Tutorial Note - Week 08 - Solution
Double Integrals
1. (a)
i. Integrate x first: (Figure 1a)
Z 3Z
ZZ
2
3
2y − 3xy dA =
2
R
ii. Integrate y first: (Figure 1b)
Z 2Z
ZZ
2
3
2y − 3xy dA =
3
Z
2
y
3
y
3
1
Figure 1
1
Figure 1a
y=3
y=0
π
6
0
15 2 −
15
sin y dy =
2
4
R
ii. Integrate y first: (Figure 2b)
ZZ
Z 4Z π
Z
6
x sin y dA =
x sin y dy dx =
1
4
1
0
R
y=
y
π
6
y=0
1
Figure 2
Figure 2a
1
4
π
6
x=4
4
x
0
y=
x=1
1
y=0
y
π
6
x=4
x
0
√ 3
√ √ !
15 2 − 3
3
x dx =
x−
2
4
π
6
x=1
x=4
x=1
R
y=
π
6
2
Figure 1b
1
0
x
2
i. Integrate x first: (Figure 2a)
Z
Z πZ 4
ZZ
6
x sin y dx dy =
x sin y dA =
y
dy
1
x=2
x 0
y=0
2
x=1
x=2
x=1
x=2
x=1
2
Z
y=3
x 0
y=0
1
2
3
R
0
3
y
y=3
3
π
6
2
3
2 3 3 4
y − xy
dx
2y − 3xy dy dx =
3
4
1
0
1
0
2
Z 2
243
243 2
585
18 −
=
x dx = 18x −
x
=−
4
8
8
1
1
R
(b)
3
3
2xy − x2 y 3
2y − 3xy dx dy =
2
0
1
0
3
Z 3
2 3 9 4
585
9 3
2
=−
=
2y − y dy = y − y
2
3
8
8
0
0
2
x
0
1
y=0
Figure 2b
4
2. (a)
i. Integrate x first:
D = {(x, y) | y ≤ x ≤ 1, 0 ≤ y ≤ 1} (Figure 3a)
ZZ
Z 1Z 1
Z 1
e2 − 2e + 1
x+y
x+y
ey+1 − e2y dy =
e
dA =
e
dx dy =
2
0
y
0
D
ii. Integrate y first:
D = {(x, y) | 0 ≤ x ≤ 1, 0 ≤ y ≤ x} (Figure 3b)
ZZ
Z 1Z x
Z 1
e2 − 2e + 1
x+y
x+y
e2x − ex dx =
e
dA =
e
dy dx =
2
0
0
0
D
y
y
x
=
x
0
1
y
=
x
=
y
0
y=0
y=0
0
Figure 3
1
x=1
x
0
x=1
x=1
D
1
y
1
x
1
y
x
0
0
Figure 3a
y=0
1
Figure 3b
(b) Before finding the limits for x and y, we have to find the intersection points of
y = x − 6 · · · (1)
.
y2 = x
· · · (2)
Substitute (1) into (2):
(x − 6)2 = x =⇒ x2 − 13x + 36 = 0 =⇒ (x − 4) (x − 9) = 0 =⇒ x = 4 or x = 9.
Hence, the intersection points are (4, −2) and (9, 3).
i. Integrate x first:
D = {(x, y) | y 2 ≤ x ≤ y + 6, −2 ≤ y ≤ 3} (Figure 4a)
Z 3
Z 3 Z y+6
ZZ
500
3
3
4y 3 y + 6 − y 2 dy =
4y dx dy =
4y dA =
3
−2
−2 y 2
D
ii. Integrate y first:
√
√
D1 = {(x, y) | 0 ≤ x ≤ 4, − x ≤ y ≤ x} (Figure 4b)
Z 4
ZZ
Z 4 Z √x
3
3
4y dy dx =
0 dx = 0
4y dA =
√
0
D1
− x
0
√
D2 = {(x, y) | 4 ≤ x ≤ 9, x − 6 ≤ y ≤ x} (Figure 4b)
Z 9
Z 9 Z √x
ZZ
2
500
3
3
x − (x − 6)4 dx =
4y dy dx =
4y dA =
3
4
x−6
4
D2
Therefore,
ZZ
D
3
4y dA =
ZZ
D1
2
3
4y dA +
ZZ
D2
4y 3 dA =
500
.
3
y
(9, 3)
3
x
2
y =
y
D
0
y
3
b
x
3
2
x=
y
x
0
9
b
−3
−3
y =x−6
Figure 4a
x
Figure 4b
i. Integrate x first:
D1 = {(x, y) | 0 ≤ x ≤ 2y, 0 ≤ y ≤ 1} (Figure 5a)
ZZ
Z 1 Z 2y
Z
xy dA =
xy dx dy =
0
D2
9
y = −√
x
y =x−6
−3
x=y+6
√x
D1
0
9
(4, −2)
Figure 4
3. (a)
y=
0
1
2y 3 dy =
0
1
2
D1
D2 = {(x, y) | 0 ≤ x ≤ 4 − 2y, 1 ≤ y ≤ 2} (Figure 5a)
ZZ
Z 2 Z 4−2y
Z 2 3
4y − 16y 2 + 16y
5
xy dA =
xy dx dy =
dy =
2
6
1
0
1
D2
Therefore,
ZZ
xy dA =
ZZ
xy dA +
4
xy dA = .
3
D2
D1
D
ZZ
ii. Integrate
y first:
D = (x, y) | 0 ≤ x ≤ 2, x2 ≤ y ≤ 2 − x2 (Figure 5b)
ZZ
Z 2 Z 2− x
Z 2
2
4
2x − x2 dx =
xy dA =
xy dy dx =
x
3
0
0
2
D
y
2
y
A
b
y
x=
2
D2
D1
O
x
0
1
2
0
2y
2y
x=
0
x
2
Figure 5
(b)
y=
x=0
b
b
2
x=0
0
B
D
1
4−
Figure 5a
2−
y=
0
x
2
x
2
0
2
Figure 5b
i. Integrate
x first:
D = (x, y) | 0 ≤ x ≤ 3 − y2 , 0 ≤ y ≤ 2 (Figure 6a)
Z 2 3
Z 2 Z 3− y
ZZ
2
11
y − 12y 2 + 36y
dy =
xy dx dy =
xy dA =
4
2
0
0
0
D
3
x
ii. Integrate y first:
D1 = {(x, y) | 0 ≤ x ≤ 2, 0 ≤ y ≤ 2} (Figure 6b)
Z
Z 2Z 2
ZZ
xy dy dx =
xy dA =
2x dx = 4
0
0
0
2
D1
D2 = {(x, y) | 2 ≤ x ≤ 3, 0 ≤ y ≤ 6 − 2x} (Figure 6b)
ZZ
Z 3 Z 6−2x
Z 3
3
2x3 − 12x2 + 18x dx =
xy dA =
xy dy dx =
2
2
0
2
D1
Therefore,
ZZ
xy dA =
ZZ
xy dA =
Q
y=2
2
y=
D1
6−
y
2x
3−
x=0
x=
x=0
D
1
y
y=2
2
b
11
.
2
D2
y
y
P
b
xy dA +
D1
D
2
ZZ
b
R x
O
b
0
1
2
x
0
3
y=0
0
Figure 6
4. (a)
0
3
0
3y
y=0
0
Figure 6a
3
Figure 6b
R=
{(x, y) | 3y ≤ x ≤ 3, 0 ≤ y ≤ 1}
(Figure 7a)
x
(Figure 7b)
R = (x, y) | 0 ≤ x ≤ 3, 0 ≤ y ≤ 3
Z 3Z x
Z 1Z 3
Z
3
x2
x2
e dxdy =
e dy dx =
0
D2
2
0
0
0
3
e9 − 1
1 x2
xe dx =
3
6
Note: use substitution, let u = x2 .
y
1
y
x
1
3
=
y
3y
=
x
0
y=0
0
Figure 7a
(b)
x
0
3
y=0
y2
0
3
Figure 7b
R = {(x, y) | y 2 ≤ x ≤ 9, 0 ≤ y ≤√3}
(Figure 8a)
R = {(x, y) | 0 ≤ x ≤ 9, 0 ≤ y ≤ x} (Figure 8b)
Z
Z 9 Z √x
Z 3Z 9
2
2
y cos x dy dx =
y cos x dxdy =
0
x=3
x=3
x
0
0
0
Note: use substitution, let u = x2 .
4
9
1
sin 81
x cos x2 dx =
2
4
x
y
y
3
3
2
y=0
0
x
0
9
y=0
0
Figure 8a
y
0
9
Figure 8b
√
(Figure 9a)
R = (x, y) | y ≤ x ≤ 1, 0 ≤ y ≤ 1
(c)
2
R = {(x, y) | 0 ≤ x ≤ 1, 0 ≤ y ≤ x }
(Figure 9b)
Z 1Z 1 √
Z 1 Z x2 √
Z
x3 + 1 dxdy =
x3 + 1 dy dx =
√
0
x=9
x
0
√ x
y=
y
x=9
x=
0
0
Note: use substitution, let u = x3 + 1.
y
1
√
2 3
22 − 1
x2 x3 + 1dx =
9
y
1
x2
y=
x=
x
0
0
y=0
x
0
1
0
Figure 9a
x=1
x=1
√
y
1
y=0
1
Figure 9b
Change of Variables in Multiple Integrals
1.
∂x
J = ∂u
∂y
∂u
∂x
∂v
∂y
∂v
1 1 1
3 3 = 2 1 =
−
3
3
3
7
5
1
1
(u + v) + 4
(v − 2u) = v − u
f (x, y) = 3x + 4y =⇒ f (x (u, v) , y (u, v)) = 3
3
3
3
3
(
1
1
x = 3u + 3v
u= x−y
=⇒
2
1
v = 2x + y
y = −3u + 3v
Region R is bounded by four straight lines, see Figure 10. Solving for the
intersection
1
5
points, we know that R is a rectangle with vertices (0, 0), (1, 1), 3 , − 3 , and 32 , − 43 .
For each point in the xy-plane, we have to use the above system of equations to find the
corresponding point in the uv-plane:


(x, y) = (0, 0) =⇒ (u, v) = (0, 0)


 (x, y) = (1, 1) =⇒ (u, v) = (0, 3)
(x, y) = 35 , − 13 =⇒ (u, v) = (2, 3) .



 (x, y) = 2 , − 4 =⇒ (u, v) = (2, 0)
3
3
5
Or, by rewriting the equations:

y=x



y =x−2
y
= −2x



y = 3 − 2x
=⇒
=⇒
=⇒
=⇒
x−y =0
x−y =2
2x + y = 0
2x + y = 3
=⇒
=⇒
=⇒
=⇒
u=0
u=2
.
v=0
v=3
So, region S is a rectangle in the uv-plane and
S = {(u, v) | 0 ≤ u ≤ 2, 0 ≤ v ≤ 3} (Figure 10a).
Hence,
ZZ
3
Z
(3x + 4y) dA =
0
R
2
Z
0
Z 3
1
7
11
14
5
10
1
dv = .
v − u du dv =
v−
3
3
3
3 0
3
3
3
v
y
1
x
b
3−
1
b
y=
x
−2
y=
x−
u=0
x
2
u=2
2x
R
b
−1
b
y=
y=
v=3
3
b
S
2
b
b
b
v=0
Figure 10
2.
Figure 10a
2 3
J = 3 −2
Or,

y




y

y




y
= − 32 x
= −13
f (x (u, v) , y (u, v)) = 5u + v
(
2
3
u = 13
x + 13
y
x = 2u + 3v
=⇒
3
2
y = 3u − 2v
v = 13 x − 13 y

