VinnyCee
- 486
- 0
Here is the problem:
First Part (already done): Find the volume of the solid that is bounded above by the cylinder z = 4 - x^2, on the sides by the cylinder x^2 + y^2 = 4, and below by the xy-plane.
Answer: \int_{-2}^{2}\int_{-\sqrt{4 - x^2}}^{\sqrt{4 - x^2}}\int_{0}^{4 - x^2}\;dz\;dy\;dx\;=\;12\pi
Using the integral worked out above, and assuming that f\left(x, y, z\right) = \sqrt{x\;y\;z}. Setup the integral to find the average value of the function within that solid.
Here is what I have:
\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
Does that look right?
First Part (already done): Find the volume of the solid that is bounded above by the cylinder z = 4 - x^2, on the sides by the cylinder x^2 + y^2 = 4, and below by the xy-plane.
Answer: \int_{-2}^{2}\int_{-\sqrt{4 - x^2}}^{\sqrt{4 - x^2}}\int_{0}^{4 - x^2}\;dz\;dy\;dx\;=\;12\pi
Using the integral worked out above, and assuming that f\left(x, y, z\right) = \sqrt{x\;y\;z}. Setup the integral to find the average value of the function within that solid.
Here is what I have:
\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
Does that look right?
