Mathematics 205
Solutions for HWK 21b
§5.4 p 326
Problem 1abc. §5.4 p326. For the given integral, sketch the region of integration, change the
order of integration, and evaluate the integral either way. See the page of sketches for HWK 21
for the sketches. Note that it’s not feasible to do these problems without the sketches!
(a)
Z
Z
1
0
1
xy dy dx
x
See the sketch. The original integral corresponds to using vertical arrows. The new integral, with
dx and dy reversed, uses horizontal arrows. For fixed y, a horizontal arrow shot to the right will
enter the region D where x = 0 and leave where x = y. These arrows must vary over the interval
[0, 1] on the y-axis. Therefore
Z
1
0
Z
1
xy dy dx =
x
Z Z
xy dA =
D
Z
1
0
Z
y
xy dx dy.
0
For evaluating the integrals, the calculations, in brief, are as follows:
I1 =
Z
1
0
I2 =
Z
Z
π
2
0
Z
xy dy dx =
x
Z
0
(b)
1
1
Z
Z
1
(
0
y
xy dx dy =
0
Z
x x3
1
− ) dx =
2
2
8
1
0
y3
1
dy =
2
8
cos θ
cos θ dr dθ
0
See the sketch, for which I’ve put θ on the horizontal axis and r on the vertical. The original integral,
with θ fixed at the initial stage, corresponds to shooting vertical arrows. The new integral, with
the order reversed, corresponds to shooting horizontal arrows. Such an arrow, shot to the right at
some level r, will enter the region D when θ = 0 and leave when θ satisfies r = cos θ. In other
words it will leave the region when θ = arccos r. The interval on the r-axis that must be covered
by the horizontal arrows is [0, 1]. Therefore
Z
0
π
2
Z
0
cos θ
cos θ dr dθ =
Z Z
cos θ dr dθ =
D
Page 1 of 3
Z
0
1
Z
arccos r
cos θ dθ dr
0
A. Sontag
April 17, 2000
Math 205 HWK 21b Solns continued
Evaluating the original integral, we find
Z
Z
π
2
0
cos θ
cos θ dr dθ =
0
Z
π
2
0
=
Z
·
¸r=cos θ
r cos θ
dθ
r=0
π
2
cos2 θ dθ
0
1
=
2
Z
π
2
(1 + cos 2θ) dθ
0
µ
¶¯θ= π
1
sin 2θ ¯¯ 2
=
θ+
¯
2
2
θ=0
π
=
4
(c)
Z
1
0
Z
2−y
(x + y)2 dx dy
1
See the sketch. The original integral, with y fixed at the initial stage, corresponds to shooting
horizontal arrows. The new integral, with the order reversed, corresponds to shooting vertical
arrows. With x fixed, an arrow shot straight up at position x will enter the region where y = 0
and leave the region where y = 2 − x. The interval on the x axis that’s needed for these arrows is
[1, 2]. Therefore
Z
1
0
Z
2−y
2
(x + y) dx dy =
1
Z Z
2
(x + y) dA =
D
Z
1
2
Z
2−x
(x + y)2 dy dx
0
Here’s one way to evaluate:
Z
1
2
Z
0
2−x
¯y=2−x
(x + y)3 ¯¯
(x + y) dy dx =
dx
¯
3
1
y=0
¶
Z 2µ
8 x3
=
−
dx
3
3
1
µ
¶¯2
8x x4 ¯¯
=
−
3
12 ¯
2
Z
2
1
17
=
12
Page 2 of 3
A. Sontag
April 17, 2000
Math 205 HWK 21b Solns continued
Problem 2c. §5.4 p326.Find
Z
0
4
Z
2
y
2
2
ex dx dy.
Solution. Refer to the sketch, since we’ll need it shortly. As it stands, this integral can’t be
2
evaluated in closed form, because ex has no elementary antiderivative. We’ll reverse the order
of integration to see if that helps. With the order reversed we’ll be fixing x at the initial stage,
consequently shooting vertical arrows. For fixed x, an arrow shot straight upward at that x will
enter the region D where y = 0 and leave where y = 2x. The interval on the x-axis that’s needed
at the second stage is [0, 2]. Therefore
Z 4Z 2
Z 2 Z 2x
2
x2
e dx dy =
ex dy dx
0
y
2
0
=
=
Z
Z
0
2
0
·
¸y=2x
x2
ye
dx
y=0
2
2
2xex dx
0
¯2
2¯
= ex ¯¯
0
4
=e −1
Problem 3. §5.4 p326. If f (x, y) = esin(x+y) and D = [−π, π] × [−π, π], show that
Z Z
1
1
≤
f (x, y) dA ≤ e.
e
4π 2
D
Solution. The region of integration, D, is a square that is 2π units long on each side and
therefore has area 4π 2 . Now examine the integrand, f (x, y) = esin(x+y) . No matter what x and y
are, sin(x + y) will satisfy −1 ≤ sin(x + y) ≤ 1. Since the exponential function is increasing, we
therefore have
e−1 ≤ esin(x+y) ≤ e1 .
So if f (x, y) is integrated over D, the resulting value must be at least as large as e−1 times the
area of D and no larger than e times the area of D. In other words, we have
Z Z
1 2
4π ≤
f (x, y) dA ≤ 4π 2 e.
e
D
Now divide by 4π2 to obtain
as required.
1
1
≤
e
4π 2
Z Z
D
f (x, y) dA ≤ e
Page 3 of 3
A. Sontag
April 17, 2000