# 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.

samhouston | Middle School Teacher | (Level 1) Associate Educator

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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

quantatanu | Student, Undergraduate | (Level 1) Valedictorian

Posted on

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 ?

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]