Swinburne University of Technology
School of Science, Computing and Engineering Technologies
MTH20011
Solution to Tutorial 11
Topic: double integrals using Cartesian and polar coordinates.
1. Evaluate the volume bounded by z = x2 + y 2 and z = 0 over the rectangular region
1 ≤ x ≤ 2 and 0 ≤ y ≤ 3.
Solution
The volume V is given by the double integral
Z Z
V =
(x2 + y 2 )dA,
D
where D denotes the region in the xy plane over which the volume is calculated and
dA represents the differential of the area of the region D.
Using Cartesian coordinates we set dA = dxdy or dA = dydx. In case of the rectangular
region D we may choose any order of integration: either
Z 2
Z 2Z 3
Z 2
y 3 y=3
27
2
2
2
2
V =
(x + y )dydx =
x y+
3x +
dx =
dx
3 y=0
3
1
1
0
1
27x 2
= x3 +
= 8 + 18 − (1 + 9) = 16.
3
1
or
Z 3
Z 3 3
x=2
8 1
x
2
2
2
+ xy
− + 2y − y dy
dy =
=
(x + y )dxdy =
3
3 3
x=1
0
0
1
0
7
y3 3
=
y+
= 7 + 9 = 16.
3
3 0
Z 3Z 2
V
2
2
2. Evaluate the volume under z = x2 + 3y + 2 and above z = 0 over the region in the xy
plane bounded by y = x and y = x2 between x = 0 and x = 1.
Solution
The volume V is given by the double integral
Z Z
V =
(x2 + 3y + 2)dA,
D
where D denotes the region in the xy plane over which the volume is calculated and dA
represents the differential of the area of the region D. Using Cartesian coordinates we
set dA = dxdy or dA = dydx. The upper and the lower bounds of each integral depend
on the order of integration and must be determined from the sketch below
Mathematics 4A
Page 1 of 6
Swinburne University of Technology
School of Science, Computing and Engineering Technologies
MTH20011
Solution to Tutorial 11
1.5
x2
x
y
1
0.5
0
0
0.2
0.4
0.6
0.8
x
1
1.2
1.4
In case of dA = dydx
Z 1 Z y=x
V
=
=
=
=
=
Z 1
y=x
3 2
(x + 3y + 2)dydx =
x y + y + 2y
dx
2
y=x2
0
y=x2
0
Z 1
3 2
4
2
2
2
x (x − x ) + (x − x ) + 2(x − x ) dx
2
0
Z 1
3 2
3
4
4
2
x − x + (x − x ) + 2x − 2x dx
2
0
4
5
x
3 x3 3 x5
2 3 1
x
2
−
+
−
+x − x
4
5
2 3
2 5
3
0
3
2
15 − 12 + 30 − 18 + 60 − 40
35
7
1 1 1
− + −
+1− =
=
= .
4 5 2 10
3
60
60
12
2
2
In case of dA = dxdy
Z 1 Z x=√y
V
=
0
(x2 + 3y + 2)dxdy
x=y
Z 1 3
√
x= y
x
=
+ 3yx + 2x
dy
3
x=y
0
#
Z 1 " 3/2
y − y3
√
√
=
+ 3y( y − y) + 2( y − y) dy
3
0
"
#
1
2 y 5/2
y4
y 5/2
4
2
1
6
4
7
=
=
−
+6
− y 3 + y 3/2 − y 2
−
+ −1+ −1= .
5 3
12
5
3
15 12 5
3
12
0
3. Determine the area of the planar region that can be described in polar coordinates (r, θ)
by r ≤ 1 + cos(θ) and 0 ≤ θ ≤ π/2.
Solution
The region is shown in the figure below.
Mathematics 4A
Page 2 of 6
Swinburne University of Technology
School of Science, Computing and Engineering Technologies
MTH20011
Solution to Tutorial 11
2
1.5
1
0.5
0
0
0.5
1
1.5
2
The area A of any planar region D can be determined as
Z Z
dA
A=
D
In polar coordinates we set
dA = rdr dθ
r=1+cos θ
The inner integral is over r. The limits for r are r
Z π/2 Z r=1+cos θ
A =
Z π/2 rdrdθ =
0
=
=
=
=
r=0
Z π/2
. So we have
r=0
0
1 2
r
2
r=1+cos θ
r=0
dθ
Z
1 π/2
(1 + cos θ) dθ =
(1 + cos2 θ + 2 cos θ)dθ
2 0
0
Z
1 π/2
1 + cos (2θ)
1+
+ 2 cos θ dθ
2 0
2
π/2
1 3
1
θ + sin(2θ) + 2 sin θ
2 2
4
0
3π
+1
8
1
2
2
4. Use polar coordinates to evaluate
Z Z
(x + 2)dA,
D
where D is the upper half of the circle x2 + y 2 = 1.
