revolutions: cylindrical hole drilled into sphere

[Clifford]

Show that if you drill a cylindrical hole of radius b vertically through the centre of a sphere of radius a, where 0 < b < a, then the reamining volume of the sphere is V = pi/6 * h^3, where hh=2*sqrt(a^2-b^2)
Using the washer method

radius sphere: x = sqrt(a^2-y^2)

Volume = Volume sphere - Volume Cylinder
V = pi * integral from -a to a (a^2-y^2 dy) - pi * integral from -a to a (dy)
V = pi (a^2*y - 1/3*y^2 - b^2*y) evaluated from -a to a
After some simplifcation I get
2pia(a^2-a/3-b^2)

This isn't anywhere near what I am suppose to get which is given in the question

 

[skeeter]

Re: revolutions

\(\displaystyle V = 2\pi \int_0^{\frac{h}{2}} (a^2 - y^2) - b^2 \, dy\)

 

[Clifford]

Re: revolutions

It works out now, but why is there a 2pi infront instead of just pi?

 

[skeeter]

pi times the integral represents the volume of that part in quad I rotated about the y-axis (note the limits of integration) ... to get the entire volume requires doubling.