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Math 210Q
Practice Exam 3
SOLUTIONS
Name:
Directions: Show all of your work and clearly indicate your answers. Correct answers with
no work may not receive credit.
1. Calculate the iterated integrals
Z 2Z 2
a.
(y + 2xey ) dx dy
1
0
=
b.
Z
1
0
Z
c.
π
0
2
1
(2y + 4ey ) dy = 3 + 4e2 − 4e
x
cos(x2 ) dy dx
0
=
Z
Z
Z
1
0
Z √1−y2
Z
1
cos(x2 ) x dx =
0
1 1
1
sin(x2 ) = sin(1)
0
2
2
y sin(x) dz dy dx
0
=
Z
π
0
Z
1
0
Z
p
sin(x) y 1 − y 2 dy dx =
=
2. Evaluate
π
0
sin(x) dx ·
Z
1
y
0
p
1 − y 2 dy
π 1 2 3/2 1 2
− cos(x) · −
1 − y2
=
2 3
3
0
0
Z Z
xy dA, where D = {(x, y) : 0 ≤ y ≤ 1, y 2 ≤ x ≤ y + 2}.
D
Z 1 Z y+2
Z 1 41
1 3
2
5
=
xy dx dy =
y + 4y + 4y − y dy =
24
0
y2
0 2
√
y
dA, where D is bounded by y = x, y = 0, x = 1.
2
D 1+x
Z 1 Z √x
Z
1 1
y
x
1 1
1
2 =
dy dx =
dx = ln(1 + x ) = ln(2)
1 + x2
2 0 1 + x2
4
4
0
0
0
3. Evaluate
Z Z
1
Z
1
Z
1
4. Calculate the iterated integral:
cos(y 2 ) dy dx. [Change order of integration.]
0
x
Z 1Z y
Z 1
1 1
1
2
=
cos(y ) dx dy =
cos(y 2 ) · y dy = sin(y 2 ) = sin(1)
2
2
0
0
0
0
1
2
1
yex
dx dy. [Change order of integration.]
5. Calculate the iterated integral:
√
x3
0
y
Z 1 Z x2
Z 1 x2
Z 1 x2
2
yex
y 2 x2
x4
e
e
=
dy
dx
=
·
·
dx
dx
=
x3
x3 2 0
x3 2
0
0
0
0
Z
1 2 1 1
1 1 x2
e x dx = ex = (e − 1)
=
2 0
4
4
0
Z
Z
6. Find the volume of the solid under the paraboloid z = x2 + 4y 2 and above the rectangle
R = [0, 2] × [1, 4].
Z 4
Z 4Z 2
8
528
2
2
2
(x + 4y ) dx dy =
+ 8y
= 176
dy =
=
3
3
1
0
1
7. Calculate the volume of the solid under the surface z = x2 y that lies above the triangle
in the xy-plane with vertices (1, 0), (2, 1) and (4, 0). [Warning: tedious algebra ahead.]
Z
Z 1 Z −2y+4
53
1 1
2
4
3
2
=
x y dx dy =
− 9y + 45y − 99y + 63y dy =
3 0
20
0
y+1
Alternatively,
Z 4 Z − x +2
Z 2 Z x−1
2
31 32
53
2
x y dy dx +
=
x2 y dy dx =
+
=
60 15
20
2
0
1
0
8. Evaluate
Z Z
p
x2 + y 2 dA, where D is the region in the first quadrant bounded by the
D
circle x2 + y 2 = 2 and the x− and the y−axes.
=
9. Evaluate
x2 + y 2 = 9.
Z Z
Z
π
2
0
Z
√
2
0
√
√
√
π 2
r 3 2 π2
2 2 π
· =
r · r dr dθ = · θ =
3 0
3
2
3
0
x dA, where D is the region above the x-axis bounded by the circle
D
=
Z
π
0
Z
3
0
r cos(θ) · r dr dθ =
2
π
r 3 3
· sin(θ) = 9 · 0 = 0
3 0
0
10. Calculate the volume of the solid bounded by the cylinder x2 + y 2 = 4 and the planes
z = 0 and y + z = 3.
=
Z
2π
0
Z
2
0
3−r sin(θ) · r dr dθ =
=
Z Z Z
11. Evaluate
=
Z
3
0
Z
x
Z
0
12. Evaluate
2π
Z
0
8
12π +
3
2
0
2
3r−r sin(θ) dr dθ =
8
− 0+
3
Z
2π
0
8
6 − sin(θ)
3
dθ
= 12π
xy dV , where E = {(x, y, z) : 0 ≤ x ≤ 3, 0 ≤ y ≤ x, 0 ≤ z ≤ x + y}).
E
x+y
xy dz dy dx =
0
Z Z Z
Z
Z
3
0
Z
x
2
2
(x y + xy ) dy dx =
0
Z
3
1 3 81
5 4
x dx = x5 =
0
6
6
2
0
z dV , where E is the solid region bounded by the planes z = 0,
E
z = 1 + x + y, and the cylinder x2 + y 2 = 1 in the first octant. [ Warning : highly involved
and technical algebra here.]