Solution
Using
x = r cos θ, y = r sin θ.
Mathematics 4A
Page 3 of 6
Swinburne University of Technology
School of Science, Computing and Engineering Technologies
MTH20011
Solution to Tutorial 11
For the upper half of the circle x2 + y 2 = r2 = 1 we have
Z Z
Z π Z r=1
(x + 2)dA =
(r cos θ + 2)rdrdθ
D
0
r=0
Z π 3
r=1
r
2
=
dθ
cos θ + r
3
r=0
0
Z π
1
=
cos θ + 1 dθ
3
0
π
1
=
sin θ + θ = π.
3
0
5. Write each of the following double integrals with the order or integration reversed.
Sketch the region of integration. Evaluate both double integrals and verify that the
same result is obtained in each case.
(a)
Z 2 Z ex
dydx
0
1
Solution
The region bounded by y = 1 and y = ex between x = 0 and x = 2 is shown in
the graph below.
1
ex
6
4
2
0
0
0.5
1
x ylabel
1.5
2
Evaluate the original integral
Z 2 Z ex
Z 2
Z 2
y=ex
2
dydx =
y
dx =
(ex − 1)dx = (ex − x) = e2 − 3.
0
1
y=1
0
0
0
To reverse the order of integration we find the inverse function
y = ex ⇒ x = ln(y).
Then
Z e2 Z x=2
Z 2
dxdy =
1
x=ln(y)
x=2
x
1
x=ln y
dy
Z e2
(2 − ln y)dy = (2y − y ln y + y)
=
1
Mathematics 4A
e2
1
= 3e2 − 2e2 − 3 = e2 − 3.
Page 4 of 6
Swinburne University of Technology
School of Science, Computing and Engineering Technologies
MTH20011
Solution to Tutorial 11
(b)
Z 1 Z 3x+2
dydx
−2
x2 +4x
Solution
The region bounded by y = x2 + 4x and y = 3x + 2 between x = −2 and x = 1 is
shown in the graph below.
x2 + 4x
3x + 2
4
2
0
−2
−4
−2
−1.5
−1
−0.5
0
x ylabel
0.5
1
Evaluate the original integral
Z 1 Z 3x+2
Z 1
3x+2
dydx =
y 2
dx
−2
x2 +4x
x +4x
−2
Z 1
2
1
x3
3x
+ 2x −
− 2x2
(3x + 2 − (x2 + 4x))dx =
2
3
−2
−2
1
12
8
3
+2− −2−
−4+ −8
2
3
2
3
7
10
9
− −
= .
6
3
2
=
=
=
To reverse the order of integration we find the inverse functions
p
y = x2 + 4x ⇒ x = −2 + 4 + y
and
y = 3x + 2 ⇒ x =
Z 5 Z x=√4+y−2
dxdy =
−4
x=(y−2)/3
=
=
=
Mathematics 4A
y−2
3
Z 5 p
y−2
4+y−2−
dy
3
−4
5
2
y2 2
3/2
(4 + y) − 2y −
+ y
3
6
3
−4
2 3/2
25 10
2
42 8
3/2
(9) − 10 −
+
−
(4 − 4) + 8 −
−
3
6
3
3
6
3
20 − 25
32
9
18 − 10 +
−8+
= .
6
6
2
Page 5 of 6
Swinburne University of Technology
School of Science, Computing and Engineering Technologies
(c)
MTH20011
Solution to Tutorial 11
Z 1 Z 1+√1−y2
dxdy
y
0
Solution
2
1.5
1
0.5
0
0
0.5
1
1.5
2
To integrate we use the identity
Z p
xp
1
1 − x2 dx =
1 − x2 + arcsin(x).
2
2
Z 1 Z 1+√1−y2
dxdy =
0
Z 1h
y
1+
i
p
1 − y 2 − y dy
0
=
yp
y2
1
y+
1 − y 2 + arcsin(y) −
2
2
2
1
0
=
1 π
+ .
2 4
In the reversed order
Z 2 Z √1−(x−1)2
Z 1Z x
dydx +
0
0
Z 1
dydx =
1
0
=
=
Mathematics 4A
Z 2p
xdx +
1 − (x − 1)2 dx
0
1
2
1
x − 1p
1
+
1 − (x − 1)2 + arcsin(x − 1)
2
2
2
1
1 π
+ .
2 4
Page 6 of 6