Quiz 9
Name:
SOLUTIONS
Maths 114 - Calculus II
November 16, 2010
Note: In order to receive full credit, you must show work that justifies your answer.
1. Suppose
(
Cxy
f (x, y) =
0
if 0 ≤ x ≤ 1 and 0 ≤ y ≤ 1
otherwise
(1)
is a joint density function. What is the value of the constant C which appears
in equation (1)?
RR
To be a joint density function, we need to have R2 f (x, y) dA = 1. Now
because f (x, y) is zero outside the region D = [0, 1] × [0, 1],
ZZ
ZZ
f (x, y) dA =
f (x, y) dA
R2
D
Z 1Z 1
=
Cxy dx dy
0
0
x=1
Z 1
C 2 x y
dy
=
0 2
x=0
Z 1
C
y dy
=
0 2
y=1
C 2 y
=
4 y=0
C
=
.
4
For this to be 1, we need C = 4.
2. Use cylindrical coordinates to evaluate the integral
is the region
RRR p
x2 + y 2 dV , where E
E
E = {(x, y, z) ∈ R3 | x2 + y 2 ≤ 1, y ≥ 0, 4x ≤ z ≤ 6}.
The constraints x2 + y 2 ≤ 1 and y ≥ 0 are equivalent to 0 ≤ r ≤ 1 and
0 ≤ θ ≤ π. In cylindrical coordinates, the constraint 4x ≤ z ≤ 6 becomes
4r cos(θ) ≤ z ≤ 6, so we can describe E in cylindrical coordinates as
{(r, θ, z) | 0 ≤ r ≤ 1, 0 ≤ θ ≤ π, 4r cos(θ) ≤ z ≤ 6}.
Therefore
ZZZ p
x2 + y 2 dV
Z
π
1
Z
Z
6
r × r dz dr dθ
=
0
E
Z
0
π
4r cos(θ)
1
Z
=
Z0 π Z0 1
=
0
Z
z=6
r2 z z=4r cos(θ) dr dθ
r2 (6 − 4r cos(θ)) dr dθ
0
π
=
Z0 π
r=1
2r3 − r4 cos(θ) r=0 dθ
(2 − cos(θ)) dθ.
=
0
= (2θ − sin(θ))|θ=π
θ=0
= 2π.
Quiz 9
Name:
SOLUTIONS
Maths 114 - Calculus II
November 18, 2010
Note: In order to receive full credit, you must show work that justifies your answer.
RRR
1. Evaluate the triple integral
3x dV , where V is bounded by the parabolic
E
cylinders y = x2 , x = y 2 and the planes z = 0 and z = 2y.
The parabolic cylinders intersect in the lines x = y = 0 and x = y = 1. So if we
project the
the
√ region where 0 ≤ x ≤ 1 and
√ region E onto the xy-plane, we get
2
2
x ≤ y ≤ x (note that if 0 ≤ x ≤ 1, then x ≤ x). This gives us the bounds
on the two outer integrals. The bounds on the inner integral (with respect to z)
are from z = 0 to z = 2y.
Therefore
ZZZ
1
Z
3x dV
√
Z
x
Z
2y
=
3x dz dy dx
E
x2
0
Z
1
= 3
Z
0
√
x
(2y − 0)dy dx
x
x2
0
Z
1
√
2 y= x
y y=x2
= 3
x
dx
0
Z 1
=
3x(x − x4 ) dx
0
x=1
1 6 3
=
x − x 2
x=0
1
.
=
2
Note: This is not the only way you can evaluate this integral. For example,
ZZZ
Z 2 Z 1 Z √y
3x dV =
3x dx dy dz.
E
0
z/2
y2
As a challenge, try evaluate the integral using dy dx dz.
2. Use spherical coordinates to evaluate the integral
solid hemisphere x2 + y 2 + z 2 ≤ 1, x ≥ 0.
RRR
E
x dV , where E is the
The bounds are 0 ≤ ρ ≤ 1, −π/2 ≤ θ ≤ π/2, 0 ≤ φ ≤ π, as you can see by
sketching or visualizing the solid hemisphere1 . Therefore
ZZZ
Z
x dV
π/2
Z
π
Z
=
−π/2
E
=
=
=
=
=
=
(ρ sin φ cos θ)ρ2 sin φ dρ dφ dθ
0
ρ=1
ρ4 dφ dθ
sin φ cos θ 4 ρ=0
−π/2 0
Z
Z π
1 π/2
cos θ
sin2 φ dφ dθ
4 −π/2
0
Z π/2
Z π
1
1
cos θ
(1 − cos(2φ)) dφ dθ
4 −π/2
0 2
φ=π
Z
1 π/2
1
cos θ φ − sin(2φ) dθ
8 −π/2
2
φ=0
Z
1 π/2
cos(θ)π dθ
8 −π/2
π
θ=π/2
sin(θ)|θ=−π/2
8
π
.
4
Z
=
0
1
π/2
Z
π
2
Note: Because the bounds are constants, we could have integrated in any order
without changing the bounds.
1
Alternatively, you can reason as follows: x2 + y 2 + z 2 ≤ 1 is equivalent to 0 ≤ ρ ≤ 1, and x ≥ 0
is equivalent to r cos θ ≥ 0, and since r is positive (except on the z-axis), we must have cos θ ≥ 0, so
−π/2 ≤ θ ≤ π/2. There are no restrictions on φ except for the natural ones, so 0 ≤ φ ≤ π.