volume of a solid with a cylindrical hole

Question

A sphere of radius 'a' has a cylindrical hole of radius 'b' bored from it. The axis of the cylinder coinciding with the diameter of the sphere. find the volume of the remaining part of the sphere. I am having problem in calculating the volume of the cylinder.

Answer 1

In order to go about this problem, we need to look at it in terms of two parts: the top and the bottom of the sphere where the cylinder intersects the sphere.

Here is what the sphere and cylinder hole would look like on a cartesian plane

These interception points will be your bounds as they define where exactly the cylinder will be drilling into the sphere

Keeping those bounds in mind, you can then construct the two equations (the sphere and the cylinder). Remember which one is on top, which in this case is the sphere.

Top function (sphere/circle) $$ f(y) = \pm \sqrt{(a^2 - y^2)} $$

Bottom function (cylinder) $$ g(y) = x = b $$

Now you have all you need in order to create your integral! Since the volume also includes an extrusion, you are doing the difference of two equations!

$$ V = \int_{\sqrt{(a^2 - b^2)}}^{-\sqrt{(a^2 - b^2)}}π[f^2(y) - g^2(y)] dy $$

Just to make it look less messy, I'm going to replace the square root of a^2 - b^2 as u

$$ V = \int_{u}^{-u}π[a^2 - y^2 - b^2] dy $$

With this integral, just integrate it like you normally would and you should get the following equation:

$$ V = (4/3)(π)(a^2 - b^2)^{3/2} $$

And if you think about it, if there was no cylinder to begin with (b = 0), you get the equation for the volume of a sphere!

Please let me know if anything needs clarifying!