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Homework Help: Tricky Calculus III Problem

  1. Sep 8, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the volume of the solid that lies in between both of the spheres:


    2. Relevant equations
    This is the first chapter of the calculus III material so no double or triple integrals are needed to solve this problem.

    3. The attempt at a solution

    I completed the square on the first equation and obtained:


    So I just have 2 intersecting spheres. Finding the volume of a "slice" of a sphere can be done using a solid of revolution, however, I would need know how far one sphere is intersecting into the other in order to calculate the volume.

    My idea is to minimize the distance between (x+2)2+(y-2)2+(z+2)2=4 and the point (0,0,0). Then I will be able to figure out how "thick" the 2 "slices" of the spheres are.

    However, I'm not quite sure how to do this. I know how to find the distance between points in space but solving for each variable I will wind up with 2 solutions because of the radical. Is there an easier way to do this that I am missing?

    Last edited: Sep 8, 2011
  2. jcsd
  3. Sep 8, 2011 #2
    Well, the easiest way to solve this is by calculating a double integral with spherical coordinates, but you said that this is not needed.

    Your method is also nice, but a little more work. Let's see if we can do it right.

    We need to find the shortest distance between (0,0,0) and the equation [itex](x+2)^2+(y-2)^2+(z+2)^2=4[/itex].
    We can do that by calculus by minimizing a certain function, but that might get tedious. Here's another method: the line which has the shortest distance connect (0,0,0) with (-2,2,-2) (the center of the circle). So you only need to set up the equation of the line and see where it intersects the sphere.
  4. Sep 8, 2011 #3
    Genius! I love it. Can't believe I didn't think of that.

    so for the line I I created the equation x=-2t,y=t,z=-2t

    Next I plugged the equation into the sphere with it's center at (-2,1,-2) obtaining...


    The first t value is the one that will help me so I plugged it back into my equation for a line, obtaining[tex](\frac{-2}{3},\frac{1}{3},\frac{-2}{3})[/tex]

    The distance from (0,0,0) to this point turns out conveniently be 1. So one half of the solid section has a distance of 1 from the edge of the sphere.

    Now I integrate to find the volume of the whole solid.

    [tex]2\pi\int^{2}_{1}4-y^2 dy = 2\pi[4y-\frac{y^3}{3}]^{2}_{1}=\frac{10\pi}{3}[/tex]

    Does this look correct?
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