Triple Integral bounded by a cylinder and a plane Solve $\iiint{z} dV$
The region is defined as E bounded by $y^2 + z^2 = 4$ and the planes $x = 0$, $y = x$, and $z =0$ in the first octant. So it is a cylinder with radius 2 that goes along the yz plane and is intersected by the plane $y = x$. 
Since it is a cylinder, I should use cylindrical coordinates. 
For $\theta$:
$ rcos(\theta)=rsin(\theta)$ => $tan(\theta) = 1$ => $\theta = \pi/4$
so $\pi/2<\theta<\pi/4$
For z:
$x^2 + y^2 + z^2 = 4$ => $z = \sqrt{4-r^2}$
so $0<z<\sqrt{4-r^2}$
I am not entirely sure if my r bound is correct, but I said that it should go from $0<r<2$ since the cylinder has a radius of 2.
Anyway, since the z bound is kinda gross, I just plugged it into symbolab and I found out that my answer was wrong. Also this was a practice exam problem, and I obviously cannot use symbolab on my practice exam, so I was wondering if there is a better way to go about doing this problem. Although I am not sure how spherical would work on a cylindrical problem, and the cartesian coordinates are just as nasty.
So TL;DR: Are my bounds correct? Can I do this problem in another, easier way?
 A: I would recommend using rectangular/Cartesian coordinates here. Cylindrical coordinates are going to be a bit of a hassle when we get to working with finding the bounds for $r$ and $\theta$ on the $yz$-plane.
So for this problem, we know that since we're in the first octant, we know that each of $x,y,z$ are nonnegative. You have the right idea of establishing bounds for $z$ here, but remember, the equation that you're given is simply $y^2+z^2=4$, not $x^2+y^2+z^2=4$, so you'll want to use $\sqrt{4-y^2}$ as your upper bound. (so that $0\le z\le \sqrt{4-y^2}$)
For your other two bounds, you'll want to project your given planes as well as your bounds for $z$ onto the $xy$-plane (where $z=0$). From there, since you'll want to have that $\sqrt{4-y^2}$ is nonnegative (or, if you prefer, $y$ must satisfy the equation $y^2+z^2=4$), we have that $-2\le y\le 2$, and since we're working in the first octant, this means we can further restrict our bounds for $y$ to $0\le y\le 2$. With this and your bounds for $x$ as given by the problem statement (so that $0\le x\le y$), we have that we can obtain the following integral:
$$\int^2_0\int^y_0\int^{\sqrt{4-y^2}}_0z\text{ }dz\text{ }dx\text{ }dy$$
which, after going through the Calculus, should evaluate to 2.
A: You can do it in cylindrical coordinates.
The integral is
$$\int_{\theta =\frac\pi4}^\frac{\pi}2 \iint_{r^2\sin^2\theta + z^2=4 \\r,z\ge 0} zr\,dr\,dz\,d\theta$$
Notice that the inner integral is over the elliptical sector $r^2 + \left(\frac{z}{\sin\theta}\right)^2\le \left(\frac{2}{\sin\theta}\right)^2$ in the first quadrant. Hence we can introduce polar coordinates $(\rho, \phi)$ with $$\begin{cases} r = \rho\cos\phi \\ z = \rho\sin\phi\sin\theta\end{cases}$$
and bounds $0 \le \rho \le \frac{2}{\sin\theta}$ and $\phi \in \left[0, \frac{\pi}2\right]$. The Jacobian of this transformation is $\rho\sin\theta$ so the integral is equal to
$$\int_{\theta =\frac\pi4}^\frac{\pi}2 \int_{\phi=0}^{\frac\pi2}\int_0^{\frac2{\sin\theta}} \rho\sin\phi\sin\theta \cdot \rho\cos\theta \cdot \rho\sin\theta\,d\rho\,d\phi\,d\theta = 2$$
A: This should help in setting your limits:

