A ball of radius 11 has a round hole of radius 5 drilled through its center. Find the volume of the resulting solid. Find it using integrals.
2 Answers  Add Yours
The ball is a sphere.
Volume of a sphere: 4/3 * `pi` * r^3
The radius of the ball is 11.
4/3 * `pi` * 11^3 = 5324/3 * `pi`
The volume of the ball is 5324/3 * ``
The hole drilled through the ball is a cylinder.
Volume of a cylinder: `pi` * r^2 * h
The radius of the hole is 5. The height of the cylinder is the same as the diameter of the ball, so the height is 22.
`pi` * 5^2 * 22 = 550 * `pi`
To find the volume of the resulting solid, subtract the volume of the cylinder from the volume of the sphere.
(5324/3 * `pi` )  (550 * `pi` ) = 3674/3 `pi`
`~~` 1224.67`pi`
`~~` 3847.4
I am sorry I am not allowed to insert graphics so it will be difficult for me to explain you easily, but please try to follow.
What volume of the sphere of radius R = 11 unit, we have lost by this drilling ?
Answer:
1) 2 slices of the sphere of base radius r = 5 unit, one from each side of the drilling.
2) A cillynder of length L = 2*Sqrt [R^2  r^2]
We proceed part by part:
Part 1) Volume of the slices:
Take a disc perpendicular to the cillyndrical drill, of thickness dz and radius x ( where x^2 = R^2  z^2 ), and sum it from the top of the cillynder up to the full radius R, so we need to integrate from z= R  L/2 to z = R
So volume of one slice = Integral[ Pi*x^2 dz] from (R  L/2) to R
= Integral[ Pi*{R^2  z^2} dz] from (R  L/2) to R
= Pi*[R^2 * z  (z^3)/3 ] from (R  L/2) to R
= Pi * L^2 * (R  L/6) /4 [after simplification]
= Pi * 4(R^2  r^2) * [R  2 Sqrt{R^2  r^2}/6]/4
Volume of cyllinder = Pi * r^2 * L
= Pi * r^2 * 2 Sqrt[R^2  r^2]
So volume lost by drilling = 2 * slice volume + cyllinder volume
= Pi * 4(R^2  r^2) *
[R  2 Sqrt{R^2  r^2}/6]/2
+
Pi * r^2 * 2 Sqrt[R^2  r^2]
And finally
Volume of the drilled sphere = volume of original sphere
 volume lost by drilling
= (4/3) Pi * R^3

Pi * 4(R^2  r^2) *
[R  2 Sqrt{R^2  r^2}/6]/2
 Pi * r^2 * 2 Sqrt[R^2  r^2]
= (Pi*R^3)/3 + Pi * R * r^2
+ Pi*(R^2  r^2)^{3/2}
 2 Pi*r^2 * Sqrt[R^2  r^2]
Putting r = 5 and R = 11, we get:
Required volume = 3673.69 cubic unit [approximately]