersect at (1, 1) and
(–2, –2),
Clearly, the area need to be required is ABCDA.
1  2− x
1

A = ∫  ∫ dy dx = ∫ ( 2 − x2 − x ) dx

−2  x
−2
2
∴
1
9

x3 x2 
=  2x −
−  = units.
3
2  −2 2

Example 25: Find by double integration, the area lying between the parabola y = 4x – x2
and the line y = x.
[KUK, 2001]
x=0
x=1
x=2
x=3
x=4
⇒
⇒
⇒
⇒
⇒
y = 0
y = 2

y = 4
y = 3
y = 0
i.e. it passes through the points (0, 0), (1, 2), (3, 3) and
(4, 0).
Likewise, the curve y = x passes through (0, 0) and
(3, 3), and hence, (0, 0) and (3, 3) are the common points.
Otherwise also putting y = x into y = 4x – x2, we get
x = 4x – x2 ⇒ x = 0, 3.
=
Y
Solution: For the given curve y = 4x – x2;
B(2, 4)
x
y
(1, 2)
C(3, 3)
A
(0, 0)
O
(4, 0)
x=3
x=0
Fig. 5.35
X
380
Engineering Mathematics through Applications
See Fig. 5.35, OABCO is the area bounded by the two curves y = x and y = 4x – x2
∴ Area
3
4 x − x2
0
x
OABCO = ∫ ∫ dy dx
3
4x − x2
= ∫ y x
dx
0
=
 x
∫0 ( 4x − x2 − x) dx = 3 2
3
2

3
−
9
x3 
= units

3 0 2
Example 26: Calculate the area of the region bounded by the curves y =
3x
and 4 y = x2
x2 + 2
[JNTU, 2005]
Solution: The curve 4y = x2 is a parabola
where
y = 0 ⇒ x = 0, 
i.e., it passes through (–2, 1), (0, 0), (2, 1).
y = 1 ⇒ x = ±2
Likewise, for the curve y =
3x
x2 + 2


y = 1 ⇒ x = 1, 2 

x = –1 ⇒ y = –1 
y=0 ⇒ x=0
Hence it passes through points (0, 0), (1, 1), (2, 1), (–1, –1).
Also for the curve (x2 + 2) y = 3x, y = 0 (i.e. X-axis) is an asymptote.
For the points of intersection of the two curves y =
we write
3x
x2
=
x +2
4
2
3x
and 4y = x2
x +2
2
or x2 (x2 + 2) = 12x
x=0 ⇒ y=0
x=2 ⇒ y=1
i.e. (0, 0) and (2, 1) are the two points of intersection.
Then
Y
y =
Q
x2
4
(2, 1)
A
P
O
X
x=2
Fig. 5.36
Multiple Integrals and their Applications
381
The area under consideration,
A=
2
3x
x2 + 2
x2
y=
4
∫ ∫
0


y=

dy dx =

∫
2
0
 3x
x2 
 x2 + 2 − 4  dx
2
x3 
3
=  log ( x2 + 2) − 
12  0
2
=
3
3
2
2
log 6 − log 2) − = log 32 − .
(
2
3
3
Example 27: Find by the double integration, the area lying inside the circle r = a sinθ
θ and
θ).
outside the cardiod r = a(1 – cosθ
[KUK 2005; NIT Kurukshetra 2007]
Soluton: The area enclosed inside the circle r = a sinθ and the cardiod r = a(1 – cosθ) is shown
as doted one.
For the radial strip PQ, r varies from r = a(1 – cosθ) to r = a sin θ and finally θ varies in
π
θ = π /2
between 0 to .
2
For the circle r = a sin θ
θ = 0 ⇒ r = 0

π
θ = ⇒ r = a
2

θ = π ⇒ r = 0
A
Q
P
θ
θ=π
r = a(1 – cosθ )
θ=0 ⇒ r=0 

π
θ= ⇒ r=a
2

θ = π ⇒ r = 2a 
Thus, the two curves intersect at θ = 0 and θ =
∴
θ=0
O
(0, 0)
Likewise for the cardiod r = a(1 – cosθ);
π
2
r = a sinθ
Fig. 5.37
π
.
2
a sin θ
A= ∫ ∫
rdrdθ
0 a(1− cos θ )
=
∫
π /2 2 a sin θ
r
dθ
2 a(1− cos θ)
0
π /2
=
∫
=
a2
2
1 2
sin θ − (1 + cos2 θ − 2 cos θ) dθ
2
0
∫
π /2
0
[ − cos 2θ − 1 + 2 cos θ] dθ,
since (sin2 θ − cos2 θ) = − cos 2θ
382
Engineering Mathematics through Applications
=
π /2
a2  − sin 2θ
− θ + 2 sin θ 
2  2
0
π
= a2  1 −  .

4
θ + cosθ
Example 28: Calculate the area included between the curve r = a(secθ
θ ) and its
θ.
asymptote r = a secθ
[NIT Kurukshetra, 2007]
 1 + cos θ
 contains cosine terms
Solution: As the given crave r = a(secθ + cosθ) i.e., r = a 
cos θ
only and hence it is symmetrical about the initial axis.
Further, for θ = 0, r = 2a and, r goes on decreasing above and below the initial axis as θ
π
π
and at θ = , r = ∞.
approaches to
2
2
Clearly, the required area is the doted region in which r varies along the radial strip from
π
π
to θ = .
r = a secθ to r = a(secθ + cosθ) and finally strip slides between θ = −
2
2
π
2
a(sec θ+ cos θ)
0
a sec θ
A = 2∫
∴
∫
=2
= a2
=a
2
r dr dθ
∫
π /2
0
∫
Y
a(sec θ+ cos θ )
 r2 
dθ
 2 
a sec θ
π /2 
2
Q
2
 1 + cos θ
 1 

  dθ
 − 

θ
cos
cos θ  

0
π /2
2
∫ (cos θ + 2) dθ
θ
r = a (secθ + cosθ )
P
X
r = a secθ
2
0
π /2
( 5 + cos 2θ) dθ
= a2
∫
a2
=
2
 5θ + sin 2θ 

2 0
0
2
π /2
5π a2
=
.
4
Fig. 5.38
ASSIGNMENT 4
1. Show by double integration, the area bounded between the parabola y2 = 4ax and x2 =
16 2
a.
[MDU, 2003; NIT Kurukshetra, 2010]
3
2. Using double integration, find the area enclosed by the curves, y2 = x3 and y = x.
[PTU, 2005]
4ay is
θ.
Example 29: Find by double integration, the area of laminiscate r2 = a2 cos2θ
[Madras, 2000]
Solution: As the given curve r 2 = a 2 cos2θ contains cosine terms only and hence it is
symmetrical about the initial axis.
Multiple Integrals and their Applications
Further the curve lies wholly inside the circle r = a,
since the maximum value of |cos θ| is 1.
Also, no portion of the curve lies between
383
π
3π and the extended axis.
to θ =
4
4
See the geometry, for one loop, the curve is
π
π
bounded between θ = −
to
4
4
π/
3π
=
=
/4
θ
θ
4
θ = π /2
θ=
π
4
∴
θ
A
θ=0
O
rdr dθ
∫
π
4
∫
=4
θ
r = a2 cos 2θ
Area = 2 ∫
−
B
r=0
Fig. 5.39
π /4 2 a cos2θ
r
2
0
= 2a2
∫
π /4
0
dθ
0
cos 2θ dθ = 2a2 

