Homework 10 Standard problem solutions
2. Consider the solid formed by ρ = sin(2φ) for 0 ≤ φ ≤ π/2 with constant density ρ
y
0.6
0.4
0.2
0.2
0.4
0.6
x
Figure 1: ρ = sin(2φ) for fixed θ
(a) Set up the integrals to determine the z-coordinate of the center of mass using spherical
coordinates.
Recall that the z coordinate of center of mass is given by
RRR
δzdV
z̄ = RRRD
D δdV
where D is the domain of integration. Writing the above in spherical coordinates we
have
Z 2π Z π/2 Z sin(2φ)
δρ3 cos(φ) sin(φ) dρdφdθ
0
0
z̄ = 0 Z 2π
Z π/2 Z sin(2φ)
δρ2 sin(φ) dρdφdθ
0
0
0
(b) We now repeat the above setup using the order dφdρdθ. To determine the integration
bounds in φ we fix θ, which allows us to view a cross section of the surface and fix ρ
which allows us to consider circles of varying radii centered at the origin. An example
of these paths of fixed ρ and θ is given in Figure 2.
0.6
0.4
0.2
0.2
0.4
0.6
Figure 2: ρ = sin(2φ) for fixed θ
1
Thus our integration in the desired order is
2π
Z
0
z̄ =
1 Z π/2− 12 sin−1 (ρ)
Z
1
2
0
2π
Z
1 Z π/2− 12 sin−1 (ρ)
Z
0
δρ3 cos(φ) sin(φ)
dφdρdθ
sin−1 (ρ)
1
2
0
δρ2 sin(φ) dφdρdθ
sin−1 (ρ)
(c) Here we will evaluate the integral from part (a) by evaluating the numerator and denominator separately.
Z 2π Z π/2 Z sin(2φ)
δρ3 cos(φ) sin(φ) dρdφdθ
Mz =
0
0
2π
Z
0
Z
π/2
4 sin5 (φ) cos5 (φ)dφdθ
=δ
0
0
Z 2π
δ
=
dθ
15 0
2δπ
=
15
and
2π
Z
π/2 Z sin(2φ)
Z
m=
0
0
δρ2 sin(φ) dρdφdθ
0
Z
Z
δ 2π π/2
=
sin(φ) sin3 (2φ)dφdθ
3 0
0
32δπ
=
105
which yields
z̄ = 7/16
3.
(a) The equation for work is
ZZZ
δ(x, y, z)h(x, y, z)g dV
D
where g is the acceleration due to gravity (9.8 N/kg). The equation for the conical
mountain is
2
R
(z − H)2 = x2 + y 2
H
So the work is
Z
R
W =
−R
√
Z
R2 −x2
√
− R2 −x2
Z
H− H
R
√
x2 +y 2
δ(x, y, z)h(x, y, z)g dzdydx
0
(b) Now we are given the following values
R = 62, 000f t = 18.9 × 103 m
H = 12, 400f t = 3.78 × 103 m
δ = 2.7g/cm3 = 2.7 × 103 kg/m3
2
To calculate the work, it’s much simpler to consider the integral in cylindrical coordinates
which is
Z 2π Z R Z H− H r
R
W =
δzgr dzdrdθ
0
0
0
Z 2π Z R 1
H 2
= δg
H−
r drdθ
R
0
0 2
R
Z 2π 2 4
2H 2 r3 H 2 r2 1 H r
−
+
= δg
2 4R2
3R
2
0
0
Z 2π 2 2
H R
= δg
dθ
24
0
δgH 2 R2 π
=
12
plugging in the given values we have
W = 3.54 × 1019 J
4.
(a) We first need to consider some arbitrary region A which we can integrate. The locations
of center of mass are given by
ZZ
rdσ
r̄ = Z ZA
dσ
A
ZZ
rdσ
A
=
A
it is helpful to note here that this yields the relation
ZZ
Ar̄ =
rdσ
A
(b) Volume of revolution is given by
ZZ Z
2π
V =
dθdrdz
A
0
(c) Using the integral in part (b) we can integrate in θ to arrive at
ZZ
V = 2π
drdz
A
(d) Now using the result from part (a) we can rewrite the double integral giving Pappus’
Theorem for volumes of revolution
V = 2πAr̄
3
(e) We now want to use this result to determine calculate the volume of a cone with height
H and radius R. We set up a similar integral in problem 2, so the details of the setup
are omitted here. Via Pappus’s Theorem, the volume is given by
R Z H−(H/R)r
Z
r dzdr
V = 2π
0
0
H
= 2π Hr − r2 dr
R
1
= πHR2
3
which is the result we expect.
(f) Now we verify the result of the theorem using a sphere of radius R. For simplicity, we
will only consider the sphere for z > 0 and multiply our result by 2. So we have
V = 2(2π)Ar̄
Z RZ
V = 4π
0
0
R2 − z 2 dz
= 2π
0
2R3
3
4
= πR3
3
(g) Newton: 1642 - 1727
Leibniz: 1646 - 1716
(h) Pappus: 290 - 350
4
!
R2 −z 2
R
Z
= 2π
√
r drdz