- #1

VinnyCee

- 489

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__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?