π /4
sin 2θ 
2 0
= a2
5.7 CHANGE OF VARIABLE IN DOUBLE INTEGRAL
The concept of change of variable had evolved to facilitate the evaluation of some typical
integrals.
Case 1: General change from one set of variable (x, y) to another set of variables (u, v).
If it is desirable to change the variables in double integral ∫ ∫ f ( x, y ) dA by making
R
x = φ(u, v) and y = Ψ(u, v), the expression dA (the elementary area δxδy in Rxy) in terms of u
and v is given by
 x, y 
dA = J 
du dv,
 u, v 
 x, y 
J
≠0
 u, v 
J is the Jacobian (transformation coefficient) or functional determinant.
∴
 x, y 
∫ ∫ f ( x, y ) dx dy = ∫ ∫ F (u, v ) J  ,  du dv
uv
R
R
Case 2: From Cartesian to Polar Coordinates: In transforming to polar coordinates by means
of x = r cosθ and y = r sinθ,
∂x
 x, y 
= ∂r
J
 r, θ  ∂y
∂r
∴
∂x
sin θ
∂θ = cos θ
∂y
−r sin θ r cos θ
∂θ
dA = r dr dθ and
∫ ∫ f ( x, y ) dx dy = ∫ ∫ F (r, θ ) r dr dθ
R
R´
384
Engineering Mathematics through Applications
Example 30: Evaluate ∫ ∫ ( x + y ) dx dy where R is the parallelogram in the xy plane with
2
R
vertices (1, 0), (3, 1), (2, 2), (0, 1) using the transformation u = x + y, v = x – 2y.
[KUK, 2000]
Solution: Rxy is the region bounded by the parallelogram ABCD in the xy plane which on
transformation becomes Rúv i.e., the region bounded by the rectangle PQRS, as shown in the
Figs. 5.40 and 5.41 respectively.
V
Y
C (2, 2)
P (1, 1)
O
U
(0, 0)
u=4
B (3, 1)
u=1
D
(0, 1)
X
A (1, 0)
S (1, – 2)
Fig. 5.40
With
Fig. 5.41
}
u=x+y
v = x − 2y ,
A (1, 0) transforms to
C(2, 2) transforms to R(4, – 2)
D(0, 1) transforms to S(1, – 2)
and
∂x
∂v = − 1
∂y
3
∂v
Hence the given integral
∫∫ u
2
R
1
du dv
3
=
∫∫
=
4
1
× (1 + 2 ) ∫1 u2 du
3
4
1
1
1 2
1
u du dv =
−2 3
3
∫ [ v]
4
1
 u3 4  63
I=
 = 3 = 21units
 3 1
1
−2
u2 du
}
u = 1+ 0 = 1
v =1−0 =1
B(3, 1) transforms to Q(4, 1)
∂x
∂ ( x, y ) ∂ u
J
=
∂ (u, v ) ∂ y
∂u
Q (4, 1)
i.e., P(1, 1)
R (4, – 2)
Multiple Integrals and their Applications
385
Example 31: Using transformation x + y = u, y = uv, show that
∫∫
1
0
1 − x  y 
 x+y
0
e
dxdy =
1
(e − 1) .
2
[PTU, 2003]
Now point O(0, 0) implies 0 = u(1 – v)
…(1)
and
0 = uv
…(2)
From (2), either u = 0 or v = 0 or both zero. From (1), we get
u = 0, v = 1
Hence (x, y) = (0, 0) transforms to (u, v) = (0, 0), (0, 1)
O
(0, 0)
x=1
x=0
Solution: Clearly y = f(x) represents curves y = 0 and y = 1 – x, and x which is an independent
variable changes from x = 0 to x = 1. Thus, the area OABO bounded
Y
between the two curves y = 0 and x + y = 1 and the two ordinates
x = 0 and x = 1 is shown in Fig. 5.42.
B (0, 1)
On using transformation,
x+y=1
x+y=u ⇒
x = u(1 – v)
Q
y = uv
⇒
y = uv
P
A
(1, 0)
Fig. 5.42
Point A(1, 0), implies 1 = u(1 – v)
…(3)
and
0 = uv
…(4)
From (4) either u = 0 or v = 0, If v = 0 then from (3) we have u = 1, again if u = 0, equation
(3) is inconsistent.
Hence, A(1, 0) transforms to (1, 0), i.e. itself.
From Point B(0, 1), we get 0 = u(1 – v)
…(5)
and
…(6)
1 = vu
From (5), either u = 0 or v = 1
If u = 0, equation (6) becomes inconsistent.
If v = 1, the equation (6) gives u = 1.
Hence (0, 1) transform to (1, 1). See Fig. 5.43.
Hence
 y 
1 1− x 
 x + y 
∫∫
0 0
e
=
dxdy =

u
0 
1 1
1
v
0 0

ev dv du =

0
∫ ∫
1
∫ ∫ ue du dv
∫
1
0
where
J=
u ⋅ ( e − 1) du = ( e − 1)
Example 32: Evaluate the integral
∫ ∫
4a
0
y
2
y
4a
∂ ( x, y )
=u
∂ (u, v )
u2
2
1
=
0
O´´
(0, 1)
Q´
B´ (1, 1)
O
(0, 0)
P´
A (1, 0)
Fig. 5.43
1
( e − 1)
2
x2 − y2
dxdy by transforming to polar coordinates.
x2 + y2
386
Engineering Mathematics through Applications
Y
y2
or y 2 = 4ax is
4a
parabola passing through (0, 0), (4a, 4a).
Likewise the curve x = y is a straight line passing
through points (0, 0) (4a, 4a).
Hence the two curves intersect at (0, 0), (4a, 4a).
In the given form of the integral, x changes (as a
=
Solution: Here the curves x =
x
θ = π/2
y
P
y2 = 4a
A (4a, 4a)
θ
O
(0, 0)
y2
to x = y and finally y as an
4a
independent variable varies from y = 0 to y = 4a.
For transformation to polar coordinates, we take
θ=0
B
(4a, 0)
X
function of y) from x =
x = r cosθ, y = r sinθ and J =
Fig. 5.44
∂ ( x, y )
=r
∂ ( r, θ)
The parabola y2 = 4ax implies r2 sin2θ = 4a r cosθ so that r(as a function of θ) varies from
4a cos θ
π
π
and θ varies from θ =
to θ =
sin2 θ
4
2
Therefore, on transformation the integral becomes
r = 0 to r =
I=
=
=
π /2
∫ ∫
π /4
∫
r=
4a cos θ
sin2 θ
r2 ( cos2 θ − sin2 θ)
r2
0
4a cos0
π /2
 r2  sin2 θ
cos 2θ ⋅  
dθ
 2 0
π /4
π /2
16a
∫π/4 (1 − 2 sin2 θ) 2
2
∫
π /2
= 8a2
∫
π /2
= 8a2
∫
π /2
∫
π /2
= 8a2
= 8a2
⋅ r dr dθ
π /4
π /4
π /4
π /4
cos2 θ
dθ
sin4 θ
(1 − 2 sin2 θ)(1 − sin2 θ) dθ
sin4 θ
1 − 3 sin2 θ + 2 sin4 θ
dθ
sin 4 θ
cosec2θ (1 + cot2 θ) − 3cosec2θ + 2  dθ
 cot2 θ cosec2θ − 2 cosec2θ + 2 dθ
Multiple Integrals and their Applications

= 8a2 

π /2
∫
π /4
π

π /2
cot2 θ cosec2θ dθ + 2 (cot θ)π/4 + ( 2θ) π2 

4
π
, t = 1

4

π
θ = , t = 0
2

Limits for θ =
Let cot θ = t so that – cosec2 θ dθ = dt.

= 8a 

2
∫
0
1
387
0
π  = 8a  − t3 − 2 + π 
−t dt + 2 (0 − 1) + 


2 
2
 3 1
2
π 5
= 8a 2  −  .
2 3
a
x/a
Example 33: Evaluate the integral ∫0 ∫x / a
( x2 + y2 ) dxdy by changing to polar coordinates.
Solution: The above integral has already been discussed under change of order of integration
in cartesian co-ordinate system, Example 7.
For transforming any point P(x, y) of cartesian coordinate to polar coordinates P(r, θ), we
take x = r cosθ, y = r sinθ and J =
⇒
cos θ 
r cos θ
x

implies r2 sin 2 θ =
i.e., r  r sin2 θ −
 =0

a 
a
a
either r = 0 or
Limits, for the curve y =
θ = tan
and for the curve y =
−1
r=
cos θ
a sin2 θ
x
,
a
Y
y
BA
1
= tan−1
= tan−1
x
OB
a
x
a
x
a
P
A (a, 1)
y=
x
a
θ=0
B (a, 0)
θ
θ = tan−1
O
(0, 0)
0 π
=
a 2
Here r (as a function of θ) varies from 0 to
−1
and θ changes from tan
y =
θ = π/2
The parabola y2 =
∂ ( x, y)
= r.
∂ (r, θ )
1
π.
to
2
a
cos θ
a sin2 θ
Fig. 5.45
X
388
Engineering Mathematics through Applications
Therefore, the integral,
a x/ a
∫ ∫
0 x/ a
( x2 + y2 )

I=

1
tan −1   
 a 
transforms to.
∫
π /2
I=
∫
π /2
=
1
4
0
cos θ 
r = 
 a sin2 θ 
cot−1(a) 0
dr dθ
π/2
cos4 θ
dθ
cot−1 a a4(sin4 θ)2
∫
π
∫−1 cot4 θ (1 + cot2 θ) cosec2 θ dθ
1
I= 4
4a
⇒
∫
∫

r dr  dθ

 cos θ 


a sin2 θ  3
2
cot a
−1
Let cot θ = t so that cosec2θ dθ = dt (– 1) and θ = cot a ⇒ t = a
π
θ=
⇒ t = 0 

2
1 0 4
I = 4 ∫ t (1 + t2 ) ( − dt )
∴
4a a
1
I =4 4
a
∫
a
0
1
t 4 + t 6  dt = 4
4a
a
 t 5 t7 
 5 + 7 
0
 a
a3 
I=
+ .
 20 28 
n
Example 34: Evaluate ∫ ∫ xy(x2 + y2 ) 2 dxdy over the positive quadrant of x2 + y2 = 4,
supposing n + 3 > 0.
[SVTU, 2007]
Solution: The double integral is to be evaluated over the area enclosed by the positive
quadrant of the circle x2 + y2 = 4, whose centre is (0, 0) and radius 2.
Y
Let x = r cosθ, y = r sinθ, so that x2 + y2 = r2 .
π
θ=
Circle r = 2
Therefore on transformation to polar co-ordinates,
2
I=
=
=
∫
θ=π /2
θ= 0
∫
r =2
r =0
π /2
2
0
0
∫ ∫ (r
∫
π /2
0
P
r cos θ r sin θ rn J dr dθ,
n+ 3


∂ ( x, y )
= r
dr ) sin θ cos θ dθ,  J =
∂ (r, θ )


