Triple Integral over cylindrical volume I am trying to calculate the triple integral shown below. I've also computed the rectangular limits for x, y and z. I've attempted to compute the integral while staying in rectangular, and it ended up being several pages of wild integrals that just kept getting larger and larger. I tried to set this up to do as a cylinder, but I am getting lost. I feel like this problem should be easier, but I just can't wrap my head around it:
Problem statement and my calculation of the limits here.
In the above image, I've set up the integral correctly for rectangular (I think?) but actually computing it has proven to be overly labor intensive. Can someone please help me set up this integral in a better way? Thank you in advance!
P.s. I apologize for not using MathJax; I wanted to be able to write some nicely formatted math, but the learning curve looks a little high for me right now.
 A: METHOD 1:  Cartesian Coordinates
We can exploit symmetry to evaluate this integral.  First, we note that we can write the integral of interest $I$ as
$$I=\int_{-2}^2\int_{-2}^2 \int_{-\sqrt{4-y^2}}^\sqrt{4-y^2} \left(4+5x^2yz^2\right)\,dx\,dy\,dz$$
We proceed to evaluate the integral as
$$\begin{align}
I&=\int_{-2}^2\int_{-2}^2  \left(8\sqrt{4-y^2}+\frac{10}3 (4-y^2)^{3/2}yz^2\right)\,dy\,dz \tag 1\\\\
&=64\int_0^2\sqrt{4-y^2}\,dy \\\\
&+\frac{160}{9}\int_0^2y(4-y^2)^{3/2}\,dy \tag 2\\\\
&=64\int_0^2\sqrt{4-y^2}\,dy \tag 3\\\\
&=64\left.\left(\frac{y\sqrt{4-y^2}}{2}+2\arctan\left(\frac{y}{\sqrt{4-y^2}}\right)\right)\right|_{0}^{2} \tag 4\\\\
&=64\pi \tag 4
\end{align}$$
In arriving at $(1)$, we carried out the integration over the inner integral (i.e., $x$).
In going from $(1)$ to $(2)$, we carried out the integral over $z$.
In going from $(2)$ to $(3)$ we exploited the fact that the integrand of the second integral in $(2)$ was an odd function of $y$ and the limits of integration were symmetric around $y=0$.  Therefore, this integral is zero.
In arriving at $(4)$, we used standard trigonometric substitution $y\to 2\sin \theta$ to derive the anti-derivative of $\sqrt{4-y^2}$.
And finally, we used the fact that $\lim_{x\to 0^+}\arctan (1/x)=\pi/2$ to obtain $(5)$.

METHOD 2:  Cylindrical Coordinates
We can express $I$ in terms of cylindrical coordinates as
$$I=\int_{-2}^2\int_0^{2\pi}\int_0^2\left(4+5\rho^3\cos^2\phi \sin \phi z^2\right)\,\rho \,d\rho\,d\phi\,dz$$
We see immediately that the integral of the second term of the integrand vanishes since $\int_0^{2\pi}\cos^2\phi \sin \phi \,d\phi=0$.  Therefore, we have
$$I=\int_{-2}^2\int_0^{2\pi}\int_0^2\,4\,\rho \,d\rho\,d\phi\,dz=64\pi$$
as expected!
