Calculate Volume of Partial Torus (a=10, r=5)

[RollingThundr]

I have a real world problem. To design a calculator for a special machining process, I need to determine the volume of material removed when making a groove in the shape of a semicircle inside a tube on a lathe. The volume removed looks like this:

[PLAIN]http://img683.imageshack.us/img683/9854/halftorus.jpg

I've made many attempts, but I've been unsuccessful in creating an equation that works. This is mostly because I don't have the integration skills that are required. I've tried online integration calculators, but they haven't produced good results.

To check the result, a SolidWorks tells me that when a = 10 and r = 5, the volume is 2991.00 units cubed.

Thank you for your help.

Homework Equations

The Attempt at a Solution

 

[Dick]

Use the second theorem of pappus. Then look up the centroid of a semicircle. It's located 4*r/(3*pi) from the center of the hemisphere. So the circle swept by the centroid has radius a+4*r/(3*pi). The area of the semicircle is pi*r^2/2. So the total volume is (pi*r^2/2)*(4*r/(3*pi)+a)*2*pi.

 

[RollingThundr]

Use the second theorem of pappus. Then look up the centroid of a semicircle. It's located 4*r/(3*pi) from the center of the hemisphere. So the circle swept by the centroid has radius a+4*r/(3*pi). The area of the semicircle is pi*r^2/2. So the total volume is (pi*r^2/2)*(4*r/(3*pi)+a)*2*pi.

Brilliant!

(pi*5^2/2)*(4*5/(3*pi)+10)*2*pi = 2991

I must have spent 3 hours trying to remember how to figure out how to integrate a modified equation for a circle. Your answer is short, sweet, and accurate.

Thanks Dick