# Cylindrical hole with radius r is drilled symmetrically through the center of a sphere with radius R.?

Cylindrical hole with radius r is drilled symmetrically through the center of a sphere with radius R. Where r ≤ R. What is the volume of the remaining material?

### 3 Answers

- DambarudharLv 78 years agoBest Answer
Radus of the Sphere = R

Volume of the Sphere = V = (4/3) π R³

Radius of the cylinder = r

Height of the cylinder = h

Draw the figure. From the figure it is observed = R² - r² = (h/2)²

=> h² = 4(R² - r²)

=> h = 2√(R² - r²)

Volume of the cylindrical hole = v = π r² { 2√(R² - r²) }

Volume of the remaning material = V - v = π{ (4/3) R³ - 2 r²√(R² - r²)}

= 2π { (2/3)R³ - r²√(R² - r²) }

- Anonymous8 years ago
You can use the 'Washer Method' to find the volume.

http://mathworld.wolfram.com/images/eps-gif/Spheri...

(Just assume the hole goes all the way through)

(Also suppose the axes are x and y, and z is out of the screen)

It's a bit difficult to explain, but you can rotate the 'area' outside the cylinder and inside the sphere, about the y axis to get the answer.

Finding this area in terms of y:

It's the shape of a washer, outer radius (let it be r') varying with y (height), and inner radius constant as r.

The outer radius varies with y as r' = √(R² - y²)

So, the area of the 'washer' would be π ( r'² - r² ) = π (R² - r² - y²)

Integrating this with respect to y:

y varies from -√(R² - r²) to +√(R² - r²)

of:

∫ π (R² - r² - y²) dy

Integrating and substituting limits, we get

Volume = (4/3) π (R² - r²) (√(R² - r²) )

Source(s): Google images/Wolfram for the image. - 8 years ago
you can easily derive the formula the formula of the remaining material.

V = volume of the spherical segment - volume of the cylinder

the formula is,

V = 8pi/6[(R^2 - r^2)^(3/2)]

note:

x = multiplication sign

sqrt = square root