Math 53 Worksheet Solutions - Multivariable Integration
Note: Due to time constraints, these solutions are unusually abbreviated. Please see me in
office hours if you would like more details.
1. Find the volume of the solid in the first octant bounded by the cylinder z = 16 − x2
and the plane y = 5.
Solution. Region is D = [0, 4] × [0, 5]. V =
640
.
3
√
2. Find the average value of f (x, y) = ey x + ey over the rectangle R = [0, 4] × [0, 1].
Solution. Area of region in question is A = 4. Then
Z Z
1 1 4 1 y√
e x + ey dy dx =
(4 + e)5/2 − e5/2 − 55/2 + 1 ≈ 3.327.
Average Value =
4 0 0
15
3. Find the volume of the solid bounded by the planes z = x, y = x, x + y = 2, and z = 0.
Solution. Discussed details at length in class. V = 13 .
4. Compute
ZZ p
1 − x2 − y 2 dA
D
2
2
where D is the disk x + y ≤ 1, by first identifying the integral as the volume of a
solid.
Solution. It’s the top half of a sphere of radius one. The volume of an entire sphere
of radius one is 43 π so,
2
V = π.
3
5. Evaluate the integrals below by reversing the order of integration.
R √π R √π
(a) 0
cos(x2 ) dx dy
y
R1R1
(b) 0 x ex/y dy dx
Solution. Draw pictures!
(a)
√
Z
π
√
Z
√
π
Z
2
π
Z
cos(x ) dx dy =
0
y
0
1
0
x
cos(x2 ) dy dx = 0.
(b)
Z
1
Z
1
e
0
x/y
Z
1
y
Z
dy dx =
x
0
0
1
ex/y dx dy = (e − 1).
2
6. Sketch the solid whose volume is given by the iterated integral
Z 1 Z 1−x2
(1 − x) dy dx.
0
0
Solution. Projection in xy plane is portion of parabola y = 1 − x2 in the first
quadrant. Height function is z = 1 − x.
7. Evaluate
ZZ
(x2 tan x + y 3 + 4) dA,
D
2
2
where D = {(x, y)|x + y ≤ 2}. Hint: Exploit the fact that D is symmetric with
respect to both axes.
Solution. The first two terms in the integrand are odd functions of x and y,
respectively. As the region is symmetric in x and y (meaning when (x, y) is in D, so is
(−x, −y)), we have
ZZ
ZZ
√
2
3
(x tan x + y + 4) dA =
4 dA = 4(π( 2)2 ) = 8π,
D
D
where the area of the region is calculated using the formula for the area of a circle.
8. Find the volume of the solid bounded by the cylinders x2 + y 2 = r2 and y 2 + z 2 = r2 .
Solution. Symmetry.
V
=
8
Z r Z √r2 −y2 p
0
0
so
V =
2
2
r2 − y 2 dx dy = r3 ,
3
16 3
r .
3