(x, y) = (0, 0)
=⇒ (u, v) = (0, 0)



(x, y) = (2, 3)
=⇒ (u, v) = (1, 0)
(x,
y)
=
(5,
1)
=⇒ (u, v) = (1, 1)



(x, y) = (3, −2) =⇒ (u, v) = (0, 1)
=⇒ 2x + 3y = 0 =⇒
= − 23 x +
= 32 x
= 23 x −
2
13
2
13
3
=⇒ 2x + 3y = 13 =⇒
=⇒ 3x − 2y = 0 =⇒
=⇒ 3x − 2y = 13 =⇒
6
2
x
13
2
x
13
3
x
13
3
x
13
+
+
−
−
3
y
13
3
y
13
2
y
13
2
y
13
= 0 =⇒ u = 0
= 1 =⇒ u = 1
= 0 =⇒ v = 0
= 1 =⇒ v = 1
.
u
So, region S is a square in the uv-plane and
S = {(u, v) | 0 ≤ u ≤ 1, 0 ≤ v ≤ 1} (Figure 11a).
Hence,
ZZ
(x + y) dA =
1
Z
0
R
Z
1
(5u + v) |−13| du dv = 13
0
y
3
5
v+
2
dv = 39.
b
x
13
2
5
3
x−
4
S
2
−2
3x
3
b
y=
2
b
b
v=0
Figure 11
3.
∂u
1 ∂x
= ∂v
J
∂x
Figure 11a
∂u
∂y
∂v
∂y
x
y
= 2
y 2xy
= xy 2 = v =⇒ J = 1
v
xy 2
v2
v
=
= y =⇒ f (x, y) = y 2 = 2
u
xy
u
Since u = xy and v = xy 2 , then the boundary of R becomes
xy = u = 1, xy = u = 2, xy 2 = v = 1, and xy 2 = v = 2.
So, region S is a square in the uv-plane and
S = {(u, v) | 1 ≤ u ≤ 2, 1 ≤ v ≤ 2} (Figure 12a).
Hence,
ZZ
R
b
13
3
R
1
y=
b
u=1
−2
v=1
1
−2
3x
+
y=
2
3
x
y=
b
−1
0
u=0
1
1
v
b
2
Z
(x + y) dA =
Z
1
2
Z
1
2
7
v2
u2
Z 2
1
3
v
du dv =
dv = .
v 4
1 2
1
u
y
2
v
b
xy =
xy 2 = 1
S
b
u=2
R
b
b
u=1
2
2 =
xy
xy = 1
1
v=2
2
b
2
b
x
0
b
0
2
b
v=1
4
Figure 12
u
2
Figure 12a
4.
2 0
J = 0 3
=6
f (x (u, v) , y (u, v)) = (2u)2 = 4u2
Since x = 2u and y = 3v, then
9x2 + 4y 2 = 36 =⇒ 9 (2u)2 + 4 (3v)2 = 36 =⇒ u2 + v 2 = 1
So, region S is a unit circle in the uv-plane and
√
√
S = (u, v) | − 1 ≤ u ≤ 1, − 1 − u2 ≤ v ≤ 1 − u2 (Figure 13a).
Hence,
ZZ
(x + y) dA =
Z
1
−1
R
Z
√
48
1
u
−1
2
√
1−
u2
4u
√
− 1−u2
Let u = sin θ, then du = cos θ dθ and
θ = − π2 .
Z
1−u2
√
du =48
=6
|6| dv du = 48
Z
1
−1
√
u2 1 − u2 du.
1 − u2 = cos θ. When u = 1, θ = π2 . When u = −1,
Z
Z
2
π
2
2
2
sin θ cos θ dθ = 12
− π2
π
2
− π2
(1 − cos 4θ) dθ = 6π
8
Z
π
2
− π2
sin2 2θ dθ
y
v
9x2 + 4y 2 = 36
3
1
u2 + v 2 = 1
S
R
u
x
−2
−1
2
1
−3
−1
Figure 13
Figure 13a
9