# Math 53 Worksheet Solutions - Integration and Change of ```Math 53 Worksheet Solutions - Integration and Change of Coordinates
RRR
1. Evaluate
xy dV , where E is bounded by the parabolic cylinders y = x2 and x = y 2
and the planes z = 0 and z = x + y.
Solution.
ZZZ
1
Z
√
Z
x
Z
x+y
xy dz dy dx =
xy dV =
x2
0
0
3
.
28
2. Use a triple integral to find the volume of the solid bounded by the cylinder y = x2
and the planes z = 0, z = 4, and y = 9.
Solution.
Z
3
Z
9
4
Z
dz dy dz = 144.
V =
x2
−3
0
RRR z
3. Use cylindrical coordinates to evaluate
e dV , where E is enclosed by the
E
paraboloid z = 1 + x2 + y 2 , the cylinder x2 + y 2 = 5, and the xy-plane.
√
Solution. Bounded by z = 1 + r2 and r = 5, cylindrically symmetric.
ZZZ
Z 2π Z √5 Z 1+r2
ez r dz dr dθ = π(e6 − e − 5).
ez dV =
E
0
0
0
4. Evaluate the integral
ZZ
ex+y dA,
R
where R is given by the inequality |x| + |y| ≤ 1.
Solution. Set u = x + y and v = x − y, inspired by drawing the region and writing
the equations for each of the lines forming the four parts of the boundary. Then we
have x = (u + v)/2, y = (u − v)/2, and
1
∂(x, y) 12 21 = 1
1 = − ,
−2
∂(u, v)
2
2
so
ZZ
e
R
x+y
Z
1
dA =
−1
1
1
e du dv = e − e−1 .
2
−1
Z
1
u
5. Evaluate the integral
Z
2
Z √4−y2 Z
−
−2
√
4−y 2
2
√
xz dz dx dy,
x2 +y 2
by changing to cylindrical coordinates.
Solution.
Z
2
−2
Z √4−y2 Z
−
√
4−y 2
2
√
Z
2π
Z
2
Z
2
xz dz dx dy =
x2 +y 2
0
0
r2 cos θ dz dr dθ = 0.
r
Can also see answer immediately by a symmetry argument.
6. Let f be continuous on [0, 1] and let R be the triangular region with vertices (0, 0),
(1, 0), and (0, 1). Show that
Z 1
ZZ
uf (u) du.
f (x + y) dA =
0
R
Solution. Set u = x + y and v = y. Then
∂(x, y) 1 1
=
∂(u, v) 0 1
and so
ZZ
Z
1
Z
f (x + y) dA =
R
= 1,
u
Z
f (u) dv du =
0
0
2
1
uf (u) du.
0
```