# Thread: Surface area and Volume

1. ## Surface area and Volume

A Circle C is given by the equation x^2 +(y-1)^2 = 1. It is rotated about the x-axis to give a surface of revolution called a torus.

a) Find the total surface area of the torus.

(b) Find the volume of the torus and verify that it is 2pi times the area of C.

2. ## re: Surface area and Volume

Have you covered the theorems of Guldin?

3. ## re: Surface area and Volume

No, what we have covered are that :

the surface area is the integral of 2pi y (1+(dy/dx)^2)^1/2 dx, and the volume is the integral of pi y^2 dx. I have used these equations, integrated from -1 to 1 and got 2pi(2+pi) for the area, and 10pi/3 +pi^2. But I think I'm doing something wrong. Can you help me?

4. ## re: Surface area and Volume

Better work in parametric equations $\displaystyle C\equiv x=\cos t,\;y=1+\sin t\;(t\in[0,2\pi])$ . Then, the area is $\displaystyle A=2\pi\int_0^{2\pi} y\sqrt{(dx/dt)^2+(dy/dt)^2}\;dt=\ldots$ . Let's see what do you obtain.

5. ## re: Surface area and Volume

Originally Posted by FernandoRevilla
Better work in parametric equations $\displaystyle C\equiv x=\cos t,\;y=1+\sin t\;(t\in[0,2\pi])$ . Then, the area is $\displaystyle A=2\pi\int_0^{2\pi} y\sqrt{(dx/dt)^2+(dy/dt)^2}\;dt=\ldots$ . Let's see what do you obtain.

When I integrated and used the limits 0 and 2pi, I got an answer of 4pi^2, but when I changed to 0 and pi, I got 2pi(pi+2) which is the same answer as when I worked it out using the cartesian equation rather than the parametric. Which one is correct?

And what about the volume? I worked it out using parametric as well but got an answer of 2pi^2. The question says that it should be 2pi times the answer of the area...

Right, the volume is $\displaystyle V=2\pi^2=2\pi\cdot \pi=2\pi \cdot \textrm{Area}(C)$