WRITTEN HOMEWORK #7 SOLUTIONS (1) (Chapter 15.2, Problem ...

WRITTEN HOMEWORK #7 SOLUTIONS
(1) (Chapter 15.2, Problem #36) Find the average value of f(x, y) = e y√ x + e y
over the rectangle R = [0, 4] × [0, 1].
Solution. The average value of a function over a region R is obtained by
integrating that function over R and then dividing by the area of R. In this
case, R obviously has area 4. The iterated integral which equals the integral
of f(x, y) over R is
� 4�
1
e y√ x + ey dy dx.
0
0
Integrating the inner integral using a u-substitution u = x + e y (we suppress
the details here) gives
� 4
2
3 · (x + ey ) 3/2
�
�
0
This integral equals
� y=1
y=0
� 4
2
dx =
3 (x + e)3/2 − (x + 1) 3/2 dx.
0
2 2
·
3 5 (x + e)5/2 − (x + 1) 5/2
�
�
� 4
=
0
4
15 · ((e + 4)5/2 − 5 5/2 − e 5/2 + 1).
Recall that we need to divide this by 4 if we want to find the average value.
Therefore, the answer is
1
15 · ((e + 4)5/2 − 5 5/2 − e 5/2 + 1). �
(2) (Chapter 15.2, Problem #40b) If f(x, y) is continuous on [a, b] × [c, d], and
g(x, y) =
� x � y
a
c
f(s, t) dt ds,
for a < x < b, c < y < d, show that gxy = gyx = f(x, y).
Solution. In this problem we use the fundamental theorem of calculus, which
states that if f(x) is a continuous function, then
d
dx
where a is constant. Let
� x
a
h(s, y) =
f(t) dt = f(x),
� y
c
1
f(s, t) dt

2 WRITTEN HOMEWORK #7 SOLUTIONS
be the inner integral in the definition of g(x, y); notice that h is a function of
s and y instead of s, t. Then
∂ ∂
g(x, y) =
∂x ∂x
� x
a
h(s, y) ds = h(x, y).
We now differentiate h(x, y) with respect to y:
gxy(x, y) = ∂
� y
∂
h(x, y) = f(x, t) dt = f(x, y).
∂y ∂y c
We can repeat the same procedure on gyx to find that gyx = f(x, y) as well; in
particular, we can repeat the entire argument performed above with the roles
of x, y swapped, after interchanging the order of integration in the definition
of g:
g(x, y) =
� x � y
a
c
f(s, t) dt ds =
� y � x
c
a
f(s, t) ds dt. �
(3) (Chapter 15.3, Problem #52) Evaluate the following integral by reversing the
order of integration:
� 1
0
� 1 � 1
e x/y dy dx.
0
x
Solution. The inequalities describing the region of integration are 0 ≤ x ≤
1, x ≤ y ≤ 1. If we interchange the order of integration, we can describe the
region of integration using the different set of inequalities 0 ≤ y ≤ 1, 0 ≤ x ≤
y. Therefore, the original iterated integral equals
� y
0
e x/y � 1
dx dy = ye x/y
�
�
0
� x=y
x=0
� 1
dy =
0
y · e − y dy = y2
2
�
�
(e − 1) � 1
= e − 1
0
. �
2
(4) (Chapter 15.3, Problem #62) In evaluating a double integral over a region D,
a sum of iterated integrals was obtained as follows:
��
D
� 1 � 2y
� 3 � 3−y
f(x, y) dA = f(x, y) dx dy + f(x, y) dx dy.
0
0
Sketch the region D and express the double integral as an iterated integral
with reversed order of integration.
Solution. The region D is described by the inequalities 0 ≤ y ≤ 1, 0 ≤ x ≤
2y and 1 ≤ y ≤ 3, 0 ≤ x ≤ 3 − y. This region is a triangle with vertices at
(0, 0), (1, 2), and (3, 3):
1
0

WRITTEN HOMEWORK #7 SOLUTIONS 3
This region is evidently described by inequalities 0 ≤ x ≤ 2, x/2 ≤ y ≤
3 − x, so the original integral is equal to
� 2�
3−x
f(x, y) dy dx. �
0
x/2
(5) (Chapter 15.4, Problem #26) Use polar coordinates to find the volume of the
solid bounded by the paraboloids z = 3x 2 + 3y 2 and z = 4 − x 2 − y 2 .
Solution. The first task is to find the intersection of these two surfaces.
Equating the two functions of x, y, we obtain 4x 2 + 4y 2 = 4, or x 2 + y 2 = 1.
So the intersection of these two surfaces is x 2 + y 2 = 1, z = 1. A quick sketch
of these surfaces (which we omit here) shows that the region bounded by the
surfaces is given by inequalities 0 ≤ x 2 + y 2 ≤ 1, 3x 2 + 3y 2 ≤ z ≤ 4 − x 2 − y 2 .
Therefore, the volume of this region is given by the double integral
��
D
4 − x 2 − y 2 − (3x 2 + 3y 2 ) dA,
where D is the closed disc x 2 + y 2 ≤ 1. This disc is described by polar
inequalities 0 ≤ r ≤ 1, 0 ≤ θ ≤ 2π. Therefore the integral we want to
calculate is equal to
� 2π�
1
0
0
(4 − r 2 − 3r 2 ) · r dr dθ =
� 2π �
0
2r 2 − r 4
�
�
� r=1
r=0
�
dθ = 2π. �
(6) (Chapter 15.4, Problem #38) Let D be the disk with center the origin and
radius a. What is the average distance from points in D to the origin?
Solution. Let a particular point have polar coordinates (r, θ). Then the
distance of this point from the origin is r. Therefore, we wish to evaluate the
double integral

4 WRITTEN HOMEWORK #7 SOLUTIONS
��
D
r dA.
(Recall that at the end, we will need to divide our result by the area of
D to obtain the average distance.) In polar coordinates, D is described by
0 ≤ r ≤ a, 0 ≤ θ ≤ 2π. Therefore,
��
D
� 2π
r dA =
0
� a
0
r 2 dr dθ = 2π a3
3 .
The area of D is πa 2 , so the average distance of points from the origin is
2π a3
3
1 2a
· = . �
πa2 3