θ
O
2
 rn+ 4 

 sin θ cos θ dθ
n + 4 0
Fig. 5.46
θ=0
X
Multiple Integrals and their Applications
389
π
2n + 4 2
=
∫ sin θ cos θ dθ
n+4 0
π /2
2n+ 4
sin 2 θ
=
⋅
(n + 4 ) 2 0
=
, using
∫
f ´( x ) f ( x) dx =
f 2 ( x)
2
2n+ 3
, (n + 3 ) > 0.
(n + 4 )
πa
3
Example 35: Transform to cartesian coordinates and hence evaluate the ∫ ∫ r sin θ cos θ drdθ .
00
[NIT Kurukshetra, 2007]
Solution: Clearly the region of integration is the area enclosed by the circle r = 0, r = a
between θ = 0 to θ = π.
π
∫ ∫ r sin θ cos θ dr dθ
I=
Here
0
a
3
0
π
Y
∫ ∫ r sin θ ⋅ r cos θ⋅ r dr dθ
=
0
a
Circle r = a
or x 2 + y 2 = a 2
0
On using transformation x = r cosθ, y = r sin θ,
I=
∫ ∫
=
a
xy dx dy
−a 0

x
−a 
∫
a
θ
θ=π
O
∫
a2 − x2
0
 y2 
=
 
−a  2 
∫
P
y = a2 − x2
θ=0
X

y dy dx

a2 − x2
a
Fig. 5.47
x dx
0
1 a
∫ x ( a2 − x2 ) dx
2 −a
As x and x3 both are odd functions, therefore net value on integration of the above integral
is zero.
=
I =
i.e.
1 a 2
∫ ( a x − x3 ) dx = 0.
2 −a
ASSIGNMENTS 5
Evaluate the following integrals by changing to polar coordinates:
∫∫
a
(1)
0
0
a2 − y2
(x + y ) dxdy
2
2
(2)
∫∫
a
a
0
y
x2
dxdy
x2 + y2
390
Engineering Mathematics through Applications
a
(3) ∫
−a −
a2 − x2
∫
∞∞
− (x
(4) ∫ ∫ e
dxdy
a2 − x2
2 + y2
)dx dy
[MDU, 2001]
0 0
5.8 TRIPLE INTEGRAL (PHYSICAL SIGNIFICANCE)
The triple integral is defined in a manner entirely analogous to the definition of the double
integral.
Let F(x, y, z) be a function of three independent variables x, y, z defined at every point in
a region of space V bounded by the surface S. Divided V into n elementary volumes δV1, δV2,
…, δVn and let (xr, yr, zr) be any point inside the rth sub division δVr. Then, the limit of the
sum
Z
n
∑ F ( xr , yr , zr ) δ vr ,
…(1)
r =1
Z = f2(x, y)
if exists, as n → ∞ and δVr → 0 is called the
‘triple integral’ of R(x, y, z) over the region V, and
is denoted by
∫ ∫ ∫ F ( x, y, z) dV
…(2)
In order to express triple integral in the
‘integrated’ form, V is considered to be subdivided by planes parallel to the three coordinate
planes. The volume V may then be considered as
the sum of a number of vertical columns extending
from the lower surface say, z = f1(x, y) to the upper
surface say, z = f2(x, y) with base as the elementary
areas δAr over a region R in the xy-plance when all
the columns in V are taken.
On summing up the elementary cuboids in the
same vertical columns first and then taking the sum
for all the columns in V, it becomes


r

Z = f1(x, y )
O
Y
R
δAr
X
Fig. 5.48
∑  ∑ F ( xr , yr , zr ) δz  δAr …(3)
r
with the pt. (xr, yr, zr) in the rth cuboid over the element δAr.
When δAr and δz tend to zero, we can write (3) as
z = f2 ( x, y )
F
z = f1 (x , y )
∫ ∫
R
( x, y, z) dz dA
Note: An ellipsoid, a rectangular parallelopiped and a tetrahedron are regular three dimensional regions.
5.9. EVALUATION OF TRIPLE INTEGRALS
For evaluation purpose, ∫ ∫ ∫ F ( x , y , z ) dV
V
is expressed as the repeated integral
…(1)
Multiple Integrals and their Applications
391
∫x1 ∫y1 ∫z1 F ( x, y, z ) dz dy dx
x2 y2 z2
…(2)
where in the order of integration depends upon the limits.
If the limits z1 and z2 be the functions of (x, y); y1 and y2 be the functions of x and x1, x2 be
constant, then
x = b  y =φ2 ( x )  z = f2 ( x, y )
 
I = ∫  ∫  ∫ F ( x, y, z ) dz dy dx
…(3)
x= a 
 
 y =φ1 ( x)  z= f1 ( x, y)
which shows that the first F(x, y, z) is integrated with respect to z keeping x and y constant
between the limits z = f1(x, y) to z = f2(x, y). The resultant which is a function of x, y is
integrated with respect to y keeping x constant between the limits y = f1(x) to y = f2(x).
Finally, the integrand is evaluated with respect to x between the limits x = a to x = b.
Note: This order can accordingly be changed depending upon the comfort of integration.
a x x+ y
x + y + z dz dy dx
.
Example 36: Evaluate ∫ ∫ ∫ e
[KUK, 2000, 2009]
00 0
Solution: On integrating first with respect to z, keeping x and y constants, we get
a x
(x + y)
 e( x + y )+ z 
,
I=
[Here (x + y) = a, (say), like some constant]
 0 dy dx
0 0 
∫∫
=
∫∫
=
∫ ∫ e (
=
a
a
0
∫
e( x + y ) + ( x + y ) − e( x + y ) + 0  dydx


x
0
∫
=
0
0
a
x
2 x + y)
0
− e( x + y )  dydx

x
 e2x + 2y ex + y 
 2 − 1  dx, (Integrating with respect to y, keeping x constant)
0
a
0
 e4x e2x   e2x ex  
 2 − 1  −  2 − 1   dx


On integrating with respect to x,
a
 e4x e2x e2x ex 
=
−
−
+ 
2
4
1 0
 8
1 1 1
 e4a e2a e2a

=
−
−
+ ea  −  − − + 1


 8

2
4
8 2 4
⇒
3
 e4a 3 2a
− e + ea −  .
I=
 8
4
8
Example 37: Evaluate
π /2
a sin θ
0
0
∫ ∫
∫
a2 − r 2
a
0
r dr d θ dz . [VTU, 2007; NIT Kurukshetra, 2007, 2010]
392
Engineering Mathematics through Applications
Solution: On integrating with respect to z first keeping r and θ constants, we get
π /2
∫ ∫
I=
0
0
=
1
a
π /2
∫ ∫
=
1
a
∫
π /2
=
1
a
∫
π /2
0
∫∫
log y 
1 0

∫
ex
1
a sin θ
 2 r2 r 4 
− 
 a
2
4 0
dθ, (On integrating with respect to r)
 a2 ⋅ a2 sin2 θ a4 sin4 θ
−

 dθ
2
4
∫ 2 sin2 θ − sin4 θ dθ
0
a3  π 
3   5π a3
 1 −   =

4 2
8 
64
I=
e
0
( a2 − r2 ) r dr dθ
(p − 1) ⋅ (p − 3)…  π
×
;only if p is even 
2

(p) ⋅ (p − 2)…
Example 38: Evaluate
Solution:
a sin θ
r dr dθ
a3  1 π 3 ⋅ 1 π 
⋅ ,
2⋅ ⋅ −
4  2 2 4 ⋅ 2 2 
sinp x dx =
∴
( z)
0
0
4
=
π /2
0
a2 − r2
a
0
π
3 2
a
=
∫
a sin θ
log y
e
∫∫ ∫
1
0
ex
1
log z dz dy dx.
[MDU, 2005; KUK, 2004, 05]

log z dz dxdy

[Here z = f(x, y) with z1 = 1 and z2 = ex + 0y
=
∫∫
e
1
log y 
0

∫
ex
1

log z ⋅ 1 dz dx dy

Ist
fun.
=
∫∫
log y
=
∫∫
log y
e
1
e
1
0
0
IInd
fun.
ex
log z × z − z 1 dz  dx dy
z  1

∫
( ex log ex − 1 ⋅ log 1) − ( z )ex  dx dy
1 

Multiple Integrals and their Applications
=
e
∫ ∫

1 
log y
0
393
x
 x
 
 xe − ( e − 1) dx dy
∫ ∫ ( ( x − 1) e + 1 dx) dy
= ∫  xe − 2e + x 
dy
= ∫ ( y + 1) ⋅ log y + 2 (1 − y ) dy
=
e
log y
1
x
0
e
x
log y
x
0
1
e
1
I
function
II
function
On integrating by parts,
e

 y2


I = log y ×  + y −
 2
1


∫
e
1
e
2y2  


1  y2

⋅
+ y dy +  2y −


2  1 
y  2


1
 e2

= (log e)  + e − log 1 ⋅  + 1 −


2

2

e
 2

 y2

=  e + e −  + y + 2e − e2 − 1
 4
1
 2

y


∫  2 + 1 dy + (2e − e ) − (2 − 1)
e
2
1
1
 e2
e2

=  + e − − e + + 1 + 2e − e2 − 1
4
4
2

1
=  (1 + 8e − 3e2 ) .
4

1
z
∫ ∫∫
Example 39: Evaluate
−1 0
x+ z
x−z
(x + y + z ) dx dy dz.
[JNTU, 2000; Cochin, 2005]
Solution: Integrating first with respect to y, keeping x and z constant,
z 
x+z
 
