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Write three integrals, one in Cartesian/rectangular, one in cylindrical, and one in spherical coordinates, that calculate the average of the function $f(x, y, z) = x^2 + y^2$ on the region $E$ in the first octant inside the sphere $x^2+y^2+z^2 = 9$, and above the cone $z=\sqrt{x^2+y^2}$.

The volume of $E$ is provided, $E = \frac{9\pi}{4}(2-\sqrt{2})$. Only the setup is needed, the integrals do not need to be evaluated.

I have the spherical and cylindrical integrals but I'm not quite sure of my bounds:

$\frac{1}{Vol(E)}\int_{0}^{2\pi}\int_{0}^{\frac{\pi}{4}}\int_{0}^{3}\rho^4 \sin^3(\phi) d\rho d\phi d\theta$

$\frac{1}{Vol(E)}\int_{0}^{2\pi}\int_{0}^{\frac{3}{\sqrt{2}}}\int_{r}^{\sqrt{9-r^2}}r^3dz dr d\theta$

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Yes your work is correct except the bounds of $\theta$. Please note that the region is in the first octant so $0 \leq \theta \leq \pi/2$.

In cartesian coordinates, note that at the intersection of the sphere and the cone,

$x^2 + y^2 = 9 - z^2 = 9 - x^2 - y^2 \implies x^2 + y^2 = 9/2$

$ \displaystyle \int_0^{3/\sqrt2} \int_0^{\sqrt{9/2-x^2}} \int_{\sqrt{x^2+y^2}}^{\sqrt{9-x^2-y^2}} (x^2 + y^2) ~ dz ~ dy ~ dx$

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  • $\begingroup$ I see. So the original $\theta$ bound would’ve accounted for the entire graph, right? $\endgroup$ Commented Dec 14, 2021 at 10:25
  • $\begingroup$ Yes that is correct. It would have accounted for the entire volume bound above the cone and inside the sphere. $\endgroup$
    – Math Lover
    Commented Dec 14, 2021 at 10:27
  • $\begingroup$ Got it, thank you. $\endgroup$ Commented Dec 14, 2021 at 10:28

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