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Is it a sphere?

  1. Jul 29, 2008 #1
    1. The problem statement, all variables and given/known data
    Find the volume V of the solid bounded by the graph
    x2+y2=9 and

    2. Relevant equations

    3. The attempt at a solution
    When I started this problem, I thought it was a perfect sphere with the center points (0, 0, 0). And then I thought, "Why do I need calculus, it's 4/3*pi*r3, right?" So I plugged it into the calculator and got 36pi. Then I thought... maybe that's not such a good idea.

    Then I thought about doing the shell method, and multiplying it by two. But this is all under the assumption that I'm working with a sphere.

    The hint I was given is to do it in the first octant, then multiply by 8. I thought about integrating the areas I was given (xy, yz), and then cross them to get xz, but after that, I'm lost. Do I multiply all of them to get the area, and then multiply by 8?

    I am currently in the chapter dealing with multiple integrals and polar integration, if that is any help.
  2. jcsd
  3. Jul 29, 2008 #2


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    Well you have the equations of the solid. Note that in both cases, one variable is missing. That means that it can take on any value and the solid formed would be parallel to the axis of that particular free variable. Just sketch the two solids and see where they overlap. That is the required volume. Just to note, it's not a sphere. Now what have you learnt about finding the volume of a given region by multiple integrals?
  4. Jul 29, 2008 #3
    A rough pencil sketch gets me cube-like shape that's getting rounder.

    The volume is a triple integral. V=[tex]\int[/tex][tex]\int_D[/tex][tex]\int dV[/tex].

    I'm having trouble determining the limits, or setting up the problem for that matter.
    V = [tex]\int_0^3[/tex][tex]\int_0^{\sqrt{9-x^2}}[/tex][tex]\int_0^{\sqrt{9-y^2}}dz dy dx?[/tex]
    Last edited by a moderator: Jul 29, 2008
  5. Jul 29, 2008 #4


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    Those, separately, are cylinders. The region you are asking about is the intersection of two cylinders. It definitely is not a sphere.
  6. Jul 29, 2008 #5


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    This is the volume of one octant of your shape.
  7. Jul 29, 2008 #6
    I have this same problem on my work. I got those limits and ended up with some crazy integral by the time i got to dx that I couldn't make sense of it anymore.
  8. Jul 29, 2008 #7


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    You might want to evaluate it in cylindrical coordinates instead.
  9. Jul 29, 2008 #8


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    Fix a value of y. Then x^2=9-y^2 and z^2=9-y^2. So both x and z range from -sqrt(9-y^2) to +sqrt(9-y^2). I.e. for fixed y the x-z cross-section is a SQUARE. That's the easy way to do it. And explains your "cube-like shape that's getting rounder".
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