calculus *improper integrals*
posted by John on .
A hole of a radius of 1cm is pierced in a sphere of a 4cm radius. Calculate the volume of the remaining sphere.
I know that the volume of a sphere with an exterior radius is of:
V = pie(R^2 - r^2) dH
where R^2 is the exterior surface area and r^2 is the interior surface area. dH is the thickness.
for R and r, do I simply just plug in the values that are given to me?
The answer depends upon where the hole is drilled. Is it centered along a diameter of the sphere?
yes , the hole goes through the sphere, centered along the diameter
So far so good.
h^2 = R^2 - x^2
2h dh = -2x dx
dh = -x/h dx
so the integral becomes
pi (R^2-r^2) (-x/sqrt(R^2 - x^2)) dx
= -pi (R^2-r^2) x/sqrt(R^2-x^2) dx
now let u = R^2 - x^2
du = -2x dx
and the integral is
pi/2 (R^2-r^2) u^(-1/2) du
If I've done things right,
v = 4/3 pi (R^2-r^2)^(3/2)
makes sense, since if r=0, you have the whole sphere, and if r=R, there's nothing.
I see how it makes sense now, thanks a lot Steve!