Math 3113 - Multivariable Calculus
Homework #4 - 2008.02.06
Due Date - 2008.02.13
Solutions
Find an equation of the form r = f (θ, z) in cylindrical coordinates for the following surfaces.
1. x2 + y 2 + z 2 = 4
x2 + y 2 + z 2 = 4
r2 + z 2 = 4
Solving for r gives
r=
2. x2 − y 2 = 4
p
4 − z 2.
x2 − y 2 = 4
Solving for r gives
3. z = 3xy
r2 cos2 (θ) − r2 sin2 (θ) = 4
r2 cos2 (θ) − sin2 (θ) = 4
2
2
r= q
=p
.
2
cos(2θ)
cos2 (θ) − sin (θ)
z = 3xy
z = 3r2 cos(θ) sin(θ)
Solving for r gives
r=
r
z
=
3 cos(θ) sin(θ)
1
s
2z
.
3 sin(2θ)
2
Find an equation of the form ρ = f (θ, φ) in spherical coordinates for the following surfaces.
4. z = 2
z=2
ρ cos(φ) = 2
Solving for ρ gives
ρ=
2
.
cos(φ)
5. x = z 2
x = z2
ρ cos(θ) sin(φ) = ρ2 cos2 (φ)
cos(θ) sin(φ) = ρ cos2 (φ)
Solving for ρ gives
ρ=
cos(θ) tan(φ)
.
cos(φ)
6. x2 − y 2 = 4
x2 − y 2 = 4
ρ2 sin2 (φ) cos2 (θ) − ρ2 sin2 (φ) sin2 (θ) = 4
ρ2 sin2 (φ) cos2 (θ) − sin2 (θ) = 4
ρ2 sin2 (φ) cos(2θ) = 4
Solving for ρ gives
ρ=
1
2
p
.
sin(φ) cos(2θ)
3
Sketch the graphs of the following sets of points in spherical coordinates.
7. ρ = 8,
0 ≤ θ ≤ 2π,
0≤φ≤
π
2
This is the upper half sphere of radius 8.
10
8
6
4
-10
2
-8
-6
-4
-2
y
10
8
6
4
0
2
0
0
-6
-4
-2
x
2-2
4
6
-4
z
-6
-8
-10
8
10
-8
-10
4
8. θ =
2π
3 ,
0 ≤ ρ ≤ 4,
0≤φ≤π
This is the disc of radius 4 centered on the z-axis which forms an angle of θ =
2π
3
in the xy-plane.
-4
-2
y
4.0
-3.6
3.2
2.4
1.6
x
0.4
-1.6
0.8
0
0.0
-0.8
2.4
-1.6 z
-2.4
-3.2
-4.0
2
4
5
9. φ =
π
6,
This is
1
4
0 ≤ ρ ≤ 2,
3π
2
≤ θ ≤ 2π
portion of the upper cone located in the octant with negative y and positive x coordinates.
-2
-2
2.0
1.6
y
-1
1.2
x
-1
0.8
0.4
0
0
0.0
-0.4
1
-0.8z
-1.2
1
-1.6
2
-2.0
2
6
Sketch the graphs of the following sets of points in cylindrical coordinates.
10. r = 4,
0 ≤ θ ≤ π,
0≤z≤4
This is half a cylinder of radius 4 with positive y-coordinates and height 4 starting in the xy-plane.
5
-5
4
3
-5
-4
-3
y
-3
x
2
-2
-2
1
-1-1
00
1
-1
2
3
-4
z
-2
-3
4
-4
5
-5
0
1
2
3
4
5
7
11. θ =
π
3,
0 ≤ r ≤ 5,
−4 ≤ z ≤ 0
This is a plane of height 4 starting at z = −4 which is 5 units long and forms an angle of θ =
5
4
3
2
-5
-5
1
-4
-3
-3
-1 00
1
2
5
-1
-1
y
3
-2
0
-2
4
-4
z
-2
x
1
2
3
4
-3
-4
-5
5
π
3
with the xy-plane.
8
12. z = −2,
0 ≤ r ≤ 5, π2 ≤ θ ≤
3π
2
This is a half disc of radius 5 located 5 units below the xy-plane with negative x-coordinates.
5
4
3
2
-5
-5
-4
1
-4
-3
-3
-2
y
5
00
1
2
2
4
x
-1
-1
-1
1
3
-2
0
-2
z
3
4
-3
-4
-5
5