y2
+ yz   dx dz
  xy +
0 
2
x − z 
I=
∫ ∫
=


−1 
∫ ∫ ( 4zx + 2z ) dx dz
=
∫
 x2

4z + 2 ⋅ z2 ⋅ x  dz
−1 
2

0
=
∫
1
−1
1
2
2

0
z
1


z2
4z ⋅ + 2z2 ⋅ z  dz

−1 
2

1
∫
=4
1
−1
z3 dz = 4
z4
4
1
=0
−1
394
Engineering Mathematics through Applications
ASSIGNMENT 6
Evaluate the following integrals:
(1)
∫ ∫ ∫ x yz dxdydz
(3)
∫∫ ∫
1
2
0 0
4
0
2
2
1
2 z
0
0
4z − x2
∫∫∫
(4)
∫ ∫∫
a
b
c
− a −b −c
log 2
dy dx dz
( x2 + y2 + z2 ) dxdydz
(2)
x
0
0
x + log y
0
ex + y + zdzdydx
[VTU, 2000]
[NIT Kurukshetra, 2008]
5.10 VOLUME AS A DOUBLE INTEGRAL
(Geometrical Interpretation of the Double Integral)
One of the most obvious use of double integral is the determination of volume of solids
viz. ‘volume between two surfaces’.
Z
If f(x, y) is a continuous and single valued function
defined over the region R in the xy-plane with z = f(x, y)
S
C
as the equation of the surface. Let ¬ be the closed curve
which encloses R. Clearly, the surface R (viz. z = f(x, y))
is the orthogonal projection of S(viz z = F(x, y)) in the
xy-plane.
Divided R into elementary rectangles of area δxδy
by drawing lines parallel to the axis of x and y. On each
of these rectangles errect prisms having their lengths
O
Y
parallel to the z-axis. The volume of each such prism is
zδx δy.
(Division of R is performed with the lines x = xi (i = 1,
R
δx
2, …, m) and y = yj(j = 1, 2, …, n). Through each line
δ R, δ y
x = xi, pass a plane parallel to yz-plane, and through X
each line y = yj, pass a plance parallel to xz-plane. The
Fig. 5.49
rectangle ∆Rij whose area is ∆Aij = ∆xi ∆yj will be the
base of a rectangle prism of height f(xij, hij ), whose
volume is approximately equal to the volume between the surface and the xy-plane x = xi – 1,
n
(
)
x = xi ; y = yi – 1 y = yi. Then ∑ f ξij , ηij ∆xi ⋅ ∆yj gives an approximate value for volume V of
i =1
j =1
the prism of the cylinder enclosed between z = f(x, y) and the xy-plane.
The volume V is the limit of the sum of each elementary volume z δxδy.
V = Lt ∑ ∑ z δ x δ y = ∫ ∫ z dx dy = ∫ ∫ f ( x, y ) dA
x 0
∴
δ →
δy→0
R
R
Note: In cyllidrical co-ordinates, the equation of the surface becomes z = f(r, θ), elementary area dA = r dr dθ
and volume = ∫ ∫ f (r , θ ) r dr dθ
R
Multiple Integrals and their Applications
395
Problems on Volume of a Solid with the Help of Double Integral
Example 40: Find the volume of the tetrahedron bounded by the plane
the co-ordinate planes.
x y z
+ + = 1 and
a b c
[Burdwan, 2003]
x y z
x y

+ + = 1 ⇒ z = f ( x, y ) = c  1 − − 
…(1)

a b c
a b
If f(x, y) is a continuous and single valued function over the region R (see Fig. 5. 50) in the
xy plane, then z = f(x, y) is the equation of the surface. Let C be the closed curve that is the
boundary of R. Using R as a base, construct a cylinder having elements parallel to the z-axis.
This cylinder intersects z = f(x, y) in a curve Γ , whose projection on the xy-plane is C.
Solution: Given,
Z = C(1– x/a – y/b ) = f(x, y)
Z
(0, 0, c)
c
b
(0, b, 0)
P
Y
Q
Y
a
(a , 0, 0)
R
R
C
X
Fig. 5.50
Fig. 5.51
The equation of the surface under which the region whose volume is required, may be
x y

written in the form (1) i.e., z = c  1 − −  .

a b
Hence the volume of the region =

y
∫∫ adA = ∫∫ c  1 − a − b  dx dy
R
x
R
The equation of the inter-section of the given surface with xy-plane is
x y
+ =1
…(2)
a b
If the prisms are summed first in the y-direction they will be summed from y = 0 to the line
x
y = b  1 − 

a
Therefore,
V=
∫∫
=c
a
x
b  1− 
 a
0
0
∫
a
0

x y
c  1 − −  dy dx

a b
xy y2 
−
− 
y

2b 
a
b(1− x / a )
dx
0
396
Engineering Mathematics through Applications
=c
1
x2 
∫ b  2 − a + 2a  dx
a
x
2
0
a
 x x2
x3 
= cb  −
+ 2
 2 2a 6a  0
 a a2
a 3  abc
= bc  −
+ 2=
.
6
 2 2a 6a 
Example 41: Prove that the volume enclosed between the cylinders x2 + y2 = 2ax and
128 a2 .
z2 = 2ax is
15
Solution: Let V be required volume which is enclosed by the cylinder x2 + y2 = 2ax and the
paraboloid z2 = 2ax.
Only half of the volume is shown in Fig 5.52.
Z
Now, it is evident from that z = 2ax is to be evaluated
over the circle x2 + y2 = 2ax(with centre at (a, 0) and radius
a.
Z 2 = 2ax
Here y varies from − 2ax − x2 to 2ax − x2 on the
circle x2 + y2 = 2ax and finally x varies from x = 0 to x = 2a
∴
∫ ∫
V=2
− 2ax − x2
0
∫
=2
2a 
 2 ⋅
0
∫
=4
∫
2a
2ax − x2
∫
∫
2a
0
as z = f(x, y)
(a , 0)
2 ax − x2

dy dx

2ax − x2
dx = 4
∫
2a
0
Fig. 5.52
2ax 2ax − x2 dx
x 2a − x dx
Let x = 2a sin2θ, so that dx = 4a sinθ cosθ dθ. Further, for x = 0, θ = 0 

π 
x = 2a, θ = .
2 
∴
V = 4 2a
= 64 a3
∫
π
0
∫
π
0
2
2
2a sin2 θ 2a cos θ ⋅ 4a sin θ cos θ dθ
sin3 θ cos2 θ dθ
X
x2 + y2 = 2ax

2ax  dy dx

0
2ax y 0
0
= 4 2a
[z] dx dy
0

2ax 

2a
0
=4
∫
(2 a, 0)
(0, 0)
2ax − x2
2a
Multiple Integrals and their Applications
= 64 a3
( p − 1)( p − 3 )…(q − 1)(q − 3)… ⋅ 1,
( p + q)( p + q − 2)…
= 64 a3
( 3 − 1) 1 = 128 a3 .
5⋅3
397
p = 3, q = 2
15
Problems based on Volume as a Double Integral in Cylindrical Coordinates
Example 42: Find the volume bounded by the cylinder x2 + y2 = 4 and the hyperboloid
x2 + y2 – z2 = 1 .
Solution: In cartesian co-ordinates, the section of the given hyperboloid x2 + y2 – z2 = 1 in
the xy plane (z = 0) is the circle x2 + y2 = 1, where as at the top and at the bottom end (along
the z-axis i.e., z = ± 3 ) it shares common boundary with the circle x2 + y2 = 4 (Fig. 5.53 and
5.54).
Here we need to calculate the volume bounded by the two bodies (i.e., the volume of
shaded portion of the geometry).
Z
x2 + y2 = 4
(r = 2)
Q
P
Y
O
O
X
x2 + y2 = 1
(r = 1)
Fig. 5.53
Fig. 5.54
(Best example of this geometry is a solid damroo in a concentric long hollow drum.)
In cylindrical polar coordinates, we see that here r varies from r = 1 to r = 2 and θ varies
from 0 to 2π.
∴
V = 2


= 2

∫∫ z dxdy  = 2 ∫∫ f (r, θ) r dr dθ
2π
∫ ∫
2
2π 
2
0
1
∫  ∫
=2
0
1

r2 − 1 rdr dθ

3

1
d ( r2 − 1)2  dθ
3

(³ x2 + y2 – z2 – 1 ⇒ z = x2 + y2 − 1 )
398
Engineering Mathematics through Applications
3 2
∫
=2
(r2 − 1)2
2π
dθ
3
0
1
=2 3
∫
2π
0
d θ = 4π 3 .
Example 43: Find the volume bounded by the cylinder x2 + y2 = 4 and the planes y + z
= 4 and z = 0.
[KUK, 2000; MDU, 2002; Cochin, 2005; SVTU, 2007]
Solution: From Fig. 5.55, it is very clear that z = 4 – y is to be integrated over the circle x2 +
y2 = 4 in the xy-plane.
To cover the shaded portion, x varies from − 4 − y2 to
Hence the desired volume,
V=
∫ ∫
2
4 − y2
− 2 − 4 − y2
∫∫
2
=2
−2 0
4 − y2
( 4 − y ) dxdy

∫−2 ( 4 − y)  ∫0
∫−2 ( 4 − y)
2
∫
2
∫
2
=2
=8
4 − y2

=2
Z
z dxdy
2
=2

dx dy

O
4 − y2 dy
 4 4 − y2 − y 4 − y2  dy

−2 
−2
4 − y2 and y varies from – 2 to 2.
X
Y
Fig. 5.55
4 − y2 dy − 0
(The second term vanishes as the integrand is an odd function)
2
 y 4 − y2 4
y
= 8
+ sin −1  = 16π.
2
2
2

