Triple integral using cylindrical coordinates

[Draconifors]

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 $\rho$).

 

[Orodruin]

at which point I confused myself by the presence of the two dθdθ d\theta .

You cannot express $r$ as a function of $\theta$. You must choose a name for your substitution variable that does not already exist or you will just end up with nonsense.