(adsbygoogle = window.adsbygoogle || []).push({}); Here is the problem:

First Part (already done): 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.

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

Using the integral worked out above, and assuming that [tex]f\left(x, y, z\right) = \sqrt{x\;y\;z}[/tex]. 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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# Homework Help: Average value of f(x,y,z) = sqrt(xyz) over a solid z=4-x^2 & x^2+y^2=4 & xy-plane

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