 −2
ASSIGNMENT 7
1. Find the volume enclosed by the coordinate planes and the portion of the plane
lx + my + nz = 1 lying in the first quadrant.
y
x 
2. Obtain the volume bounded by the surface z = c  1 −   1 −  and the quadrant of



a
b
2
2
y
x
the elliptic cylinder 2 + 2 = 1
a
b
[Hint: Use elliptic polar coordinates x = a rcosθ, y = b rsinθ ]
Multiple Integrals and their Applications
399
5.11 VOLUME AS A TRIPLE INTEGRAL
Divide the given solid by planes parallel to the coordinate plane into rectangular
parallelopiped of elementary volume δxδyδz.
Then the total volume V is the limit of the sum of all elementary volume i.e.,
V = Lt ∑ ∑ ∑ δ x δ y δ z =
δx→0
δy→0
δ z→ 0
∫∫∫ dx dy dx
Problems based on Volume as a Triple Integral in cartesian Coordinate System
Example 44: Find the volume common to the cylinders x2 + y2 = a2 and x2 + z2 = a2.
Solution: The sections of the cylinders x2 + y2 = a2 and x2 + z2 = a2 are the circles x2 + y2 = a2
and x2 + z2 = a2 in xy and xz plane respectively.
Here in the picture, one-eighth part of the required volume (covered in the 1st octant) is
shown.
a2 − 1x2 − 0y2 , and x and
a2 − x2 i.e.,
Clearly, in the common region, z varies from 0 to
y vary on the circle x2 + y2 = a2.
The required volume
∴
∫∫
V=8
a
0
y2 = a2 − x2
y1 = 0
a
0
a
0
a
0
x
(z
a2 − x2
0
Z
dz dy dx
) dy dx
2
0
a
O´´
B
2
a
O
P
∫(
=8
a
0
∫
=8
a
0

a2 − x2 

)∫
a2 − x2
(
a2 − x2
0

dy  dx

)
a2 − x2 − 0 dx
a

x3  
2
= 8 ( a − x ) dx = 8  a x −  
0
3  0 

∫
a
2
D
C

a − x dy dx

a2 − x2
∫  ∫
=8
z1 = 0
2− 2
∫∫
=8
∫
z2 = a2 − x2 − 0y2
2

a 3  16a 3
.
= 8 a3 −  =
3
3

(a, 0, 0)
X
x=O
Q
x=a
O´
A
Fig. 5.56
Y
400
Engineering Mathematics through Applications
Example 45: Find the volume bounded by the xy plane, the cylinder x2 + y2 = 1 and the
plane x + y + z = 3.
Solution: Let V(x, y, z) be the desired volume enclosed laterally by the cylinder x2 + y2 = 1
(in the xy-plane) and on the top, by the plane x + y + z = 3 (= a say).
Clearly, the limits of z are from 0 (on the
Z
xy-plane) to z = (3 – x – y) and x and y vary on the
circle x2 + y2 = 1
∴
V ( x, y, z ) =
∫∫
1− x2
=
∫∫
1 − x2
=

1
− 

1
∫
3−x−y
− 1 − 1− x2 0
1
(z (
3 − x −y)
0
− 1 1 1− x2
1− x2
∫ ∫
1
−
dz dy dx
1− x2
) dy dx

(3 − x − y ) dy dx
I=
⇒
Q
x2 + y2 = 1
1− x2

y2 
3
=
−
−
y
xy

−1 
2  −
∫
x=0
O
P

1
∫ (6 ×
1
−1
dx
x+
1− x2
)
3
y=
X
Fig. 5.57
1 − x − 2x 1 − x dx
2
2
π
On taking x = sin θ, we get dx = dθ; For x = − 1, θ = − 
2
π 
For x = 1, θ =

2 
Thus,
V=
=
∫ (6
π /2
−π /2
∫
π /2
−π /2
=6×2
)
1 − sin2 θ − 2 sin θ 1 − sin2 θ cos θ dθ
(6 cos2 θ − 2 sin θ cos2 θ) dθ
∫
π /2
0
∫
cos2 θ dθ − 2
Ist
= 12
Using
∫
π /2
0
(2 − 1) π
cos3 θ
⋅ +2
2
2
3
cosp θ dθ =
π /2
−π /2
sin θ cos2 θ dθ
IInd
π /2
= 3π +
− π /2
2
× 0 = 3π
3
( p − 1)(p − 3 )… ×  π , only if p is even
2

p ( p − 2 )…
and
Y
Multiple Integrals and their Applications
∫
401
f n +1 ( x )
for Ist and IInd integral respectively
n+1
f ´( x ) f n ( x ) dx =
x2 y2 z 2
+
+
= 1.
a2 b2 c 2
[MDU, 2000; KUK, 2001; Kottayam, 2005; PTU, 2006]
Example 46: Find the volume bounded by the ellipsoid
Solution: Considering the symmetry, the desired volume is 8 times the volume of the ellipsoid
Z
into the positive octant.
The ellipsoid cuts the XOY plane in the ellipse
C
x2 y2
+
= 1 and z = 0.
a2 b2
Therefore, the required volume lies between the
ellipsoid
B
x2 y2
z = c 1− 2 − 2
a
b
O
and the plane XOY (i.e., z = 0) and is bounded on the
sides by the planes x = 0 and y = 0
Hence,
∫∫
b 1−
∫∫
b 1−
V=8
=8
a
0
0
a
0
a
0


0
c
V=8
b
α
0
x2 y2
−
a2 b2
∫ 
x2
a2
c 1−
Fig. 5.58
x2 y2
−
dy dx
a2 b2


x2  α 
taking
1
−

 =


a2  b 

α
α2 − y2 α2
y
sin−1  dx
+
2
2
α 
0

 Using formula
c
b
∫
4c
b
∫
=8
=
a
0
a
0
A
X
dz dy dx

c
α2 − y2 dy dx

b
ay
0
∫
c 1−
0
∫ ∫
=8
x2
a2
Y
∫
a2 − x2 dx =
x 2
a2
x
a − x2 + tan−1 
2
2
a

α2

0
sin−1 1 dx
+

2

2πc
π 2
α dx =
2
b
a
1 x3 

= 2πbc  x − 2 
a 3 0

4
= πabc .
3

x2 
x2
b2  1 − 2  dx, α = b 1 − 2

0
a 
b
∫
a
402
Engineering Mathematics through Applications
dxdydz
taken throughout the volume
a − x2 − y2 − z 2
[MDU, 2000]
∫∫∫
Example 47: Evaluate the integral
2
of the sphere.
Solution: Here for the given sphere x2 + y2 + z2 = a2, any of the three variables x, y, z can be
expressed in term of the other two, say z = ± a2 − x2 − y2 .
In the xy-plane, the projection of the sphere is the circle x2 + y2 = a2.
Thus,
a2 − x2
∫∫
a
I=8
0
=8
∫ ∫
a
∫  ∫
a2 − x2
a
a2 − x2
0
0
0
∫  ∫
0
π
2
a
∫  ∫
∫
= 4π
∫
a2 − x2 − y2
0
dz  
dy dx , α 2 = (a2 – x2 – y2)
α2 − z2  
Z

 sin−1 z 

  dy  dx
α 0 
α
C
a2 − x2
0
Z = a 2 − x2 − y2

(sin−1 1 − sin −1 0) dy dx

0
0


dx dy dz
a − x 2 − y2 − z2
2
0
a2 − x2
=8
=8
∫
0
a
=8 
0 
a2 − x2 − y2
a

a2 − x2 
dy dx = 4π  y 0
dx


0

∫
O
A
B
Y
Circle x 2 + y 2 = a 2
Fig. 5.60
X
a
 x a2 − x2 a2
x
+ sin− 1 
a2 − x2 dx = 4π 
2
2
a 0

a
0

a2 π 
= 4π 0 +
I = π2a2.
2 2 

Example 48: Evaluate ∫ ∫ ∫ ( x + y + z ) dx dy dz over the tetrahedron bounded by the planes
x = 0, y = 0, z = 0 and x + y + z = 1.
Solution: The integration is over the region R(shaded portion) bounded by the plane x = 0,
y = 0, z = 0 and the plane x + y + z = 1.
The area OAB, in xy plane is bounded by the lines x + y = 1, x = 0, y = 0
Hence for any pt. (x, y) within this triangle, z goes from xy plane to plane ABC (viz. the
surface of the tetrahedron) or in other words, z changes from z = 0 to z = 1 – x – y. Likewise
in plane xy, y as a function x varies from y = 0 to y = 1 – x and finally x varies from 0 to 1.
whence,
I = ∫ ∫ ∫ ( x + y + z ) dx dy dz
( over R )
=
1
1− x


