Math 2263
Practice Midterm 2A
This is the cover page that will appear on the actual exam.
Name:
You may use a scientific calculator, but you may not use books, notes,
graphing calculators, or your neighbors’ papers. Sign your name below to
certify that you followed these instructions.
Signature:
Show all work for each problem so that partial credit may be given. Answers without supporting work will receive no credit. Clearly indicate
your answers. If you need more space, use the blank sheet at the back of
your test booklet.
Make sure that you have 6 problems on your test. If any exam material
is missing, or if any questions are unclear, raise your hand or bring your
exam to the proctor.
Good luck!
1
1. Set up a triple integral to find the volume of the solid bounded by
the cylinder z = 1 − x2 and the planes z = 0, y = 0, x = 0, and y = 1,
using the specified order of integration.
(a) dy dz dx
Solution:
1
Z
1−x2
Z
1
Z
1 dy dz dx
0
0
0
(b) dz dy dx
Solution:
1
Z
1
Z
1−x2
Z
1 dz dy dx
0
0
0
(c) dy dx dz
Solution:
Z
1
√
Z
1−z
Z
1
1 dy dx dz
0
0
0
2
2. Use spherical coordinates to evaluate the integral
ZZZ
2
2
2 3/2
e(x +y +z ) dV,
E
p
where Ep
is the solid bounded by the hemispheres
z
=
4 − x2 − y 2
p
and z = 1 − x2 − y 2 , and inside the cone z = x2 + y 2 .
Solution:
Z π/4 Z 2π Z
2
Z
ρ3 2
π/4
Z
e ρ sin φ dρ dθ dφ =
0
0
1
0
e8 − e
=
3
Z
π/4
Z
2π
0
2π
sin φ dθ dφ
0
0
Z π/4
2π 8
=
sin φ dφ
(e − e)
3
0
2π 8
π/4
=
(e − e)(− cos φ)|0
3
2π 8
1
=
(e − e)(1 − √ )
3
2
3
3
(1/3)eρ sin φ|ρ=2
ρ=1 dθ dφ
3. Evaluate the triple integral
ZZZ
z dV,
E
where E is bounded by the cylinder y 2 + z 2 = 9 and the planes x = 0,
y = 3x, and z = 0 in the first octant.
Solution:
ZZZ
3
Z
Z
y/3
Z √9−y2
z dV =
E
Z
3
y/3
Z
z dz dx dy =
0
0
0
Z 3
1
3
2
=
y(9 − y ) dy =
y−
2
0 6
0
3
3 2
1 4 27 81
=
y − y =
−
4
24
4
24
0
1 3
y
6
Z
0
0
3
4
=
1
(9 − y 2 ) dx dy
2
27
8
dy
4. Use a double integral to find the area inside the curve r = 1+sin θ and
outside the curve r = 1. (It maybe be helpful to know that sin2 θ =
1
− 12 cos(2θ).)
2
Solution:
Z
π
Z
1+sin θ
Z
r dr dθ =
0
π
2
2
(1/2) (1 + sin θ) − 1 dθ
0
1
Z
=
π
sin θ + (1/2) sin2 θ dθ
0
=2+
5
π
4
5. Consider the integral
Z
0
2
√
Z
4−x2
√
− 4−x2
Z √4−x2 −y2 p
2
2
x2 + y 2 ex +y dz dy dx
0
(a) Rewrite this integral in cylindrical coordinates, r, θ, and z.
Solution:
Z
π/2
Z
−π/2
2
√
Z
0
4−r2
2
r2 er dz dr dθ
0
(b) Rewrite this integral in spherical coordinates, ρ, θ, and φ.
Solution:
Z
π/2
−π/2
Z
0
π/2
Z
2
ρ3 sin2 φ eρ
0
6
2
sin2 φ
dρ dφ dθ
6. Evaluate the integral.
√
4 π
Z
π
Z
0
sin(x2 ) dx dy
y/4
Solution: We’ll switch the order of integration, since finding an
antiderivative with respect to x looks problematic.
Z
√
4 π
√
Z
√
π
Z
2
π
Z
4x
sin(x ) dx dy =
0
y/4
0
√
Z
sin(x2 ) dy dx
0
π
y sin(x2 )|y=4x
y=0 dx
=
0
√
Z
π
4x sin(x2 ) dx
=
0
Do a substitution with u = x2 , du = 2x dx.
Z π
2 sin u du
=
0
= −2 cos u|π0 = −2(−1 − 1) = 4
7