# Rewriting triple integrals rectangular, cylindrical, and spherical coordinates

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:

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

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

• Here: math.stackexchange.com/questions/2836487/…. Here: math.stackexchange.com/questions/3396108/…. Here: math.stackexchange.com/questions/2291054/…. You may only need the last link.Another:math.stackexchange.com/questions/1606630/… Commented Dec 14, 2021 at 8:48
• How exactly do you go about writing these integrals? you need to study about it somewhere, practice simple questions and build on it. One place to get started - tutorial.math.lamar.edu/classes/calciii/tripleintegrals.aspx Commented Dec 14, 2021 at 8:52
• @MathLover Apologies, I needed to elaborate a little further. I wasn't sure if I started it off right but this is my progress so far. Commented Dec 14, 2021 at 9:28
• @Ajay Thanks again Ajay, the links helped! :) Commented Dec 14, 2021 at 9:31
• Yes your edit helps and in fact it is correct, except that you need to consider only first octant so the upper bound of $\theta$ should be $\pi/2$. Before your edit, the way your question read seemed like you were asking guidance on where to start with such integrals. Commented Dec 14, 2021 at 10:22

Yes your work is correct except the bounds of $$\theta$$. Please note that the region is in the first octant so $$0 \leq \theta \leq \pi/2$$.
$$x^2 + y^2 = 9 - z^2 = 9 - x^2 - y^2 \implies x^2 + y^2 = 9/2$$
$$\displaystyle \int_0^{3/\sqrt2} \int_0^{\sqrt{9/2-x^2}} \int_{\sqrt{x^2+y^2}}^{\sqrt{9-x^2-y^2}} (x^2 + y^2) ~ dz ~ dy ~ dx$$
• I see. So the original $\theta$ bound would’ve accounted for the entire graph, right? Commented Dec 14, 2021 at 10:25