Complicated integral using cylinder coordinates So I'm taking a calculus course and have gotten a homework problem containing this complicated triple integral:
$$I = \int_{-2}^{0} \int_{2y^2}^{-4y} \int_{-y}^{\sqrt{z-y^2}} \frac{1}{x^2+y^2} dxdzdy$$
I'm supposed to solve this using cylindrical coordinates, so using the transformation $T(r,\theta,z) = (r\cos\theta,r\sin\theta,z)$ with Jacobian equal to $r$.
The problem is that I do not know what the integration boundaries will be.
I've tried plotting the boundary functions in 3d using GeoGebra, and have acquired some visualisation for the volume I have to integrate over.
This led me to the integral
$$I = \int_{-\frac{\pi}{4}}^0 \int_{0}^{2\sqrt{2}} \int_{2r^2\sin^2\theta}^{-4r\sin\theta} \frac{1}{r} dzdrd\theta = 8\sqrt{2} - 6 - \pi$$
but I have very big doubts about the correctness of this conversion.
I would evaluate the original integral normally, but I arrive at the integral of some arctangent with some roots inside, which even Symbolab does not evaluate further, and it seems that Wolfram Alpha has trouble interpreting my inputs for an analytical solution.
Could somebody please help me? Thank you in advance.
 A: We have to figure out what the boundaries were in Cartesian in the first place. The innermost bounds will always be the most helpful because they give the direct equations for some of the surfaces involved. In this case we have
$$x^2+y^2 = z \hspace{24 pt} x \geq 0$$
$$x+y=0$$
And since $y \leq 0$ we have that $x \geq 0$ for the bottom surface as well. Moving on to the next bounds, $z=2y^2$ is simply the projection of the intersection of the above plane and paraboloid on to the $yz$ plane ($x = -y \implies z = 2y^2$). This means that $z=-4y$ is a completely independent surface boundary from the previous ones. 
Lastly as a sanity check, $-4y = 2y^2 \implies y = 0,2$, so we do not have to cut up the region any further. This gives us enough information to figure out the bounds in polar coordinates. 
$x+y=0$ is a straight vertical plane, which means it represents a $\theta = k$ surface. Given the context of the quadrant in the $xy$ plane we are in (quadrant IV), we get that $\theta = -\frac{\pi}{4},0$. The other surfaces are standard cylindrical coordinates transitions. 
Setting up the integral with $z$ first we have that
$$I = \int_{-\frac{\pi}{4}}^0 \int_0^{-4\sin\theta} \int_{r^2}^{-4r\sin\theta} \frac{1}{r}\:dz\;dr\:d\theta = \pi-2$$
which is almost the same as the expression you originally obtained.
A: So thanks to @Ninad Munshi, I understand this problem now.
I have tried to explain the bounds they found to myself, and figured out my own explanation:
The volume in question is a sort of banana shape.
When all the relevant surfaces are projected onto the $xy$-plane, we get this picture:
$xy$-plane">
The black lines are self explanatory.
The blue circle is the level curve of the paraboloid at $z=8$, which is the height of the intersection point of the paraboloid and the planes.
The red circle is the intersection of the paraboloid and the $z=-4y$ plane.
It is clear that $\theta$ ranges from $-\frac{\pi}{4}$ to $0$.
$r$ then goes from $0$ to the red circle, which has equation $x^2+(y+2)^2=4$ (simply found by setting $x^2+y^2=z=-4y$).
In polar coordinates, simply fill in $x=r\cos\theta$ and $y=r\sin\theta$ to find $r=-4\sin\theta$ as expected.
As for $z$, the lower bound is the paraboloid, which translates to $z=r^2$, and the plane $z=-4y$ becomes $z=-4r\sin\theta$, which yields 
$$ \int_{-\frac{\pi}{4}}^0 \int_0^{-4\sin\theta} \int_{r^2}^{-4r\sin\theta} \frac{1}{r} dzdrd\theta = \pi - 2, $$
just as @Ninad Munshi described.
