Math 241 Exam 3 Sample 3 Solutions
1. (a) The easiest method would be r̄(r, θ) = r cos θ ı̂ + r sin θ ̂ + (9 − r2 ) k̂ with 0 ≤ r ≤ 3 and
0 ≤ θ ≤ 2π.
(b) Since z = 4 − y 2 this shape is a parabolic sheet as shown:
−2
2
3
(c) We need to change this to polar first since the integrand is not integrable with respect to
y or x. The region R is the quarter disk of radius 1 in the first quadrant and so we have:
Z
1
0
Z
√
1−x2
sin(x2 + y 2 ) dy dx =
0
Z
=
Z
=
Z
π/2
0
π/2
0
π/2
0
π/2
Z
1
sin(r2 )r dr dθ
0
1
1
− cos(r2 ) dθ
2
0
1
1
− (cos(1) − cos(0)) dθ
2
2
1
(1 − cos(1)) dθ
2
0
π/2
1
= θ(1 − cos(1))
2
0
1
1
= (0)(1 − cos(1)) − (π/2)(1 − cos(1))
2
2
=
Z
2. (a) The picture is:
y=1/2+4
y=x
x=8
The lines meet when y = x meets y =
R 8 R 21 x+4
y dy dx
0 x
1
2x
+ 4 at x = 8. Thus the integral would be
(b) The picture is:
r=1
r=2cos(theta)
The circles meet when r = 1 meets r = 2 cos θ which is when cos θ =
R 2 cos θ
R
(r cos θ)(r sin θ)r dr dθ.
the integral is −π/3 π/3 1
1
2
or θ = ± π3 . Thus
3. (a) The picture is:
pi
x=sin(y)
(b) The picture is:
theta=pi/4
r=1
r=2sec(theta) aka x=2
theta=−pi/4
(c) We have z = 4 −
p
√
x2 + y 2 = 4 − r2 = 4 − r.
4. (a) The picture is:
D
R
1
The iterated integral is
Z
1
−1
Z
√
1−x2
√
− 1−x2
Z
9−x2
z dz dy dx
0
(b) The picture is:
The iterated integral is
Z
π/2
0
Z
π/3
π/6
Z
3
ρ2 sin φ dρ dφ dθ
0
5. We rewrite the ellipse as (2x)2 + (3y)2 = 36 and subsitute u = 2x and v = 3y. The new region
S is then inside the circle u2 + v 2 = 26.
The pictures are:
2
R
3
S
6
Then x = u/2 and y = v/3 so that
0
= 1/6
1/3
1/2
J=
0
. So we have
ZZ
x dA =
R
1
1
u|1/6| dA =
2
12
ZZ
ZZ
S
S
which we then parametrize in polar
1
=
12
Z
2π
0
Z
6
r cos θ r dr dθ
0
u dA