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Triple integral in cylindrical/spherical

  1. Apr 29, 2010 #1
    1. The problem statement, all variables and given/known data
    Use cylindrical coordinates to find the volume of the solid.
    The solid is enclosed by the paraboloid z=x2+y2 and the plane z=9

    2. Relevant equations

    3. The attempt at a solution
    So I'm getting close to the answer but not quite, and I keep getting a negative which doesn't make sense. And I think my limit on the second integration needs to be a function of theta.

    I chose z=9 as my z upper limit and z=r2 as my lower and just used rdzdrd(theta) as my integrand. Used 0 for lower limit for both dr and dtheta and started to use sqrt(9/2) as my upper limits for both dr and d(theta) but then changed d(theta's) upper limit to 2pi.

    I need help lol.
    BTW.. My notation on this is sloppy. I originally started with the z limits reversed but changed it once I integrated.. sorry.
  2. jcsd
  3. Apr 29, 2010 #2
    I believe you have the limits for z reversed, and the limits for r should be 0 and 3.
  4. Apr 29, 2010 #3
    I thought I would mention that there is a simple way to solve this problem by considering the total volume of the cylinder with radius R=3 and height H=9 and then subtracting off the excess volume. The excess volume can be expressed as an area spun around the z axis, and the result is very simple as follows.

    [tex] V=2\pi R^2 H - 2 \pi \int_0^R f(x) x dx =81\pi-2\pi \int_0^3 x^3 dx ={{81 \pi}\over{2} [/tex]

    Little tricks like this can save time, or provide a way to check your answer when you are required to demonstrate your ability to do the triple integral.
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