Download Solutions for Exam 4

S. F. Ellermeyer
MATH 2203 — Exam 4 Solutions
April 5, 2004
Name
Instructions. This exam contains seven problems, but only six of them will be graded. You
may choose any six to do. Please write DON’T GRADE on the one that you don’t want me
to grade. In writing your solution to each problem, include sufficient detail and use correct
notation. (For instance, don’t forget to write “=” when you mean to say that two things are
equal.) Your method of solving the problem must be clear to the reader (me). If I
have to struggle to understand what you have written, then you might not get full credit or
even any credit for your solution even if you get a correct answer.
1. Find three positive numbers, x, y, and z whose sum is 300 and whose product is as
large as possible. (You must use calculus methods as studied in this course to do this.
Be detailed. The correct answer is not worth any credit unless supported by valid
reasoning.)
Solution: We want to find the absolute maximum value of the function
f (x, y) = xy (300 − x − y) = 300xy − x2 y − xy 2
over the triangular domain
D = {(x, y) | 0 ≤ x ≤ 300, 0 ≤ y ≤ 300 − x} .
It is clear that the maximum of f does not occur on the boundary of D, because at all
points (x, y) on the boundary of D we have f (x, y) = 0.
Since
fx = 300y − 2xy − y 2
fy = 300x − x2 − 2xy
we see that the critical points of f must satisfy
y (300 − 2x − y) = 0
x (300 − x − 2y) = 0.
The only critical points that lie in the interior of D must satisfy
2x + y = 300
x + 2y = 300
and hence the only critical point of f in the interior of D is (x, y) = (100, 100). Let us
check that this critical point does indeed correspond to the absolute maximum of f on
1
D:
fxx = −2y
fyy = −2x
fxy = 300 − 2x − 2y
D = fxx (100, 100) fyy (100, 100) − (fxy (100, 100))2
= (−200) (−200) − (−100)2
= 30, 000 > 0
fxx (100, 100) = −200 < 0
so the Second Derivative Test tells us that we have a maximum.
In conclusion, the three positive numbers whose sum is 300 and whose product is as
large as possible are 100, 100, and 100.
2. Use a geometric argument (not an iterated integral) to evaluate the double integral
Z Z
(4 − 2y) dA
R
where R is the rectangle
R = {(x, y) | 0 ≤ x ≤ 1, 0 ≤ y ≤ 1} .
In your solution, include
• a drawing of the rectangle R
• a drawing of the solid whose volume is given by the integral (drawn to the best
of your three—dimensional drawing ability)
• a written explanation of your result.
Solution: For R = {(x, y) | 0 ≤ x ≤ 1, 0 ≤ y ≤ 1}, the integral
Z Z
(4 − 2y) dA
R
is the volume of a prism with one of its triangular bases in the yz plane and with
height 2 plus the volume of a rectangular block with dimensions 1 × 1 × 2. Thus
µ
¶
Z Z
1
(4 − 2y) dA = 1
· 1 · 2 + 1 · 1 · 2 = 3.
2
R
2
3. Use an iterated integral to compute the double integral
Z Z
xyey dA
R
where R is the rectangle
R = {(x, y) | 0 ≤ x ≤ 2, 0 ≤ y ≤ 1} .
Solution:
Z Z
y
xye dA =
Z 2Z
0
R
To evaluate the inner integral,
Z
1
xyey dy dx = 2.
0
1
xyey dy,
0
we use integration by parts:
u = xy dv = ey dy
du = x dy
v = ey
so
Z
y
xye dy =
Z
u dv
Z
= uv − v du
Z
y
= xye − xey dy
= xyey − xey + C
= xey (y − 1) + C
3
and
Z
1
xyey dy = xey (y − 1)|y=1
y=0 = x.
0
Now we compute the outer integral
¯x=2
Z 2
1 2 ¯¯
x dx = x ¯
= 2.
2 x=0
0
4. Evaluate the iterated integral
Z 1Z
0
ey
√
x dx dy.
y
Solution: First we do the inner integral:
¯x=ey
Z ey
´
√
3
2 3/2 ¯¯
2 ³ 3y
e2 − y 2 .
x dx = x ¯
=
3
3
y
x=y
Now we do the outer integral:
¶¯y=1
µ
Z 1 ³
´
3
2 3y
2 2 3 y 2 5 ¯¯
e 2 − y 2 dy =
e2 − y 2 ¯
3 3
5
0 3
y=0
µ
¶
µ ¶
2 2
2 2 3 2
=
e2 −
−
3 3
5
3 3
4 3 32
= e2 − .