=
Z Z Z
D
1+x+y
z · dz dA =
0
=
=
1
2
Z
π
2
0
Z
1
0
π
2
Z
π
2
0
Z
1
0
Z
π
2
0
Z
1
0
Z
1+r(cos(θ)+sin(θ))
z · r dz dr dθ
0
2
1
1 + r(cos(θ) + sin(θ)) · r dr dθ
2
r + 2r 2 (cos(θ) + sin(θ)) + r 3 (cos(θ) + sin(θ))2 dr dθ
1 2
1
+ (cos(θ) + sin(θ)) + (1 + cos(θ) sin(θ)) dθ
2 3
4
0
π2
2
1
3π 19
1 1
1 2 θ + (sin(θ) − cos(θ)) +
+
=
θ + sin (θ) =
2 2
3
4
2
16 24
1
=
2
Z
0
13. Find the volume of the solid inside the cylinder x2 + y 2 = 4 under the paraboloid
z = x2 + y 2 and above the plane z = 0.
=
Z Z Z
D
x2 +y 2
1 dz dA =
0
Z
2π
0
Z
2
0
Z
r2
0
4
1 · r dz dr dθ =
2π
r 2
=
· θ = 8π
4 0
0
3
Z
2π
0
Z
2
r 3 dr dθ
0
14. Evaluate
Z Z Z
(x2 + y 2 ) dV , where E is the solid region above the xy-plane between
E
the spheres x2 + y 2 + z 2 = 1 and x2 + y 2 + z 2 = 4.
=
Z
π
2
0
Z
2π
Z
0
2
1
ρ2 sin2 φ cos2 θ + ρ2 sin2 φ sin2 θ · ρ2 sin φ dρ dθ dφ
Z
=
π
2
Z
0
2π
0
Z
2
ρ4 sin3 φ dρ dθ dφ
1
2π Z π/2
ρ5 2
2
124π
31
=
· 2π · =
sin3 φ dφ =
·θ ·
5 1
5
3
15
0
0
To evaluate the last integral, do the following:
Z
π/2
3
sin φ dφ =
0
Z
π/2
2
(1 − cos φ) sin φ dφ =
0
=
15. Evaluate
Z Z Z
y2
E
Z
π/2
0
sin φ − cos2 φ sin φ dφ
1
π/2 2
3
− cos φ + cos φ =
0
3
3
p
x2 + y 2 + z 2 dV , where E is the solid region in the first octant
inside the sphere x2 + y 2 + z 2 = 1.
=
Z
π
2
Z
0
π
2
0
Z
=
=
1 6
=
ρ
6
Z
Z
1
0
π
2
0
Z
ρ2 sin2 φ sin2 θ · ρ · ρ2 sin φ dρ dθ dφ
0
π
2
Z
1
5
0
ρ dρ ·
1
ρ5 sin3 φ sin2 θ dρ dθ dφ
0
Z
π
2
1
6
To evaluate the third integral, see above.
π/2
2
sin θ dθ =
0
sin θ dθ ·
0
Z
π
2
sin3 φ dφ
0
1 π/2 π/2
1
1
1
3
· θ − sin(2θ)
· − cos φ + cos φ
2
4
3
0
0
0
2
π
=
3
36
do the second, use the half-angle identity:
1
1
1
π/2 π
1 − cos(2θ) θ dθ =
θ − sin(2θ) =
2
2
2
0
4
=
Z
2
Z
π/2
0
π
4
To
·
·
4
16. Find the area of the cardioid r = 4 + 3 cos(θ).
4
3
2
1
0
2
4
6
0
−1
−2
−3
−4
Area =
ZZ
1
=
2
1 dA =
D
Z
2π
0
Z
2π
0
Z
4+3 cos θ
r drdθ =
0
Z
2π
0
1 + cos 2θ
16 + 24 cos θ + 9 ·
2
1
(4 + 3 cos θ)2 dθ
2
dθ =
41π
.
2
17. Find the area of one petal of the polar rose r = cos(3θ).
For this problem, it is important to get your bounds right. It is easy to see that 0 ≤ r ≤
cos(3θ). To find the bounds for θ, we look at the graph of the curve.
0.75
0.5
0.25
0.0
−0.5
−0.25
0.0
0.25
0.5
0.75
1.0
−0.25
−0.5
−0.75
We want only one petal, so we need to find an interval of θ values where r starts at zero
and then goes back to zero. That is, we need to find the angles θ where r = 0 and our range
will be the interval between two consecutive such θ values.
Consider cos(t). We know that cos(t) = 0 when t = − π2 , π2 , 3π
, 5π
, · · · . In our case, t = 3θ,
2
2
π π 3π 5π
so we have θ = − 6 , 6 , 6 , 6 , · · · . The petal on the right is the one where θ goes from − π6
to π6 , so we will use this interval for θ. (You can use any consecutive θ values, however.)
Therefore,
Area =
ZZ
1 dA =
D
Z
π
6
− π6
Z
cos(3θ)
r drdθ =
0
Z
π
6
− π6
5
1
(cos 3θ)2 dθ =
2
Z
π
6
− π6
π
1
(1 + cos 6θ) dθ = .
4
12
p
18. Find the volume of the solid region bounded below by the cone z = x2 + y 2 and above
by the sphere x2 + y 2 + z 2 = 1. (This region is sometimes called an ice cream cone.)
Volume =
ZZZ
1 dV =
E
Z
π/4
0
Z
2π
0
Z
1
ρ2 sin φ dρ dθ dφ
0
π/4 2π
ρ3 1 2π =
· θ · − cos φ =
3 0
3
0
0
6
√
2
+1
−
2
!