∫ ∫ ∫

0
0
1− x − y
0
( x + y + z) dz dy dx
Multiple Integrals and their Applications
=
∫∫
=
∫∫
=
∫∫
=
∫∫
a
0
1− x
0
a
0
1 1− x
0 0
1
2
=
1
2
dy dx
2
1 − x − y) 
(
 dy dx
+
1− x 
0
1
=
1− x − y

z2 
 ( x + y ) z + 
2 0
( x + y )(1 − x − y )
2

1− x
1
(1 − x − y )(1 + x + y ) dy dx
0
2
0
∫
1

2
1
1 − ( x + y )  dy dx

2
y −
0 

Z
3 1− x
( x + y)
3

 dx ,

0
1
C (0, 0, 1)

 dx
3  
1 x
(1 − x ) −  −
∫
0
403
3
3
=
(0, 1, 0)
B
O
1
1 2
x2 x 4 
=  x−
+
2 3
2 12 0
P
(1, 0, 0)
A
1 2 1 1  1
− +
=
2  3 2 12  8
Y
Q
(1, 0, 0)
X
Fig. 5.61
ASSIGNMENT 8
1. Find the volume of the tetrahedron bounded by co-ordinate planes and the plane
x y z
[KUK, 2002]
+ + = 1, by using triple integration
a b c
2. Find the volume bounded by the paraboloid x2 + y2 = az, the cylinder x2 + y2 = 2ay and
the plane z = 0.
5.12. VOLUMES OF SOLIDS OF REVOLUTION AS A DOUBLE INTEGRAL
Let P(x, y) be any point in a region R enclosing an elementary
area dx dy around it. This elementary area on revolution
Y
R
about x-axis form a ring of volume,
δx
δV = π[(y + δy)2 – y2] δx = 2πyδxδy …(1)
δy
Hence the total volume of the solid formed by revolution
of this region R about x-axis is,
V = ∫ ∫ 2πy dx dy
R
…(2)
Similarly, if the same region is revolved about y-axis,
then the required volume becomes
V = ∫ ∫ 2πx dx dy
R
…(3)
P (x, y)
y
O
X
Fig. 5.62
404
Engineering Mathematics through Applications
Expressions for above volume in polar coordinates about the initial line and about the
2
2
pole are ∫ ∫ 2πr sin θ dr dθ and ∫ ∫ 2πr cos θ dr dθ respectively.
R
R
Example 49: Find by double integration, the volume of the solid generated by revolving
x 2 y2
the ellipse 2 + 2 = 1 about y-axis.
a
b
B
x2 y2
+
= 1 is symmetrical about the
a2 b2
y-axis, the volume generated by the left and the right halves
overlap.
Hence we shall consider the revolution of the right-half ABD
Solution: As the ellipse
V=
∫∫
b
a 2 2
b −y
b
−b 0
∫
= 2π
a
=
a2
b2
∫
b
0
Fig. 5.63
2π x dx dy
b2 − y2
 x2  b
= 2π  
− b  2 0
b
Q
A
D
y2
and y-varies from – b to b.
b2
for which x-varies from 0 to a 1 −
∴
P
C
πa2
b2
dy =
(b2 − y2 ) dy =
∫
b
−b
2π a2
b2
(b2 − y2 ) dy
b
 2
y3 
b y − 3 

0
4 2
πa b .
3
Example 50: The area bounded by the parabola y2 = 4x and the straight lines x = 1 and y
= 0, in the first quadrant is revolved about the line y = 2. Find by double integration the
volume of the solid generated.
Y
Solution: Draw the standard parabola y2 = 4x to which
the straight line y = 2 meets in the point P(1, 2), Fig. 5.64.
Now the dotted portion i.e., the area enclosed by
parabola, the line x = 1 and y = 0 is revolved about the line
y = 2.
∴ The required volume,
V=
∫∫
1
2 x
0 0
2 x
1
8 3
10π
= 2π  x 2 − x2  =
3
3
0
x=1
X´
y2 
= 2π  2y −  dx = 2π
0 
2 0
∫
1
2π ( 2 − y ) dx dy
2–y
∫ (4
1
0
y 2 = 4x
y=2
P (1, 2)
y=0
O
)
x − 2x dx
Y´
Fig. 5.64
X
Multiple Integrals and their Applications
405
Example 51: Calculate by double integration, the volume generated by the revolution of
θ) about its axis.
the cardiod r = a(1 – cosθ
[KUK, 2007, 2009]
Soluton: On considering the upper half of the cardiod, because due to symmetry the lower
half generates the same volume.
Y
∴
a(1− cos θ )
π
∫∫
= 2π∫
V=
0
0
3 a (1− cos θ )
π
r
3
0
2π a3
=
3
2π r2 sin θ dr dθ
∫
π
0
sin θ dθ
0
(1 − cos θ)
3
2πa 3 (1 − cos θ )
=
3
4
4
θ
θ=π
θ=0
X
sin θ dθ
π
=
0
θ = π/2
8πa 3 .
3
Fig. 5.65
Example 52: By using double integral, show that volume generated by revolution of
8
θ) about the initial line is πa3 .
cardiod r = a(1 + cosθ
3
Solution: The required volume
π
∫∫
0
a(1+ cos θ)
0
∫
π
∫
π
= 2π
= 2π
0
0
a (1+ cos θ )
r 
 3 
0
3
2πr2 sin θdr dθ
Y
sin θ dθ
a3 (1 + cos θ) sin θ dθ
3
4 π
2πa  (1 + cos θ )
=
−
3 
4
3
=−
2π a3
3
θ = π/2
θ=π
θ
=
θ=0
X


0
24  8π a3 .

0 − 4  = 3
Fig. 5.66
ASSIGNMENT 9
1. Find by double integration the volume of the solid generated by revolving the ellipse
x2 y2
+
= 1 about the X-axis.
a2 b2
2. Find the volume generated by revolving a quadrant of the circle x2 + y2 = a2, about its
diameter.
3. Find the volume generated by the revolution of the curve y2(2a – x) = x3, about its
asymptote through four right angles.
4. Find the volume of the solid obtained by the revolution of the leminiscate r2 = a2 cos2θ
about the initial line.
[Jammu Univ., 2002]
406
Engineering Mathematics through Applications
5.13. CHANGE OF VARIABLE IN TRIPLE INTEGRAL
For transforming elementary area or the volume from one sets of coordinate to another, the
necessary role of ‘Jacobian’ or ‘functional determinant’ comes into picture.
(a) Triple Integral Under General Transformation
Here
∫∫∫ f (x, y, z) dx dy dz = ∫∫∫ F(u, v, w)| J | du dvdw; where J =
R( x, y, z)
R '(u, v, w)
∂(x, y, z)
(≠ 0) …(1)
∂(u, v, w)
Since in the case of three variables u(x, y, z), v(x, y, z), w(x, y, z) be continuous together
with their first partial derivatives, the Jacobian of u, v, w with respect to x, y, z is
defined by
∂u
∂x
∂v
∂x
∂w
∂x
∂u
∂y
∂v
∂y
∂w
∂y
∂u
∂z
∂v
∂z
∂w
∂z
(b) Triple Integral in Cylindrical Coordinates
Here
∫ ∫ ∫ f ( x, y, z ) dx dy dz = ∫ ∫ ∫ F (r, θ, z ) J dr dθ dz , where |J| = r
R
R´
The position of a point P in space in cylindrical coordinates is determined by the
three numbers r, θ, z where r and θ are polar co-ordinates of the projection of the point
P on the xy-plane and z is the z coordinate of P i.e., distance of the point (P) from the
xy-plane with the plus sign if the point (P) lies above the xy-plane, and minus sign if
below the xy-plane (Fig. 5.67).
Z
Z
∆θ
∆z
P
N
P(x, y, z)
M
z
O
∆θ
O
Y
θ
y
r
θ
Q
Q
X
X
Fig. 5.67
Y
∆r
R
r ∆θ
Fig. 5.68
In this case, divide the given three dimensional region R' (r, θ, z) into elementary
volumes by coordinate surfaces r = ri, θ = θj, z = zk (viz. half plane adjoining z-axis,
circular cylinder axis coincides with Z-zxis, planes perpenducular to z-axis). The
Multiple Integrals and their Applications
407
curvilinear ‘prism’ shown in Fig. 5. 68 is a volume element of which elementary base
area is r ∆r∆θ and height ∆z, so that ∆v = r ∆r ∆θ ∆z.
Here θ is the angle between OQ and the positive x-axis, r is the distance OQ and z is
the distance QP. From the Fig. 5.62, it is evident that
x = r cosθ, y = r sinθ, z = z and so that,
cos θ
sin θ 0
 x, y, z 
J
=
−
r
sin
θ
r
cos θ 0 = r
 u, v, w 
0
0
1
…(2)
Hence, the triple integral of the function F(r, θ, z) over R´ becomes
V = ∫ ∫ ∫ F ( r, θ, z) r dr dθdz
R´(r , θ, z)
…(3)
(c) Triple Integral in Spherical Polar Coordinates
V = ∫ ∫ ∫ f ( x, y, z ) dxdydz = ∫ ∫ ∫ F ( r , θ, φ ) J drdθ dφ , where |J| = r2 sinθ
Here
R
R
The position of a point P in space in spherical coordinates is determined by the
three variables r, θ, φ where r is the distance of the point (P) from the origin and so
called radius vector, θ is the angle between the radius vector on the xy-plane and the
x-axis to count from this axis in a positive sense viz. counter-clockwise.
For any point in space in spherical coordinates, we have
0 ≤ r ≤ ∞, 0 ≤ θ ≤ π, 0 ≤ φ ≤ 2π.
Divide the region ‘R’ into elementary volumes ∆V by coordinate surfaces, r = constant
(sphere), θ = constant (conic surfaces with vertices at the origin), φ = constant (half
planes passing through the Z-axis ).
To within infinitesimal of higher order, the volume element ∆v may be considered
a parallelopiped with edges of length ∆r, r ∆θ, r sinθ ∆φ. Then the volume element
becomes ∆V = r2 sinθ ∆r ∆θ ∆φ.
Z
S´
Z
P´
∆φ
r sin
P (x , y,.z)
θ
r
r
O
A
Q´
r∆θ
Q
z
O
φ
x
R
θ
Y
R´
∆r
P
∆θ
z
θ
S
90°
90°
y
X
x
L
y
X
Fig. 5.69
Y
∆φ
φ
Fig. 5.70
L
408
Engineering Mathematics through Applications
For calculation purpose, it is evident from the Fig. 5.69 that in triangles, OAL and
OPL,
x = OL cos φ = OP cos (90 – θ) . cos φ = r sinθ cos φ,
y = OL sin φ = OP sin θ . sin φ = r sin θ sin φ,
z = r cos θ.
Thus,
J=
sin θ cos φ
sin θ sin φ
cos θ
∂ ( x, y, z)
= r cos θ cos φ r cos θ sin φ −r sin θ = r2 sin θ
∂ ( r, θ, φ ) −r sin θ sin φ r sin θ cos φ
0
Problems Volume as a Triple Integral in Cylindrical Co-ordinates
Example 53: Find the volume intercepted between the paraboloid x2 + y2 = 2az and the
cylinder x2 + y2 – 2ax = 0.
Z
Solution: Let V be required volume of the cylinder
x2 + y2 – 2ax = 0 intercepted by the paraboloid x2 + y2 = 2az.
Transforming the given system of equations to polarcylindrical co-ordinates.
x = r cos θ

