Volume bounded above by cylinder z = 4 - x^2, x^2 + y^2 = 4 on sides,xy-plane bottom

  1. Here is the problem:

    Find the volume of the solid that is bounded above by the cylinder [tex]z = 4 - x^2[/tex], on the sides by the cylinder [tex]x^2 + y^2 = 4[/tex], and below by the xy-plane.

    Here is what I have:

    [tex]\int_{-2}^{2}\int_{-\sqrt{4 - x^2}}^{\sqrt{4 - x^2}}\int_{0}^{4 - x^2}\;dz\;dy\;dx\;=\;12\pi[/tex]

    Is that correct? I didn't post the many steps for integration, but the integral calulation is correct, I just need to know if I set up the integral right. Thanks again :rofl:
  2. jcsd
  3. HallsofIvy

    HallsofIvy 41,264
    Staff Emeritus
    Science Advisor

    Yes, that's correct. Of course, it would be easier to do the integration in cylindrial coordinates.
  4. What if f(x, y, z) = sqrt(xyz), how to find average value?

    Thanks for the double checking! The next problem uses this same integral and assumes that [tex]f\left(x, y, z\right) = \sqrt{x\;y\;z}[/tex]. Then it says to setup the integral to find the average value of the function within that solid.

    Here is what I have:

    [tex]\frac{1}{12\pi}\;\int_{-2}^{2}\int_{-\sqrt{4 - x^2}}^{\sqrt{4 - x^2}}\int_{0}^{4 - x^2}\;\sqrt{x\;y\;z}\;dz\;dy\;dx[/tex]

    Does that look right?
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