How to set bounds in cylindrical coordinates analytically?

[cwill53]

Homework Statement

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.

I had posted this question on the Mathematics Stack Exchange:
https://math.stackexchange.com/questions/3898044/how-to-set-bounds-in-cylindrical-coordinates

Relevant Equations

$$\begin{cases}x=\rho \cos \phi \\
y= \rho \sin \phi \\
z=z'\end{cases}$$

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.

I had posted this question on the Mathematics Stack Exchange:
https://math.stackexchange.com/questions/3898044/how-to-set-bounds-in-cylindrical-coordinates

This is what one poster said. But the z-coordinate, not the y-coordinate, is what varies from 0 to 6.

So the inequality should be

$$\left\lbrace \begin{array}{a} x\leqslant \rho \sin \phi \leqslant \sqrt{1-x^{2}}\\ 0 \leqslant \rho \cos \phi \leqslant \frac{1}{\sqrt{2}} \end{array} \right\rbrace$$

However I need some assistance solving the inequality so that it would help me set the bounds.

 

[BvU]

Draw an x-y plane and shade the area over which the integral is done

 

[cwill53]

Draw an x-y plane and shade the area over which the integral is done

Can you explain how he solved the inequalities?

 

[etotheipi]

I don't understand what he did either, how he extracted those limits from the inequality with the $\text{min}()$ in it. Maybe someone clever can explain, but I wouldn't worry about it.

 

[BvU]

Dear cwill,
Forget about understanding the nonsense in stackexchange.
$$\int_0^{\sqrt 2\over 2}\;\int_x^\sqrt{1-x^2}\ ... \ dy\, dx $$ Surely you can translate this to
$\qquad\qquad x$ runs from 0 to ${1\over 2}\sqrt 2$
$\qquad\qquad y$ runs from $x$ to $\sqrt{1-x^2}$

and, once you make a sketch, the translation to bounds in $\rho$ and $\phi$ is evident.