Let y = r sin θ  sothat V(x, y, z) = V(r , θ, z)

z=z
x 2 + y 2 = 2az
Paraboloid
(a, 0) O
θ
r
Y
By above substitution the equation of the paraboloid becomes
x 2 + y 2 – 2 ax = 0
r2
cylinder
2
2
2
r = 2az ⇒ z =
and the cylinder x + y = 2ax gives
2a
X
r 2 – 2ar cosθ = 0 ⇒ r(r – 2a cosθ) = 0 with r = 0 and
Fig. 5.71
r = 2a cosθ.
r2
Thus, it is clear from the Fig. 5.71 that z varies from 0 to
and r as a function of θ varies
2a
from 0 to 2a cosθ with θ as limits 0 to 2π. Geometry clearly shows the volume covered under
1
th of the full volume.
4
the +ve octant only, i.e.
∫
V = V' = 4
( x , y , z)
(r ,θ , z)
=4
π /2 


0
∫
=4
∫
r = 2a cos θ
r=0
0
∫
=4
θ=π /2
∫
π /2 

0
1
2a
∫
z=0
2a cos θ
0
∫
∫
z = r2 /2a
r2 /2 a
r [ z]
0
2a cos θ
0
r
4
0

rdr  dθ

r3 
dr dθ
2a 
π /2 4 2a cos θ
0
r dz drdθ, as| J |= r
dθ
Multiple Integrals and their Applications
1
2a
=4
= 23 a3
=
∫
π /2
0
409
24 a4
cos4 θ dθ
4
( 4 − 1)( 4 − 3) π
4×2
2
3π a3
.
2
Example 54: Find the volume of the region bounded by the paraboloid az = x2 + y2 and
the cylinder x2 + y2 = b2. Also find the integral in case when a = 2 and b = 2.
Solution: On using the cylindrical polar co-ordinates (r, θ, z) with x = r cosθ, y = r sin θ, so
that the equations of the cylinder and that of the paraboloid are r = b and z =
See Fig. 5.72, only one-fourth of the common volume is shown.
Hence in the common region, z varies from z = 0 to z =
π
respectively.
2
∴ The desired volume
r2
respectively.
a
r2
and r and θ varies on the circle
a
from 0 to b and 0 to
π /2
b
r2 / a
0
0
∫ ∫∫
V=4
0
π /2 

 0 rdr 
∫ ∫
b
π /2 
b
=4
rdrdθ dz
0

dz  dθ

∫
r2 / a
 r2 

0
∫  ∫ r  a  dr dθ
=4
=
4
a
0
∫
0
π /2  4 b 
r

 dθ
 4 0
0
π /2
Fig. 5.72
4 b
πb
× θ
=
2a
a 4 0
As a particular case, when a = 2, b = 2, then
4
π (2)
V=
= 4π
2×2
=
4
2
Problmes on Volume in Polar Spherical Co-ordinates
Example 55: Find the volume common to the sphere x2 + y2 + z2 = a2 and the cone x2 + y2 = z2
OR
2
2
2
Find the volume cut by the cone x + y = z from the sphere x2 + y2 + z2 = a2.
[NIT Kurukshetra, 2010]
410
Engineering Mathematics through Applications
Solution: For the given sphere, x2 + y2 + z2 = a2 and the cone x2 + y2 = z2, the centre of the
sphere is (0, 0, 0) and the vertex of the cone is origin. Therefore, the volume common to the
two bodies is symmetrical about the plane z = 0, i.e. the required volume, V = 2∫ ∫ ∫ dxdydz
x = r sin θ cos φ

In spherical co-ordinates, we have y = r sin θ sin φ  ; J = r2 sin θ

z = r cos θ
Thus, x2 + y2 + z2 = a2 becomes r2 = a2 i.e., r = a
and
x2 + y2 = z2 becomes r2 sin2θ (cos2φ + sin2φ) = r2 cos2θ
i.e.,
sin2θ = cos2θ i.e. θ = π/4.
Clearly, the volume shown in the figure (Fig. 5.73) is
one-fourth, i.e. in first quadrant only and, in the common
region,
Z
x2 + y2 = z2
90°
r varies from 0 to a, 
π 
θ varies from 0 to ,
4 
π 
φ varies from 0 to
2 

O
Hence the required volume,

V = 2 4

X
π /4
π /2

∫ ∫ ∫ r sin θdr dθ dφ
0
0
π /2
∫ ∫
π /4
π /2
π /4
=8
0
0
∫ ∫
=8
0
0
∫
a
2
Fig. 5.73
0


∫
a
0
Z

r2 dr  sin θ dθ dφ

Q
a
 r3 
  sin θ dθ dφ
3 0
π /2
x2+ y2= z2
P
8 3
a
[ − cos θ]π0 /4 dφ
0
3
8 3
1  π/2
= a 1 −
dφ

3 
2 0
=
Y
φ
θ
∫
Y
X
4π a3 
1 
=
1 −

3 
2
Fig. 5.74
Alternately: In polar-cylindrical co-ordinates, intersection of the two curves x2 + y2 + z2 = a2
a2
and x2 + y2 = z2 results in z2 + z2 = a2 or z2 =
.
2
a
a
a2 a2 ⇒
, i.e. r varies from 0 to
Further, x2 + y2 = a2 − z2 = a2 −
r=
=
2
2
2
2
Hence,
∫ ∫ (
V=2
2π
0
a/ 2
0
)
a2 − r2 − r r dr dθ
Multiple Integrals and their Applications
411
|³ P lies on the cone whereas Q lies on the sphere as a function of (r, θ)
∫ (r
=2
a/ 2
0
)∫

a2 − r2 − r2 

2π
0

dθ dr

a
1
1
3
 1 2
r3  2 since r ( a2 − r2 )2 = −1  −3r ( a2 − r2 )2  = −1 d ( a2 − r2 )2 
2 3/2


= 4π  − ( a − r ) − 

3 
3
3 0 

 3
 1 a3
a3 
1 a3
= 4π  −
−
+ 
3
 32 2 32 2
3
4π a 
1 
=
1−

3 
2 
Example 56: By changing to shperical polar co-ordinate system, prove that
∫∫∫
1–
V
2
2
2
x2 y2 z2
π
– 2 – 2 dx dy dz = abc where V = ( x, y,z ) : x + y + z ≤ 1 
2
4
a
b
c
a2 b2 c2


x2 y2 z2
Solution: Taking x = u, , so that 2 + 2 + 2 ≤ 1

a
b
c
a

y
= v, 
b

z
= w

c
∂x ∂x
Now transformation co-efficient,
∂u ∂v
∂y ∂y
J =
∂u ∂v
∂z ∂z
∂u ∂v
V =
∴
∫ ∫∫
V ( x , y , z)
1−
⇒ u2 + v2 + w2 ≤ 1
∂x
∂w
a 0 0
∂y
= 0 b 0 = abc
∂w
0 0 c
∂z
∂w
x2 y2 z2
−
− dx dy dz
a2 b2 c2
= ∫ ∫ ∫ 1 − u2 − v2 − w2 ( abc) du dv dw
V´(u, v, w )
To transform to polar spherical co-rodinate system, let u = r sin θ cos φ,

v = r sin θ sin φ, 

w = r cos θ
Then
V´(u, v, w) = {(u, v, w): u2 + v2 + w2 ≤ 1, u ≥ 0, v ≥ 0, w ≥ 0} reduces to
2
V”
(r, θ, φ) = {r ≤ 1 i.e., 0 ≤ r ≤ 1, 0 ≤ θ ≤ π, 0 ≤ φ ≤ 2π}
∴
= ∫ ∫ ∫ 1 − u2 − v2 − w2 abc du dv dw
V´(u, v, w )
412
Engineering Mathematics through Applications
= ∫ ∫ ∫ abc 1 − r2 J dr dθ dφ
where |J| = r2 sin θ
V" (r ,θ, φ )
∫
V”( r,θ,φ) = abc
⇒
φ= 2π 

