A ball of radius 13 has a round hole of radius 6 drilled through its center. Find the volume of the resulting solid.

Question

anonymous · Answer

What do I use to find the volum of the second structure?

anonymous · Answer

oh so pi*r^2*h

anonymous · Answer

Pi*12^2*26

anonymous · Answer

And 4/3 pi r^3 for the ball

anonymous · Answer

(4/3) pi 6^3

anonymous · Answer

that is of this cone

zarkon · Answer

now I don't know what you are doing :)

anonymous · Answer

1/3 pi r^2 h so (1/3)pi 12^2 * 13

zarkon · Answer

I would start with $$x^2+y^2+z^2=13^2$$

then $$z=\pm\sqrt{13^2-x^2-y^2}$$

then I would integrate between these two function over the region defined by the hole.

then subtract this volume from the volume of the sphere

zarkon · Answer

to integrate one should switch to polar coordinates...

I get $$\frac{532\pi\sqrt{133}}{3}$$

anonymous · Answer

That worked! Thanks a lot!

zarkon · Answer

just so everyone can see this is what I did...

$$\frac{4}{3}\pi 13^3-2\int\limits_{0}^{2\pi}\int\limits_{0}^{6}\sqrt{13^2-r^2}r\,dr\,d\theta$$