# Triple integral using cylindrical coordinates

## Homework Statement

The first part of the question was to describe E the region within the sphere ##x^2 + y^2 + z^2 = 16## and above the paraboloid ##z=\frac{1}{6} (x^2+y^2)## using the three different coordinate systems.

For cartesian, I found ##4* \int_{0}^{\sqrt{12}} \int_{0}^{12-x^2} \int_{\frac{1}{6} (x^2 + y^2)}^{\sqrt{16-x^2-y^2}} 1 dz dy dx## by splitting the domain into quarters because I found it easier.
For cylindrical I got ##\int_{0}^{2\pi} \int_{0}^{\sqrt{12}} \int_{\frac{r^2}{6}}^{\sqrt{16-r^2}} r dz dr d\theta##.
Spherical gave me ##\int_{0}^{2\pi} \int_{0}^{\frac{\pi}{3}} \int_{0}^{4} \rho^2 sin(\phi) d\rho d\phi d\theta + \int_{0}^{2\pi} \int_{\frac{\pi}{3}}^{\frac{\pi}{2}} \int_{0}^{\frac{6 cos(\phi)}{sin^2(\phi)}} \rho^2 sin(\phi) d\rho d\phi d\theta## by splitting the domain into two with respect to ##\rho##'s boundaries.
(I'm including these mostly as a reference point, without my calculations, but I can go through my calculations if necessary.)

The second part of the question -- the one I'm having trouble with -- is then computing ##\iiint \limits_E (x^2+y^2)^2 dV## in my coordinate system of choice.

## Homework Equations

##\cos^2\theta = 1 - \sin^2\theta ##

## The Attempt at a Solution

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So to compute ##\iiint \limits_E (x^2+y^2)^2 dV##, I chose cylindrical because it reduced the inside of the integral to ##r^5##. However, I then get stuck after I integrate with respect to z, because I get $$\int_{0}^{2\pi} \int_{0}^{\sqrt{12}} \left(r^5 * \sqrt{16-r^2} - \frac{r^7}{6} \right) dr d\theta$$.

I first tried u-sub, which doesn't work on the equation as-is. I also attempted trig sub, but I don't know if it's simply because I'm out of practice or whether it's not the system to use, but I can't finish my calculations because of the complexity of the equations I remain with.

I used trig sub with ##r = 4\sin\theta## , ##dr = 4\cos\theta = \sqrt{16-r^2}##.
This gave me $$\int_{0}^{2\pi} \int_{0}^{\sqrt{12}} \left[ (4\sin\theta)^5 * \sqrt{16-16\sin^2\theta} \right) d\theta dr d\theta$$ at which point I confused myself by the presence of the two ## d\theta ##.
Even simply calculating the indefinite integral of the above, I ended up with ##-16384 \left[\frac{u^3}{3} - \frac{2u^5}{5} + \frac{u^7}{7} \right] ##, after having used a u-sub with ##u = \cos\theta## and ##du = -\sin\theta d\theta##. Considering that ##u = \sqrt{1 - (\frac{r}{4})^2}##, I don't see how I'll be able to then integrate with respect to ##\theta##.

I can put up all my calculations if necessary, but what I really want is a technique or strategy hint so that I can solve this, because, from where I'm standing, the boundaries for the three coordinate systems are quite hard to compute with respect to the first integral (i.e. z and ##\rho##).