2

Evaluate the following triple integrals as a repeated integral using an appropriate coordinate systems: $$\iiint\limits_R ze^{-(x^2+y^2+z^2)} \, \, dV ,$$ where $$R=\{ (x,y,z): \, x,y \in (-\infty, \infty), \, 0 \leq z \leq 1 \}.$$

It is simple to integrate after using cylindrical coordinates but how do you figure out the limits of $r$ and $\theta$?

Travis Willse
  • 99,363
snowman
  • 3,733
  • 8
  • 42
  • 73

3 Answers3

3

Hint Before changing coordinates, factor $$e^{-(x^2 + y^2 + z^2)} = e^{-(x^2 + y^2)}e^{-z^2}$$ and separate the integral in $z$: $$\iiint_R z e^{-(x^2 + y^2 + z^2)} dV = \int_0^1 ze^{-z^2} dz \iint_{{\Bbb R}^2} e^{-(x^2 + y^2)} dx\,dy .$$

Additional hint We can integrate over ${\Bbb R}^2$ in polar coordinates simply by integrating over $$0 \leq r < \infty, \quad 0 \leq \theta \leq 2 \pi.$$

Travis Willse
  • 99,363
  • +1 very nice! Without this writing the limit of integration in spheric coordinates would be unpleasant :) – Ant Apr 26 '15 at 12:36
  • @Travis the integration is easy. how did you just know he limits of r and theta though? – snowman Apr 26 '15 at 12:40
  • @Ant Thanks! It's not so different using cylindrical coordinates---the point of the first hint is (1) that, since the same coordinate $z$ occurs in both rectangular and cylindrical systems, we might as well isolate the coordinates that are transformed, and (2) the double integral on the r.h.s. is famous in its right, and isolating it emphasizes its evaluation. – Travis Willse Apr 26 '15 at 12:43
  • @snowman The idea is to find a region in the $r\theta$-plane that covers each point in the $xy$-plane once, and this region does this. (Actually, the given ranges for $r, \theta$ cover the points on the closed positive $x$-axis twice, but this ray has area zero, so this redundancy does not affect the value of the integral.) Does that help? – Travis Willse Apr 26 '15 at 12:47
  • @Travis Yes kinda. So it seems like of obvious that r is going to range over from 0 to infinity since it cant be negative. And putting theta from 0 to 2pi will definitely ensure no points are missed. Am I correct in saying this? – snowman Apr 26 '15 at 12:52
  • @snowman Yes, that's right. Have you used polar coordinates to set up a double integral over a disc of radius $R$ (centered at the origin)? – Travis Willse Apr 26 '15 at 12:54
  • @Travis Yes I have. Why do you ask? – snowman Apr 26 '15 at 12:58
  • @snowman Then as you know, you just integrate over $[0, R] \times [0, 2 \pi]$ in $r\theta$-space. Heuristically, the plane is the limit of these discs as $R \to \infty$, giving the claimed subset $[0, \infty) \times [0, 2\pi]$. – Travis Willse Apr 26 '15 at 13:12
1

It's like in polar coordinates, where for the plane we have

$0<r<\infty,0<\theta<2\pi$

or

$-\infty<r<\infty,0<\theta<\pi$

davyjones
  • 641
  • We are in space, we should use spheric coordinates; so you need another angle $\psi$, but calculating where it varies it's unpleasant :) Better to just separate the integral and then use polar coordinates as @Travis answer shows :) – Ant Apr 26 '15 at 12:38
  • @Ant, that is just cylindrical coordinates as is pointed by OP. – davyjones Apr 26 '15 at 12:42
  • ops! You're right, totally had a moment there :) +1 then ;) – Ant Apr 26 '15 at 12:43
0

In cylinder coordinates with $x=r\cos\phi$, $y=r\sin\phi$ and Jacobian $r$ $$ \int_0^1 dz z e^{-z^2}\int_{-\infty}^\infty dx \int _{-\infty}^\infty dy e^{-x^2-y^2} $$ $$ =\int_0^1 dzz e^{-z^2}\int_0^{2\pi} d\phi \int _0^\infty r dr e^{-r^2} $$ $$ =2\pi\int_0^1 dzz e^{-z^2} \int _0^\infty r dr e^{-r^2} $$ $$ =\pi\int_0^1 dzz e^{-z^2} \int _0^\infty du e^{-u} $$ $$ =\pi\int_0^1 dzz e^{-z^2} $$ $$ =\frac12 \pi(1-\frac{1}{e}) $$

R. J. Mathar
  • 2,324