Volume enclosed by a spherical coordinate surface

[forestmine]

Homework Statement

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

Homework Equations

dV = ∫∫∫(ρ^2)sin∅dρd∅dθ

The Attempt at a Solution

(Sorry about my notation!)

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!

 

[HallsofIvy]

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.

 

[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...

 

[LCKurtz]

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?

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...

 

[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.

 

[LCKurtz]

I get

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

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

 

[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.

 

[LCKurtz]

Hehe fair enough. I get 2pi^2.

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

Yes, that's correct.

 

[forestmine]

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