How to set bounds in cylindrical coordinates? I'm trying to evaluate the following integral in cylindrical coordinates
$$\int_0^6 \int_0^{\frac{\sqrt{2}}{2}}\int_x^{\sqrt{1-x^2}}e^{-x^2-y^2} \, dy \, dx \, dz$$
After attempting to set the bounds in cylindrical coordinates, I got
$$\int_0^6 \int_0^{\frac{\sqrt{2}}{2}}\int_{\rho \cos\varphi }^{\sqrt{1-\rho^2 \cos^2\varphi }}e^{-\rho ^2}d\varphi \rho \, d\rho \, dz$$
But I know this doesn't make sense. Can someone explain how to switch the bounds analytically? I don't understand how to transform the bounds.
 A: Analytically and without drawing: in short, solving inequalities. You have set
$$\left\lbrace \begin{array}{a}
0 \leqslant y \leqslant 6 \\
0 \leqslant x \leqslant \frac{1}{\sqrt{2}} \\
x \leqslant z \leqslant \sqrt{1-x^2}
\end{array} \right\rbrace$$
taking cylindrical coordinates
$$\begin{cases}x=\rho \cos \phi \\
y= \rho \sin \phi \\ 
z=z'\end{cases}$$
we obtain restrictions for Oxy
$$\left\lbrace \begin{array}{a}
0 \leqslant \rho \sin \phi \leqslant 6 \\
0 \leqslant \rho \cos \phi \leqslant \frac{1}{\sqrt{2}} 
\end{array} \right\rbrace$$
Solving this inequalities we obtain $0 \leqslant \rho \leqslant \min \left\lbrace \frac{1}{\sqrt{2}\cos \phi }, \frac{6}{\sin \phi} \right\rbrace$. From here we obtain angle $\tan\phi_1 = 6\sqrt{2}$. So integral will be
$$\int\limits_{0}^{\phi_1}\int\limits_{0}^{\frac{1}{\sqrt{2}\cos \phi }}\int\limits_{\rho\cos \phi}^{\sqrt{1-(\rho \cos \phi)^2}}+\int\limits_{\phi_1}^{\frac{\pi}{2}}\int\limits_{0}^{\frac{6}{\sin \phi }}\int\limits_{\rho\cos \phi}^{\sqrt{1-(\rho \cos \phi)^2}}$$
Addition:
As pointed in comments below I proceed from that sequence of limits in OP integral matches integration variables. Taking reverse direction we obtain set
$$\left\lbrace \begin{array}{a}
0 \leqslant z \leqslant 6 \\
0 \leqslant x \leqslant \frac{1}{\sqrt{2}} \\
x \leqslant y \leqslant \sqrt{1-x^2}
\end{array} \right\rbrace$$
now we have restrictions
$$\left\lbrace \begin{array}{a}
0 \leqslant \rho \cos \phi \leqslant \frac{1}{\sqrt{2}} \\
\rho \cos \phi \leqslant \rho \sin \phi \leqslant \sqrt{1-(\rho \cos \phi)^2} \\
\end{array} \right\rbrace$$
first inequality in second line gives restriction $\tan \phi \geqslant 1$, which together with $\cos \phi \geqslant 0, \sin \phi \geqslant 0$ from first line gives $\phi \in \left[\frac{\pi}{4},\frac{\pi}{2}\right]$. At last second inequality from second line gives $\rho \leqslant 1$, so integral have limits
$$\int\limits_{0}^{6}\int\limits_{\frac{\pi}{4}}^{\frac{\pi}{2}}\int\limits_{0}^{1}$$
At end let me say, that although drawing is most often  recommended solution for multi variable integrals, but it is not exact formal proof. We can consider it as good tool for creating intuition about solution, but correct proof should come from solving inequalities system.
A: As indicated by @HansLundmark, drawing a picture clarifies. The $2$-d section is

The limits indicate $y$ varies from $x$ to $\sqrt{1-x^2}$. This means region of interest is between the curves $y=x$ and $y=+\sqrt{1-x^2}$. $y=+\sqrt{1-x^2}$ is just part of unit circle $x^2+y^2=1$ in first quadrant.
As $x$ varies from $0$ to $1/\sqrt{2}$, we obtain the shaded area in the diagram. From this we see, $0 \le \rho \le 1$ and $\pi/4 \le \varphi \le \pi/2$ and $z$ is independent.
Hence our integral is separable as
$$ \int_0^6 dz \int_{\pi/4}^{\pi/2} d \varphi  \int_0^{1} \rho e^{-\rho ^2}d\rho $$
which is easy to evaluate.
