Q 25
So'S
4-4(x + y) dxdy is
(A) 35
(B) 45
(C)241/60
(D) 60
Answers
Answer:
Problem 1 (16.1.15). Use symmetry to evaluate Z Z
R
x
3
dA for R = [−4, 4] × [0, 5].
Solution. For each y, the x-integral goes from −4 to 4, so is symmetric about the x-axis, while the
function x
3
is odd, meaning that (−x)
3 = −x
3
. This means that the negative and positive portions
of the integral cancel, so each x-integral is 0, hence the entire area integral is 0.
Problem 2 (16.1.17). Use symmetry to evaluate Z Z
R
sin x dA for R = [0, 2π] × [0, 2π].
Solution. For each y, the x-integral is the integral of sin x over a whole period, so is 0. Thus the
entire area integral is 0.
Problem 3 (16.1.24). Evaluate Z 1
−1
Z π
0
x
2
sin y dx dy.
Solution.
Z y=1
y=−1
Z x=π
x=0
x
2
sin y dx dy =
Z y=1
y=−1
x
3
3
sin y
x=π
x=0
dy
=
Z y=1
y=−1
π
3
3
sin y dy
=
π
3
3
[− cos y]
y=1
y=−1
=
π
3
3
(cos(−1) − cos 1)
= 0.
Problem 4 (16.1.31). Evaluate Z 2
1
Z 4
0
dy dx
x + y
.
Solution.
Z x=2
x=1
Z y=4
y=0
dy dx
x + y
=
Z x=2
x=1
[ln(x + y)]y=4
y=0 dx
=
Z x=2
x=1
(ln(x + 4) − ln(x)) dx
= [(x + 4) ln(x + 4) − (x + 4)]x=2
x=1 − [x ln x − x]
x=2
x=1
= [(6 ln 6 − 6) − (5 ln 5 − 5)] − [(2 ln 2 − 2) − (1 ln 1 − 1)]
= 6 ln 6 − 5 ln 5 − 2 ln 2.
3
Homework Solutions MATH 32B-2 (18W)
Problem 5 (16.1.34). Evaluate Z 8
0
Z 2
1
x dx dy
p
x
2 + y
.
Solution.
Z y=8
y=0
Z x=2
x=1
x dx dy
p
x
2 + y
=
Z y=8
y=0
Z u=4+y
u=1+y
(1/2) du dy
√
u
(u = x
2 + y)
=
Z y=8
y=0
√
u
u=4+y
u=1+y
dy =
Z y=8
y=0
p
4 + y −
p
1 + y
dy
=
2
3
(4 + y)
3/2 −
2
3
(1 + y)
3/2
y=8
y=0
=
2
3
h123/2 − 9
3/2
−
4
3/2 − 1
3/2
i
=
2
3
h
24√
3 − 34i
=
−68 + 48√
3
3
.
Problem 6 (16.1.40). Evaluate Z Z
R
y
x + 1
dA for R = [0, 2] × [0, 4].
Solution.
Z Z
R
y
x + 1
dA =
Z x=2
x=0
Z y=4
y=0
y
x + 1
dy dx (Fubini)
=
Z x=2
x=0
1
2
y
2
x + 1y=4
y=0
dx =
Z x=2
x=0
8
x + 1
dx
= 8 [ln(x + 1)]x=2
x=0 = 8 ln 3.
Problem 7 (16.1.42). Evaluate Z Z
R
e
3x+4y
dA for R = [0, 1] × [1, 2].
Solution.
Z Z
R
e
3x+4y
dA =
Z y=2
y=1
Z x=1
x=0
e
3x+4y
dx dy (Fubini)
=
Z y=2
y=1
1
3
e
3x+4y
x=1
x=0
dy =
1
3
Z y=2
y=1
e
4y+3 − e
4y
dy
=
e
3 − 1
3
Z y=2
y=1
e
4y
dy =
e
3 − 1
3
1
4
e
4y
y=2
y=1
=
(e
3 − 1)(e
8 − e
4
)
12
.
4
MATH 32B-2 (18W) (L) G. Liu / (TA) A. Zhou Calculus of Several Variables
Problem 8 (16.1.47). Evaluate Z 1
0
Z 1
0
y
1 + xy
dy dx.
Solution.
Z x=1
x=0
Z y=1
y=0
y
1 + xy
dy dx =
Z y=1
y=0
Z x=1
x=0
y
1 + xy
dx dy (Fubini to change order)
=
Z y=1
y=0
Z u=1+y
u=1
1
u
du dy =
Z y=1
y=0
[ln u]
u=1+y
u=1 dy (u = 1 + xy)
=
Z y=1
y=0
ln(1 + y) dy = [(1 + y) ln(1 + y) − (1 + y)]y=1
y=0
= [2 ln 2 − 2] − [1 ln 1 − 1] = 2 ln 2 − 1.
Problem 9 (16.1.49). Using Fubini’s theorem, argue that the solid in Figure 1 has volume AL,
where A is the area of the front face of the solid.
Figure 1
Solution. The surface bounding the solid from above is the graph of a positive function z = f(y)
that does not depend on x. (Here a is the largest value that y can take, which is not labeled in the
diagram.) The volume of the solid is
Z Z
[0,L]×[0,a]
f(y) dA =
Z y=a
y=0
Z x=L
x=0
f(y) dx dy (Fubini)
=
Z y=a
y=0
[xf(y)]x=L
x=0 dy =
Z y=a
y=0
Lf(y) dy
= L
Z y=a
y=0
f(y) dy = LA,
where in the last step, we have used the fact that A is the area under the graph of z = f(y) in the
(y, z)-plane (which is equal to the area of the front face of the solid), which is then given by the
integral of f over the interval [0, a].
5
Homework Solutions MATH 32B-2 (18W)
Problem 10 (16.2.10). Sketch the region D between y = x
2 and y = x(1 − x). Express D as a
simple region and calculate the integral of f(x, y) = 2y over D.
Solution.
y = x
2
y = x(1 − x)
(1/2, 1/4)
The region D is both vertically simple and horizontally simple, but the bounds for y in terms of x
are simpler than the bounds for x in terms of y, so when we use Fubini’s theorem to evaluate the
integral, we take the y-integral on the inside and the x-integral on the outside. This gives us
Z Z
D
f(x, y) dA =
Z x=1/2
x=0
Z y=x(1−x)
y=x2
2y dy dx (Fubini)
=
Z x=1/2
x=0
[y
2
]
y=x(1−x)
y=x2 dx
=
Z x=1/2
x=0
x
2
(1 − x)
2 − x
4
dx
=
Z x=1/2
x=0
x
2
(1 − 2x) dx
=
1
3
x
3 −
1
2
x
4
x=1/2
x=0
=
1
24
−
1
32
=
1
96
.
Step-by-step explanation:
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