Math 221 - 501
Exam 2 Solutions
Nov. 1, 2010
1. (10) Define the following:
(a) The spherical coordinates of a point in R3 whose Cartesian coordinates are (x, y, z).
The spherical variables are commonly denoted as ρ, θ, and φ, and in terms
of the Cartesian variables they equal
p
y
z
ρ = x2 + y 2 + z 2
θ = arctan( )
φ = arccos( ) ,
x
ρ
with the following constraints: 0 ≤ ρ, 0 ≤ θ < 2π, and 0 ≤ φ ≤ π. That is, ρ equals
the distance from the origin to the point with Cartesian coordinates (x, y, z), and if
L denotes the line segment from the origin to this point, then θ is the angle between
the positive x-axis and the projection of L onto the x-y plane, and φ is the angle
between the positive z-axis and the line L.
In terms of the spherical variables, the Cartesian variables are:
x = ρ sin φ cos θ,
y = ρ sin φ sin θ,
z = ρ cos φ .
RR
(b) The double integral
f (x, y) dA, where R = [a, b] × [c, d], and f is a real valued
R
continuous function whose domain includes the rectangle R.
The double integral is defined to be the limit of a sequence of Riemann sums
where the norms of the partitions used to form the sums go to zero. A Riemann sum
is any sum of the form:
Σm,n
i=1,j=1 f (ξi , ωj )∆xi ∆yj ,
n
where the numbers {xi }m
i=0 and {yj=0 }0 form partitions of the intervals [a, b] and
[c, d] p
respectively. The norm of this partition is defined to be the maximum of the
set { (xi − xi−1 )2 + (yj − yj−1 ) : 1 ≤ i ≤ m, 1 ≤ j ≤ n}.
2. (20) Find the cylindrical and spherical coordinates of the point in R3 whose Cartesian
coordinates are (3, −4, 12).
√
Cylindrical
r = 9 + 16 = 5, θ = 2π − arctan (4/3), z = 12.
√
Spherical
ρ = 9 + 16 + 144 = 13, θ = 2π − arctan (4/3), φ = arccos (12/13)
Note that θ is supposed to lie between 0 and 2π. Since our point lies in the fourth quadrant
θ is between 3π/2 and 2π. Hence θ = 2π − arctan (4/3).
3. (40) Set up the integrals needed to calculate each of the following. Be sure to express
these integrals as iterated integrals, and sketch the region of integration.
(a) The area of the level surface z + y 2 − 2x2 = 5, which lies above the rectangle
R = [−1, 1] × [0, 1].
The surface area is given by:
Z Z
1/2
Surface area =
1 + zx2 + zy2
dA
R
Z
1
Z
1
1 + (4x)2 + (−2y)2
dx
=
−1
Z
0
1
=
1/2
Z
dx
−1
0
2
1
1 + 16x2 + 4y 2
1/2
dy
dy
(b) The volume of the solid bounded by the surfaces x + 2y + z = 6, x = 0, z = 0,
and x − 2y = 0.
The solid has its base in the x-y plane, and this base is bounded by x = 0,
2y = x, and x + 2y = 6. The volume is equal to
Z Z Z
Z
Volume =
0
3
Z
(6−x)/2
Z
x/2
6−x−2y
dz
dy
dx
dV =
R
3
0
4. (30) Sketch the curve (described in polar coordinates) r = 2 + sin θ for 0 ≤ θ ≤ 2π, and
then find the area enclosed by this curve.
The area is
Z Z
Z
Area =
dA =
R
=
9π
2
The curve looks like
4
2π
Z
dθ
0
2+sin θ
r dr
0