Module MA1132 (Frolov), Advanced Calculus
Tutorial Sheet 7
To be solved during the tutorial session Thursday/Friday, 10/11 March 2016
Name:
You may use Mathematica to sketch the integration regions and solids, and to check the
results of integration.
1. Evaluate the double integral
ZZ p
4x2 − y 2 dxdy ,
I=
R
where R is the region enclosed by the lines y = 0, y = x, and x = 1.
Solution: The region R is shown below
1.0
0.8
0.6
0.4
0.2
0.0
0.0
0.2
0.4
0.6
0.8
1.0
We consider R as a region of type I. Then I is given by
Z 1 Z x p
Z 1 p
y
y y=x
2
2
dx
I=
4x − y dy dx =
4x2 − y 2 + 2x2 arcsin
2
2x y=0
0
0
0
√
(1)
Z 1 √
3 π 2
1 3 π =
+
x dx =
+
≈ 0.637741 .
2
3
3 2
3
0
Rp
Here the integral
4x2 − y 2 dy can be evaluated by integrating by parts
Z p
Z
Z
p
p
p
y2
2
2
2
2
2
2
2
2
4x − y dy = y 4x − y − yd 4x − y = y 4x − y + p
dy
4x2 − y 2
Z
Z p
p
4x2
2
2
p
= y 4x − y +
dy −
4x2 − y 2 dy .
4x2 − y 2
(2)
1
Comparing the l.h.s with the r.h.s, one finds
Z
Z p
yp
4x2
1 p 2
y
2
2
2
4x − y dy =
y 4x − y + p
4x2 − y 2 + 2x2 arcsin
.
dy =
2
2
2x
4x2 − y 2
(3)
One can add any function of x to this expression which does not change the definite
integral.
2. Sketch the integration region R and reverse the order of integration
(a)
4
Z
12x
Z
f (x, y)dydx
(4)
3x2
0
Solution: The region R is shown below
40
30
20
10
0
0
1
2
3
Reversing the order of integration one gets
Z
Z 4 Z 12x
f (x, y)dydx =
0
3x2
4
48
Z √y/3
f (x, y)dxdy .
0
(5)
y/12
(b)
1
Z
Z
3x
f (x, y)dydx
0
2x
Solution: The region R is shown below
3.0
2.5
2.0
1.5
1.0
0.5
0.0
0.0
0.2
0.4
0.6
2
0.8
1.0
(6)
Reversing the order of integration one gets the sum of two repeated integrals
1
Z
Z
3x
2
Z
y/2
Z
f (x, y)dydx =
0
3
Z
Z
1
f (x, y)dxdy +
2x
0
y/3
f (x, y)dxdy .
2
(7)
y/3
3. Sketch the integration region R and write the expression as one repeated integral by
reversing the order of integration
Z 3Z 1
Z 1Z y
f (x, y)dxdy
(8)
f (x, y)dxdy +
0
y 2 /9
1
y 2 /9
Solution: The region R is shown below
3.0
2.5
2.0
1.5
1.0
0.5
0.0
0.0
0.2
0.4
0.6
0.8
1.0
Reversing the order of integration one gets
Z
1
Z
y
Z
3
Z
1
Z
f (x, y)dxdy +
0
y 2 /9
1
Z
f (x, y)dxdy =
1
y 2 /9
√
3 x
f (x, y)dydx .
0
(9)
x
4. Find the volume V of the solid bounded by
(a) the planes x = 0, y = 0, z = 0, the cylinder x2 + y 2 = a2 and the hyperbolic
paraboloid z = xy in the first octant.
Solution: The solid, and its projection R onto the xy-plane are shown below (a = 2)
3
2.0
1.5
1.0
0.5
0.0
0.0
0.5
1.0
1.5
2.0
Thus, the volume is
ZZ
Z
V =
a
√
Z
xydxdy =
R
0
0
a2 −x2
1
xydydx =
2
Z
a
0
1
1
1
x(a2 − x2 )dx = a4 − a4 = a4 .
4
8
8
(10)
2
2
(b) the planes y = 0, z = 0, y = ab x, and the elliptic cylinder xa2 + zc2 = 1.
Solution: The solid, and its projection R onto the xy-plane are shown below
(a = 1, b = 2, c = 3)
2.0
1.5
1.0
0.5
0.0
0.0
0.2
0.4
0.6
0.8
1.0
Thus, the volume is
r
ZZ r
Z a Z bx/a r
Z
x2
x2
bc a
x2
V =
c 1 − 2 dxdy =
c 1 − 2 dydx =
x 1 − 2 dx
a
a
a 0
a
R
0
0
(11)
Z 1 √
1
abc
= abc
x 1 − x2 dx = −abc (1 − x2 )3/2 |10 =
.
3
3
0
4