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March 1, 2015

Posted by **John** on Sunday, September 16, 2012 at 11:57pm.

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?

Thank you.

- calculus *improper integrals* -
**drwls**, Monday, September 17, 2012 at 2:44amThe answer depends upon where the hole is drilled. Is it centered along a diameter of the sphere?

- calculus *improper integrals* -
**John**, Monday, September 17, 2012 at 7:51amyes , the hole goes through the sphere, centered along the diameter

- calculus *improper integrals* -
**Steve**, Monday, September 17, 2012 at 10:44amSo far so good.

h^2 = R^2 - x^2

2h dh = -2x dx

or,

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.

- calculus *improper integrals* -
**John**, Monday, September 17, 2012 at 10:54amI see how it makes sense now, thanks a lot Steve!

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