Mass of ball given by x^2+y^2+z^2<=a^2 if density at any point is proportional to distance from z-axis

[aron101782]

Find the mass of a ball B given by x^2+y^2+z^2<=a^2 if the density at any point is proportional to its distance from the z axis.

I have solved this problem seemingly correctly to the same solution in the manual. However the solutions manuals method of solving the problem is unknown to me.

Here is what is in the solutions manual:

Since density is proportional to the distance from the z axis, we can say p(x,y,z) = K(x^2+y^2)^1/2. Then
m=2I(0 2pi)I(0 a)I(0 (a^2-r^2)^1/2) Kr^2 dzdrdt
=2KI(0 2pi)[1/8 r(2r^2-a^2)(a^2-r^2)^1/2 + 1/8 a^4 arcsin(r/a)][0 a]
=2KI(0 2pi)[(1/8 a^4)(pi/2)]dt
=1/4 a^4 pi^2 K

I have no idea what is going on there.

Here is my method of solving the problem.

x^2+y^2+z^2<=a^2
D= Kr
z=(a^2-r^2)^1/2

8I(0 pi/2)I(0 a)I(0 (a^2-r^2)^1/2) Kr^2 dzdrdt

4Kpi I(0 a) r^2(a^2-r^2)^1/2 dr
Trigonometric Substitution
r=asint
dr=acost

4Ka^4pi I(0 pi/2) sin^2t cos^2t dt
Ka^4pi I(0 pi/2)(1-cos2t)(1+cos2t)
Ka^4pi I(0 pi/2)1-cos^2(2t) dt
Ka^4pi(t/2-cos4t/2)[0 pi/2]
Ka^4pi^2/4

This is what I believe to in fact be the correct way to solve this integral. What I would like to know is what the **** is written in the solutions manual.

 

[HallsofIvy]

Your solutions manual is using "cylindrical coordinates"- polar coordinates for the projection of the point to the xy-plane plus the z-axis. Since the "differential of area" in polar coordinates is \(\displaystyle rdrd\theta\) (does your textbook use "t" for "$theta$" or did you do that?), the "differential of volume" in cylindrical coordinates is \(\displaystyle rdrd\theta dz\). To cover a sphere of radius a, r must go from 0 to a, \(\displaystyle \theta\) must go from 0 to \(\displaystyle 2\pi\), and z must go from -a to a. With density "Kr" the integral is \(\displaystyle \int_{z= -a}^a\int_{\theta= 0}^{2\pi}\int_{r= 0}^a (Kr) r drd\theta dz= K\int_{z= -a}^a\int_{\theta= 0}^{2\pi}\int_{r= 0}^a r^2 drd\theta dz\)
\(\displaystyle = \left(\int_{z= -a}^a dz\right)\)\(\displaystyle \left(\int_{\theta= 0}^{2\pi} d\theta\right)\)\(\displaystyle \left(\int_{r= 0}^a r^2 dr\right)\)\(\displaystyle =(2a)(2\pi)(\frac{1}{3}a^3)= \frac{4}{3}\pi a^4\).

 

[aron101782]

What Im asking is what method has been used in the solutions manual to solve the integral

2I(0 2pi)I(0 a)I(0 (a^2-r^2)^1/2)Kr^2dzdrdt

 

[HallsofIvy]

That isn't at all the question you asked in your first post!
Assuming your "I" is the integral sign,
2I(0 2pi)I(0 a)I(0 (a^2-r^2)^1/2)Kr^2dzdrdt is
\(\displaystyle 2\int_0^{2\pi}\int_0^a\int_0^{(a^2-r^2)^{1/2}} Kr^2 dzdrdt= 2K\left(\int_0^{2\pi} dt\right)\left(\int_0^a \left(\int_0^{(a^2- r^2)^{1/2}} dz\right) r^2 dr\right)\).

The first parentheses on the left is just \(\displaystyle \int_0^{2\pi} dt= 2\pi\).

The "inner" of the two parentheses on the right is \(\displaystyle \int_0^{(a^2- r^2)^{1/2}}dz= (a^2- r^2)^{1/2}\).

So the final parentheses, and the final integral is \(\displaystyle \int_0^a r^2(a^2- r^2)^{1/2} dr\). IF that were just "\(\displaystyle r (a^2- r^2)dr\)" it would be easy- just make the substitution \(\displaystyle u= a^2- r^2\). But since we have "\(\displaystyle r^2\)" rather than "\(\displaystyle r\)" we will need a trig substitution. Let \(\displaystyle r= asin(u)\). Then \(\displaystyle (a^2- r^2)^{1/2}= (a^2- a^2 sin^2(u))^{1/2}= a(1- sin^2(u))^{1/2}= a cos(u)\). Of course, \(\displaystyle r^2= a^2sin^2(u)\) and \(\displaystyle dr= a cos(u)du\). When r= a sin(u)= 0, sin(u)= 0 so we can take u= 0. When r= a sin(u)= a, sin(u)= 1 so we can take \(\displaystyle u= \pi/2\). The integral becomes \(\displaystyle \int_0^{\pi/2} (a sin(u))^2 (a cos(u)) (a cos(u) du)= a^4\int_0^{\pi/2} sin^2(u)cos^2(u)du\). Can you integrate that?

(You might need the "double angle identities":
\(\displaystyle sin^2(u)= \frac{1- cos(2u)}{2}\) and
\(\displaystyle cos^2(u)= \frac{1+ cos(2u)}{2}\).)

 

[aron101782]

Yes that is correct and what I believe to be the best method of solving this integral. But the author of the book looks to have gone down a different path.

2I(0 2pi)I(0 a)I(0 (a^2-r^2)^1/2) Kr^2 dzdrdt

I dont understand the next line

2KI(0 2pi)[1/8 r(2r^2-a^2)(a^2-r^2)^1/2 + 1/8 a^4 arcsin(r/a)][0 a]

 

[aron101782]

Does anyone know how that second Iine was derived?