# Homework Help: Volumes question, volume of a torus

1. Oct 23, 2011

Problem solved

I'm not sure how easy this will be to understand without a diagram, but I don't know how to upload one :(

1. The problem statement, all variables and given/known data
Let a and b be constants, with a > b > 0.A torus is formed by rotating the
circle (x - a)^2 + y^2 = b^2 about the y-axis.

The cross-section at y = h, where –b ≤ h ≤ b, is an annulus. The annulus has
inner radius x1 and outer radius x2 where x1 and x2 are the roots of
(x - a)^2 = b^2 + y^2

(i) Find x1 and x2 in terms of h.

(ii) Find the area of the cross-section at height h, in terms of h.

(iii) Find the volume of the torus.

2. Relevant equations
I think they're all up there.

3. The attempt at a solution

I did part (i) and got x1 = a - $\sqrt{b^2 - h^2}$
and x2 = a + $\sqrt{b^2 - h^2}$, which is correct according to the answers.

I did part (ii) and got A = $\pi$a$\sqrt{b^2 - h^2}$, which again is correct.

But for part 3 I thought V = $\pi$a$\int^{b}_{-b}[itex]\sqrt{b^2 - h^2}dh$ (sorry I don't know how to get rid of that itex in the integral)

But in the answers it is V = $\pi$a$\int^{a}_{-a}[itex]\sqrt{b^2 - h^2}dh$, and im not sure what it's between -a and a. The answers are from a different source then the question so I'm not sure if they have made a mistake or not.

Any help will be much appreciated, thanks.

Last edited: Oct 23, 2011
2. Oct 23, 2011