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Homework Help: calculus *improper integrals*

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

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?

Thank you.

• calculus *improper integrals* - drwls, Monday, September 17, 2012 at 2:44am

  The 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 7:51am

  yes , the hole goes through the sphere, centered along the diameter
  

• calculus *improper integrals* - Steve, Monday, September 17, 2012 at 10:44am

  So 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 10:54am

  I see how it makes sense now, thanks a lot Steve!
  

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