## What's the area of this volume?

Consider a 3d coordinate system with axis x,y and z.

We are given a circle on the x-z plane with function $$z^2 + (x-a)^2 = a^2$$. We rotate this circle 90 degrees around the z-axis. What's the volume of the resulting surface?
 Recognitions: Gold Member Homework Help Science Advisor Well, shouldn't that be a quarter of a torus? I'll opt for that and say: $$V=\pi{a}^{2}*\frac{\pi}{2}a=\frac{\pi^{2}}{2}a^{3}$$ Edit: The area function for a given angle measured relative to the x-axis (and with the origin as the pole) is $$f(\theta)=\pi{a}^{2}$$ By Cavalieri's principle, we have: $$V=\int_{0}^{\frac{\pi}{2}}\pi{a}^{2}ad\theta$$