# Volume enclosed by a spherical coordinate surface

1. Jul 14, 2012

### forestmine

1. The problem statement, all variables and given/known data

Find the volume enclosed by the spherical coordinate surface ρ = 2sin∅

2. Relevant equations

dV = ∫∫∫(ρ^2)sin∅dρd∅dθ

3. The attempt at a solution

Alright, here's what I've done so far...

Since the region is a torus, centered around the z-axis, I began by finding my limits of integration for ρ, which I think would simply be from 0 to 2sin∅.

For my limits for ∅, I started at the utmost point on the positive z-axis, which I believe is 0 in regards to ∅, and it covers the entire ∅ "region" down to the negative z axis, so the limits are 0 to pi.

And for θ, I got simply 0 to 2pi.

Given those limits, integrating ((ρ^2)sin∅)dρd∅dθ, I wind up with 0, which on the one hand, I've convinced myself makes sense, since it is a torus, and it is symmetrical about the axes (like saying the area under the cosine from 0 to 2pi is 0). On the other hand, a volume of 0 for a physical object simply doesn't make sense.

I have a feeling I'm simply going about the limits wrong. If I've got those right, I'll post my integration work in case someone can spot the problem.

Thanks guys!

2. Jul 14, 2012

### HallsofIvy

Staff Emeritus
So what you are saying is that your integral is
$$\int_{\theta= 0}^{2\pi}\int_{\phi= 0}^\pi\int_{\rho= 0}^{2\sin(\phi)} \rho^2sin(\phi) d\rho d\phi d\theta$$
Clearly, the $\theta$ integral gives a factor or $2\pi$. Also, the integral of $\rho^2$ is $(1/3)\rho^3$ so, evaluating that integral we have
$$\frac{16}{3}\int_0^\pi sin^4(\phi)d\phi$$
I do not get 0 when I integrate that.

3. Jul 14, 2012

### forestmine

Ok, so as far as my limits of integration go, those are ok?

I also get (1/3)$\rho$^3, and evaluating at 2sin$\phi$ I get 8/3∫∫sin^4($\phi$) d$\phi$d$\theta$. I'm not sure where I ought to be getting 16 from?

As for ∫∫sin^4($\phi$) d$\phi$d$\theta$, I think I'm simply screwing up some basic calculus. I said that integral of sin^4($\phi$) is -cos^4(($\phi$). Should I be using some trig substitutions rather than making that leap?

Then I evaluated that from 0 to pi, which gives me -1 + 1, so I can't even make it to my dθ integral...

Thinking this is a trig substition I'm forgetting...

4. Jul 14, 2012

### LCKurtz

Yes, you should. Start by writing $\sin^4\phi$ as $(\sin^2 \phi)^2$ and using $\sin^2\phi = \frac {1-\cos(2\phi)} 2$. Then after a little algebra you will need a double angle formula again...

5. Jul 15, 2012

### forestmine

Ok, so when I expand (1-cos(2$\phi$)/2)^2, I get 1/4 - (cos(2$\phi$))/2 + (cos^2(2$\phi$))/4

Replacing the last term with a half-angle formula, I have

1/4 - (cos(2$\phi$))/2 +1/8 + (cos(2$\phi$))/8

At that point, I take the integral of that with respect to $\phi$.

I wind up with 1/4*$\phi$ - 1/2*sin($\phi$)cos($\phi$) + 1/8*$\phi$ + 1/8*sin($\phi$)cos($\phi$).

Evaluating from 0 to pi, I wind up with:

1/4pi +1/8pi.

Integrate that with respect to $\theta$ and evaluating from 0 to 2pi, I get

8/3(1/2*pi^2 + 1/4*pi^2)

where 8/3 was from the very first integration of $\rho$.

Hope this is easy enough to follow along in this format. If you could, let me know if the work looks ago.

Thanks so much by the way, I really appreciate it.

6. Jul 15, 2012

### LCKurtz

Hey, it isn't my job to simplify your answer. You simplify it, then I will tell you whether it agrees with mine.

7. Jul 15, 2012

### forestmine

Hehe fair enough. I get 2pi^2.

Looks like a nice and neat enough answer -- hopefully that's what you got!

I went through the problem a second time, though, and now I'm finding myself confused about one of the trig substitutions I made. Once I've expanded (1-cos(2ϕ)/2)^2, I'm left with a few terms + (cos^2(2ϕ))/4.

I know that cos^2(x) = (1 + cos(2x))/2. But since mine is essentially (2x), does it become 1+cos(4x)...or am I beginning to overthink this. I think I am...my 2x = x = theta in that trig sub, so I should be alright.

Oh jeez, this is my problem, haha. Over-thinking.

8. Jul 15, 2012

### LCKurtz

Yes, that's correct.

9. Jul 15, 2012

### forestmine

Thank you so much. I really appreciate all the help.