# Integral to find the volume of a torus

1. Aug 25, 2013

### togame

1. The problem statement, all variables and given/known data
Find the volume of the torus formed when the circle of radius 2 centered at (3,0) is revolved about the y-axis. Use geometry to evaluate the integral.

2. Relevant equations
Formula for the semi-circle: $y=\sqrt{4-x^2}$
Solving for x give $x=\pm\sqrt{4-y^2}$

3. The attempt at a solution
I was thinking of solving the integral from -2 to 2 using both the positive and negative sides, then evaluate the integral and then multiply by 2, but I'm not so sure about my equation.

$$2\int_{-2}^2\pi\big( (3-\sqrt{4-y^2})^2 - (3+\sqrt{4-y^2})^2 \big) \mathrm d y$$

I just seem lost on this one :(

2. Aug 25, 2013

### janhaa

I believe the equation is:

$$V=\int_{-2}^2\pi\big( (3+\sqrt{4-y^2})^2 - (3-\sqrt{4-y^2})^2 \big) \mathrm d y$$

$$V=\int_{-2}^2\pi\cdot 12(\sqrt{4-y^2}) \mathrm d y$$

then trigonometric substitution...

3. Aug 25, 2013

### Zondrina

Edit : Got beat to it.

4. Aug 25, 2013

### mathwonk

the way to do this is to use pappus' theorem. the product of the distance traveled by the center of the circle by the area of the circle.