# [SOLVED] Find a volume

• June 12th 2009, 02:05 PM
arbolis
[SOLVED] Find a volume
Find the volume inside the sphere $x^2+y^2+z^2=4a^2$ and outside the cylinder $x^2+y^2=2ax$.
My attempt : I visualized the volume I must calculate. I can first calculate the volume of the sphere ( $\frac{32\pi a^3}{3}$ ) and then calculate the volume of the cylinder inside the sphere. This is where I'm having problems.
I think its volume is the following one, but I'm unsure : $\int_{-\pi}^{\pi} \int_{0}^{2a\cos \theta} \int_{-\sqrt{4a^2-r^2}}^{\sqrt{4a^2-r^2}} r dz dr d \theta = \int_{-\pi}^{\pi} \int_{0}^{2a\cos \theta} 2r \sqrt{4a^2-r^2} dr d\theta =$...
(I've used cylindrical coordinates for the volume of the cylinder inside the sphere, a better said, the part of the cylinder.)
I'd like to know if what I did until now is right. If so I'd have to remember how to solve the iterated integral I got.
• June 12th 2009, 06:00 PM
Calculus26
For the circle r = 2acos(theta) The upper half is generated as theta varies from 0 to pi/2 and the lower half as theta varies from -pi/2 to zero.

your outer limits should then be -pi/2 to pi /2

more simply go 0 to pi/2 and double it.

As for the integral let u = 4a^2 -r^2

after solving you'll end up with an integral involving sin^3(theta)

which can easily be solved as (1-cos^2(theta))sin(theta)
• June 13th 2009, 10:16 AM
arbolis
I didn't reach the result yet and I probably made an error somewhere.
Calculus of the volume of the cylinder inside the sphere : $I=2\int_{0}^{\frac{\pi}{2}} \int_{0}^{2a\cos (\theta)} \int_{-\sqrt{4a^2-r^2}}^{\sqrt{4a^2-r^2}} rdzdrd\theta$.

$\int_{-\sqrt{4a^2-r^2}}^{\sqrt{4a^2-r^2}} rdz=2r\sqrt{4a^2-r^2}$.

So $I=2\int_{0}^{\frac{\pi}{2}} \int_{0}^{2a\cos (\theta)} 2r\sqrt{4a^2-r^2} dr d\theta$.

I call $J=\int_{0}^{2a\cos(\theta)} 2r\sqrt{4a^2-r^2}dr$.
Let $u=4a^2-r^2 \Rightarrow du=-2rdr$, hence $J=-\int_{0}^{2a\cos (theta)} \sqrt{u} du=-\frac{2}{3} (2a\cos(\theta))^{\frac{3}{2}}$.
Thus $I=-\frac{4}{3} \int_{0}^{\frac{\pi}{2}} (2a\cos (\theta))^{\frac{3}{2}}d\theta$.
I don't know how to solve this integral. I didn't reach $\sin ^3 (\theta)$ yet so I think I made an error.
Could you tell me if I made an error? If no, I'd like to know which method to use to solve the integral because I've no idea.
• June 13th 2009, 10:25 AM
Amer
Quote:

Originally Posted by arbolis
I didn't reach the result yet and I probably made an error somewhere.
Calculus of the volume of the cylinder inside the sphere : $I=2\int_{0}^{\frac{\pi}{2}} \int_{0}^{2a\cos (\theta)} \int_{-\sqrt{4a^2-r^2}}^{\sqrt{4a^2-r^2}} rdzdrd\theta$.

$\int_{-\sqrt{4a^2-r^2}}^{\sqrt{4a^2-r^2}} rdz=2r\sqrt{4a^2-r^2}$.

So $I=2\int_{0}^{\frac{\pi}{2}} \int_{0}^{2a\cos (\theta)} 2r\sqrt{4a^2-r^2} dr d\theta$.

I call $J=\int_{0}^{2a\cos(\theta)} 2r\sqrt{4a^2-r^2}dr$.
Let $u=4a^2-r^2 \Rightarrow du=-2rdr$, hence ${\color{red}J=-\int_{0}^{2a\cos (theta)} \sqrt{u} du=-\frac{2}{3} (2a\cos(\theta))^{\frac{3}{2}}}$.
Thus $I=-\frac{4}{3} \int_{0}^{\frac{\pi}{2}} (2a\cos (\theta))^{\frac{3}{2}}d\theta$.
I don't know how to solve this integral. I didn't reach $\sin ^3 (\theta)$ yet so I think I made an error.
Could you tell me if I made an error? If no, I'd like to know which method to use to solve the integral because I've no idea.

you just did not change the boundaries of the integral when you let
Let $u=4a^2-r^2 \Rightarrow du=-2rdr$,

$u = 4a^2 - (2acos(\theta))^2 \Rightarrow u = 4a^2(1-cos^2(\theta))$

$u=4a^2 - (2(0))^2 = 4a^2$
• June 13th 2009, 11:48 AM
arbolis
Oh... thanks a lot.
So $J=\int_{4a^2 \sin ^2{\theta}}^{4a^2} u^{\frac{1}{2}} du = \frac{2}{3} \big( (4a^2)^{\frac{3}{2}}-(4a^2 \sin ^2 {\theta})^{\frac{3}{2}} \big)$ ?
And I still don't see any integral involving $\sin^3 (\theta)$. Again, I made an error.
• June 13th 2009, 12:54 PM
Calculus26
In general (x^2)^(3/2) = x^3

sin^2(t)^(3/2) = sin^3(t)

Of course you also have to take care of the constant 4a^2 part also
• June 13th 2009, 12:56 PM
arbolis
Lol, I missed this! Ok, I'll try to finish it up.

Ok, I've done it. My result for the volume of the cylinder inside the sphere is $\frac{64-192 \pi a^3}{54}$. I don't know if I made an arithmetic error but at least I know how to do it.
So the final answer would be $\frac{32\pi a^3}{3}+\frac{192\pi a^3-64 }{54}=\frac{384\pi a^3-64}{54}$ which is not beautiful.(Doh)