φ= 0
π

∫  ∫
0


1 − r2 r2dr sin θ dθ dφ


1
0
Now put r = sin t so that dr = cost dt and for r = 0, t = 0,

π
r = 1, t = 
2 


2π
π
0
2π 
0
∫ ∫
2π
π
= abc


0
0


∫ ∫
0
=
πabc
16
=
πabc
16
∫
2π
0
∫
0
0


cost sin2 t cos t dt sin θ dθ dφ


 1 1 π  sin θ dθ d φ

 4 2 2
[ − cos θ]0π dφ
2π
0
π/2
( 2 − 1) ⋅ ( 2 − 1) π  sin θ dθ dφ




 ( 2 + 2 )( 4 − 2 ) 2 
π
= abc


∫ ∫ ∫
V"
( r,θ ,φ ) = abc
∴
2 dφ =
πabc
8
∫
2π
0
dφ =
π2 abc
.
4
Example 57: By change of variable in polar co-ordinate, prove that
∫∫
1
0
1– x2
0
1– x2 – y 2
∫
0
dz dy dx
π2 .
=
8
1 – x2 – y2 – z2
OR
Evaluate the integral being extended to octant of the sphere x2 + y2 + z2 = 1.
OR
Evaluate above integral by changing to polar spherical co-ordinate system.
Solution: Simple Evaluation:
I=
Treating
∫ ∫
1
0
dx
1− x2
∫
dy
0
1
as
(1 − x − y2 ) − z2
2
I=
∫ ∫
1
0
dx
0
1− x2
1− x2 − y2
0
dz
1 − x2 − y2 − z2
1
a − z2
2
1− x2 − y2 

−1 z
sin

 dy
a0


Multiple Integrals and their Applications
∫ dx∫
1− x2
=
∫ ∫
1− x2
=
π
2
∫
1
=
π
2
∫
1
1
=
0
0
1
dx
0
0

z
 sin−1

1 − x2 − y2

413
1− x2 − y2 
 dy
2
2
 , as a = 1 − x − y

0
 π − 0 dy
2

 y 1− x2  dx
( )0 
0
0
1 − x2 dx
1

π  x 1 − x2 1
= 
+ sin −1 x  , using
2
2
2
0
∫
a2 − x2 dx =
x a2 − x2 a2
x
+ sin−1
2
2
a
1 π  π2
π
+
=
0
2 
2 2  8
By change of variable to polar spherical co-ordinates, the region of integration
=
V = {(x, y, z); x2 + y2 + z2 ≤ 1; x ≥ 0, z ≥ 0, y ≥ 0.}

π
π

I =  (r, θ, φ); r2 ≤ 1, i.e. 0 ≤ r ≤ 1, 0 ≤ θ ≤ , 0 ≤ φ ≤
2
2

becomes
x = r sin θ cos φ,

y = r sin θ sin φ, 

z = r cos θ
where
J=
Now
whence
∫∫∫
∂ ( x, y, z )
= coefficient of transformation = r2 sinθ.
∂ ( r , θ, φ )
dx dy dz
=
1 − x2 − y2 − z2
V
I=
∫
π /2
0
∫
dφ
π /2 
π /2
1 r 2 sin θ
0
0
0
∫ ∫ ∫

 sin θ 
0
π /2
∫
1
0
1 − r2
dr dθ dφ
r2  
 dr dθ
1 − r2  
Let r = sin t so that dr = cos t dt. Further, when r = 0, t = 0,

π
r = 1, t = 
2 
∴
}
I=
∫
=
π /2
0
∫
dφ
π /2
0
∫
π /2
0
∫
dφ
sin θ dθ
π /2
0
∫
π /2
0
sin2 t
⋅ cos t dt
cos t
1 π
dθ sin θ  ⋅  ;
2 2
414
Engineering Mathematics through Applications
=
π
4
=
π
4
∫
π /2
∫
π /2
∫
dφ
0
π /2
0
sin θ dθ
dφ ( − cos θ)
0
π /2
0
π /2
π
= φ
4 0
π2
.
=
8
Example 58: Find the volume of the ellipsoid
ordinates.
x2 y2 z2
+
+
= 1 by changing to polar coa2 b2 c2
[PTU, 2007]
Solution: We discuss this problem under change of variables.
Take
∂ ( x, y, z )
y
x
z
= X,
= Y, = Z so that J =
= abc
a
b
c
∂ (X , Y , Z )
∴ The required volume,
V = ∫ ∫ ∫ dx dy dz = ∫ ∫ ∫ J dX dY dZ
= abc∫ ∫ ∫ dX dYdZ , taken throughout the sphere X2 + Y2 + Z2 = 1.
Change this new system (X, Y, Z) to spherical polar co-ordinates (r, θ, φ) by taking
X = r sin θ cos φ,

Y = r sin θ sin φ,  so that J´= ∂ (X , Y , Z) = r2 sin θ ,

∂ ( r, θ, φ)
Z = r cos θ
V = abc∫ ∫ ∫ J dr dθ dφ = abc∫ ∫ ∫ r2 sin θ dr dθ dφ
taken throughout the sphere r2 ≤ 1, i.e. 0 ≤ r ≤ 1, 0 ≤ θ ≤ π, 0 ≤ φ ≤ 2π
On considering the symmetry,
π /2 
π /2




∫ ∫ ∫ r dr sin θ dθ dφ
V = abc ⋅ 8
= 8 abc

0
π /2 
∫  ∫
0
∫
π /2
∫
π /2
=
8
abc
3
=
8
abc
3
=
8
abcφ
3
0
0
0
π/2
0
1
2
0
1

r
sin θ dθ dφ
30

π /2 3
0
[ − cos θ]0π /2dφ
1 ⋅ dφ
=
π 4
8
abc = πabc
3
2 3
Multiple Integrals and their Applications
415
Miscellaneous Problem
3
2
2
Example 59: Evaluate the surface integral I = ∫ ∫ ( x dy dz + x y dz dx + x z dx dy) .
S
where S is the surface bounded by z = 0, z = b, x2 + y2 = a2.
OR
By transformation to a triple Integral, evaluate I = ∫ ∫ ( x3 dy dz + x2 y dz dx + x2 z dx dy) , where
S
S is the surface bounded by z = 0, z = b, x2 + y2 = a2.
Solution: On making use of Green’s Theorem,
I=
∫ ∫(
a
b
a2 − y2
−a 0
∫∫
a
+
b
−a 0
+
) dz dy − ∫ ∫ ( −
a
3
b
a2 − y2
−a 0
x2 a2 − x2 dzdx −
∫ ∫ x (−
a
a
2
−a −a
a2 − y2
( a2 − y2 ) b dx dy − ∫−a ∫−
a −y
∫∫
a
−a −
2
a
2
) dz dy
3
)
a2 − x2 dz dx
a2 − y2
a2 − y2
0 dx dy
Using Divergence Theorem,
I = ∫ ∫ ∫ ( 3x2 + x2 + x2 ) dx dy dz
V
a
∫ ∫
a2 − x2
a
a2 − x2
=4
0


0
∫ ∫
=4
0


= 20 b
=
0
∫x
a
0
2




∫ dz dy  5x dx
b
2
0

bdy  5x2 dx

a2 − x2 dx
5 4
πa b .
4
Note: As direct calculation of the integral may prove to be instructive. The evaluation of the integral can be
carried out by calculating the sum of the integrals evaluated over the projections of the surface S on the coordinate planes. Thus, which upon evaluation is seen to check with the result already obtained. It should be
noted that the angles α, β, γ are mode by the exterior normals in the +ve direction of the co-ordinate axes.
416
Engineering Mathematics through Applications
ANSWERS
Assignmet 1
1.
 π2 
 
4
2.
a4
3
3.
1
ab
6.
π
4
Assignment 2
ax

x
dy dx
1. ∫  ∫ 2
2

00 x + y
∫∫
a
2.
0
a2 − x2
− a2 − x2
a sin α
∫
3.
f ( x, y ) dy dx
a
y
m
y
l
∫ ∫
0
a
f ( x, y ) dxdy + ∫
∫
a sin α
0
ma
4.
y⋅cos α
f (x, y) dx dy +
∫
la
ma
a2 − y2
∫
0
f (x, y) dxdy
Assignment 3
1.
4a2
3
2.
3
π b4 − a 4 )
2 (
4
23
3. a  π + 
4
3
Assignment 4
2.
1
sq. units
10
Assignment 5
1.
πa4
units
8
2.
a3
( π + 2) units
12
3.
2π
units
9
4.
π
units
4
Assignment 6
1. 1
3. 8π
8 3
a bc ( 3 + 2ab2 + 2ac2 )
9
8
19
log 2 −
4.
9
9
2.
Assignment 7
1
1. 6 lmn
 π 13 
2. abc  − 
4 24
f ( x, y ) dx dy
Multiple Integrals and their Applications
417
Assignment 8
1. abc/6
2.
3πa3
2
2.
2 2
πa
3
4.
π a3  1
log

4  2
Assignment 9
1.
4π ab2
3
3. 2π2 a3
(
1
2 +1 − 
3
)