Finding The Volume Enclosed by a Torus

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Homework Statement



Find the volume enclosed by the torus rho = sin theta.


Homework Equations





The Attempt at a Solution



I tried setting the limits as phi from 0 to pi, theta from 0 to 2 pi, and rho from 0 to sin theta. However, if i do that, i get a volume of 0. How should i set up the limits?
 
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Which angle is theta? There is more than one convention for spherical coordinates. From your limits I would guess its the azimuthal angle, but from rho=sin(theta) I'd guess its the polar angle. Can you show the integral you finally got?
 
I'm an idiot. It's the azimuthal angle. I tried solving it as the polar angle. Thanks.
 
Solutions for volumes of rotation are easy using the Pappus theorem. This is attributed to Pappus of Alexandria, but was first proved by the Swiss mathematician Guldin. The theorem states that the volume is the area of the profile times the distance that the center of gravity of the profile moves. The axis of rotation cannot pass through the profile.
 
Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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