9
45
In conclusion,
Z 1Z
0
y
ey
√
4 3 32
x dx dy = e 2 −
9
45
5. Use polar coordinates to find the volume of the solid that lies inside the sphere
x2 + y 2 + z 2 = 16
and outside the cylinder
x2 + y 2 = 4.
Solution: This volume is given by
Z Z p
16 − x2 − y 2 dA
2
D
where D is the domain
©
ª
D = (x, y) | 4 ≤ x2 + y 2 ≤ 16 .
Using polar coordinates, we see that the volume is
Z 2π Z 4 √
√
2
16 − r2 r dr dθ = 32 3π.
0
2
4
Here is the computation of the integral: Letting u = 16 − r2 , du = −2r dr, we have
Z
Z 4√
1 4 √
2
16 − r r dr = −
−2 16 − r2 r dr
2 2
2
Z
1 0√
u du
=−
2 12
Z
1 12 √
u du
=
2 0
¯u=12
1 3/2 ¯¯
= u ¯
3
√ u=0
=8 3
and
2
Z
2π
√
√
8 3 dθ = 32 3π.
0
6. Find the center of mass of the lamina that occupies the part of the disk x2 + y 2 = 1 in
the first quadrant given that the density at any point of the lamina is proportional to
the square of the distance from that point to the origin.
Solution: The lamina is the region
©
ª
D = (x, y) | 0 ≤ x2 + y 2 ≤ 1, x ≥ 0, y ≥ 0
n
πo
= (r, θ) | 0 ≤ r ≤ 1, 0 ≤ θ ≤
.
2
The density function is
where k is a positive constant.
¡
¢
ρ (x, y) = k x2 + y 2
The mass of the lamina is
Z Z
¡
¢
k x2 + y 2 dA
D
Z π/2 Z 1
kr2 r dr dθ
=
0
0
Z π/2 Z 1
=
kr3 dr dθ
0
0
Z π/2
k
1 dθ
=
4 0
kπ
.
=
8
m=
5
The moment with respect to the x axis is
Z Z
¡
¢
k x2 + y 2 y dA
Mx =
D
Z π/2 Z 1
kr2 r sin (θ) r dr dθ
=
0
0
Z π/2 Z 1
=
kr4 sin (θ) dr dθ
0
0
Z π/2
1
sin (θ) dθ
= k
5 0
k
= .
5
Likewise, the moment with respect to the y axis is My = k/5. Therefore, the center of
mass of the lamina is
¶ µ
¶
µ
8 8
My Mx
(x, y) =
,
=
,
.
m m
5π 5π
7. Fubini’s Theorem states that if the function f is continuous on the rectangle
R = {(x, y) | a ≤ x ≤ b, c ≤ y ≤ d} ,
then
Z bZ
a
d
f (x, y) dy dx =
c
Z
c
d
Z
b
f (x, y) dx dy.
a
Illustrate the truth of Fubini’s Theorem for the function
f (x, y) = x2 y − 2
and the rectangle
R = {(x, y) | − 1 ≤ x ≤ 1, 1 ≤ y ≤ 2} .
Solution:
Z
1
−1
Z
1
2
¯y=2 !
¯
1 2 2
dx
x y − 2y ¯¯
2
−1
y=1
µ
¶¶
Z 1µ
¡ 2
¢
1 2
=
2x − 4 −
x −2
dx
2
−1
¶
Z 1µ
3 2
x − 2 dx
=
2
−1
¯x=1
¯
1 3
= x − 2x¯¯
2
x=−1
3 3
=− −
2 2
= −3
¢
¡ 2
x y − 2 dy dx =
Z
6
1
Ã
and
Z 2Z
1
1
−1
¯x=1 !
¯
1 3
dy
x y − 2x¯¯
3
1
x=−1
¶ µ
¶¶
Z 2 µµ
1
1
y−2 − − y+2
dy
=
3
3
1
¶
Z 2µ
2
y − 4 dy
=
3
1
¯y=2
¯
1 2
= y − 4y ¯¯
3
µ y=1 ¶
11
20
=− − −
3
3
= −3
¢
¡ 2
x y − 2 dx dy =
Z 2Ã
7