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Evaluate $$\int_0^1\int_0^{\sqrt{1-x^2}} e^{-(x^2+y^2)}\,dy\,dx$$

Sorry if the formatting is off

Is there a way to evaluate without using polar coordinates or is that the only way to integrate this?

Any help is greatly appreciated

Coop
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    What's wrong with using polar coordinates? It makes this integral much easier. – JimmyK4542 Jul 14 '14 at 02:56
  • There's nothing wrong with polar coordinates. I was just curious as to whether or not there was a way to evaluate it without using polar coordinates as the questions just states evaluate. But thank you for your help – Coop Jul 14 '14 at 03:07
  • No: the function $e^{-x^2}$ does not have an elementary antiderivative, anything you do involves shifting coordinates: keeping the standard ones will not yield the answer. – Adam Hughes Jul 14 '14 at 04:05

1 Answers1

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You could always use infinite series, and then try to determine the value to which the resulting series converges...

In all seriousness, because of the "$e^{x^2+y^2}$," I don't think you're going to get much better than polar. Recall that $e^{x^2}$ does not have an elementary anti-derivative. Perhaps an elliptical transformation (e.g. with the Jacobian) would work, but I'll lump that in the same category as polar.

tl;dr: Probably not--polar is your best bet.

apnorton
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  • One last question, what happens to my limits when I convert to polar coordinates? More specificially, the sqrt{1-x^2} upper bound is causing me some confusion – Coop Jul 14 '14 at 03:09
  • @Coop $y=\sqrt{1-x^2}$ is the top half of a circle of radius $1$. Your region only covers the part in Quadrant I. – apnorton Jul 14 '14 at 03:10
  • Ahhhh, gotcha! I was going to save this for a personal message but I have no clue how to do that. I just wanted to thank you for all of your help with my questions! :) – Coop Jul 14 '14 at 03:12
  • @Coop My pleasure. :) (And, for what it's worth, there is no private message feature on StackExchange.) – apnorton Jul 14 '14 at 03:13
  • So my integral becomes $$\int_0^{pi}\int_0^1 re^{-r^2},dr,dtheta$$ right? – Coop Jul 14 '14 at 03:22
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    $\displaystyle\int_0^{\pi/2} \int_0^1 e^{-r^2} r,dr,d\theta$, with $\pi/2$, not $\pi$. ${}\qquad{}$ – Michael Hardy Jul 14 '14 at 03:50