Question

There a sphere of radius \(R\). We then drill a cylindrical hole straight through the sphere, so that the cylindrical hole has a height \(h\) (see picture). Find the volume of the solid.

Answer 1

So we know that the volume of the sphere is 4pR^2 and the volume of the cylinder is phr^2
We have Volume(Sphere)-Volume(Cylinder) and we have the volume of the drilled solid.

Do you want the volume of the solid in relation only with h and R?

Answer 2

and the volume of the sphere would be \(\frac43\pi r^3\)

Answer 3

oops yeah excuse me :D im rusty

Answer 4

Can i ask, is the cylindrical hole at a specific location related to the sphere or it could be anywhere inside the sphere?

Answer 5

If the radius of the base of the cap is h, and the height of the cap is a, then the volume of each spherical cap is
\[ V = \frac {pi*a}{ 6} \times (3h^2 + a^2) \]
volume of the cylinder is
\[ V = pi*h^2 \times 2(r-a)\]
volume of the sphere is
\[ \frac 43 pi \times r^3 \]
so total volume = volume of sphere - volume of cylinder - 2x volume of spherical cap... yeah?

Answer 6

we drilled *straight* through the sphere, so there is no "cap" on the cylinder

Answer 7

oh

Answer 8

sorry, but in my examply im considering the bored section as 3 portions:
2 caps and one flat ended circular cylinder

Answer 9

well that problem I've ever heard before, and I doubt has as interesting an answer as this one
it is critical in this important that we cored clear through the sphere and came out the other end

Answer 10

never*

Answer 11

bored*

Answer 12

Is the center of the cylinder the same of the center of the sphere? If yes i think I can answer if not it's more complicated.

Answer 13

oh I see, you subtracted the caps. That's fine, good luck simplifying it.
I will give a hint: the answer depends on only one variable