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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.

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    $\begingroup$ Just draw a picture? (In the $xy$-plane; no need for a 3D picture since the $z$-dependence is so simple.) $\endgroup$ Commented Nov 7, 2020 at 18:20

2 Answers 2

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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.

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  • $\begingroup$ z was from 0 to 6. $\endgroup$ Commented Nov 7, 2020 at 21:07
  • $\begingroup$ @CalebWilliamsUIC. I proceed from your integral notation: in the first place you have $dy$, then $dx$, then $dz$. Accordingly, I took the boundaries. But since you mean for $z \in [0,6]$, then write also boundaries for $x$ and $y$ and I'll write analytical solution without drawing. $\endgroup$
    – zkutch
    Commented Nov 7, 2020 at 21:16
  • $\begingroup$ I'm just confused as why you wrote the inequalities $$ \left\lbrace \begin{array}{a} 0 \leqslant \rho \sin \phi \leqslant 6 \\ 0 \leqslant \rho \cos \phi \leqslant \frac{1}{\sqrt{2}} \end{array} \right\rbrace$$ when z goes from 0 to 6 and $x\leq y\leq \sqrt{1-x^2}$ $\endgroup$ Commented Nov 7, 2020 at 21:22
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    $\begingroup$ @CalebWilliamsUIC. Added this variant to answer also. Feel free to ask if/when have questions. $\endgroup$
    – zkutch
    Commented Nov 7, 2020 at 23:05
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    $\begingroup$ We have $\tan \phi \geqslant 1$, $\cos \phi \geqslant 0, \sin \phi \geqslant 0$. These inequalities gives result - which do you doubt in? $\endgroup$
    – zkutch
    Commented Nov 8, 2020 at 11:02
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As indicated by @HansLundmark, drawing a picture clarifies. The $2$-d section is

enter image description here

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.

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  • $\begingroup$ Thanks for this answer. Can you help me understand analytically how to solve the inequality given in zkutch's answer? I don't understand how to solve the inequalities. $\endgroup$ Commented Nov 8, 2020 at 21:35
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    $\begingroup$ @CalebWilliamsUIC, I have added explanation. If you have further questions, feel free to ask. Do you understand geometrical meaning of $\rho$ and $\varphi$ ? $\endgroup$
    – cosmo5
    Commented Nov 9, 2020 at 11:05
  • $\begingroup$ I think I do. $\rho$ is the radius while $\varphi$ is the positive angle from the starting axis. I was wondering if there was a way to figure out bounds without drawing a picture. I'm confused in the same way with spherical coordinates. $\endgroup$ Commented Nov 9, 2020 at 17:36

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