http://calc3.askbot.com/questions/Open source question and answer forum written in Python and DjangoenCopyright Askbot, 2010-2011.Fri, 25 Jul 2014 06:48:55 -0500http://calc3.askbot.com/question/169/setting-up-exponential-function-in-cartesian-coordinates/I am hoping for some help looking at the exponential function in Cartesian land. If we are given $e^{-(x^2+y^2)}$ and asked to set this up over the domain of a disk of radius R both in polar and Cartesian coordinates, is this what polar would look like? $$\int_0^{2\pi}\int_0^R e^{-(r^2)}rdrd\theta$$ In turn, is this what the Cartesian set-up would look like? $$\int_{-R}^R\int_0^\sqrt{R-x^2} e^{-(x^2+y^2)} dydx$$ Any help or comments would be great. Thanks! **COMMENT** - Thanks Gear Junky, that makes sense. I am struggling to remember to visualize the projection onto the xy plane. As far as evaluating it, I would definitely evaluate the polar coordinates. I was just practicing because he said we may have a problem like this on the quiz to set up in both but evaluate one. Thu, 24 Jul 2014 14:52:49 -0500http://calc3.askbot.com/question/169/setting-up-exponential-function-in-cartesian-coordinates/http://calc3.askbot.com/question/169/setting-up-exponential-function-in-cartesian-coordinates/?answer=170#post-id-170Christina, As far as the integral in polar coordinates, it looks like you know what you're doing. $$\int_0^{2\pi}\int_0^Re^{-(r^2)}rdrd\theta$$As far as the Cartesian coordinates, it seems as if you are trying to integrate only half of the disk on the $y$ - axis. I may be wrong, but I feel like it should be set up as:$$\int_{-R}^R\int_{-\sqrt{R-x^2}}^\sqrt{R-x^2}e^{-(x^2+y^2)}dydx$$ These bounds would define the area of the disk, but a question to ask now is what would be the easiest method of integrating this problem?Thu, 24 Jul 2014 17:00:30 -0500http://calc3.askbot.com/question/169/setting-up-exponential-function-in-cartesian-coordinates/?answer=170#post-id-170http://calc3.askbot.com/question/169/setting-up-exponential-function-in-cartesian-coordinates/?answer=177#post-id-177Gear Junky, I think that the easiest method of integrating this problem is using polar coordinates since there is no elementary antiderivative for $e^{-(x^2+y^2)}$. However, as you can see, in polar coordinates this turns into an extremely simple u-substitution problem: $$\int_0^{2\pi}\int_0^R e^{-r^2}r dr d\Theta$$ $$u=-r^2$$ $$du=-2rdr$$ $$\frac{du}{-2}=rdr$$ $$\int_0^{2\pi} \int_0^R e^u(-1/2) du d\Theta$$ $$-\pi \int_0^R e^e du$$ $$-\pi(e^u) \biggr|_0^R$$ $$-\pi(e^R - e^0)$$ $$\pi - \pi e^R$$ This is definitely easy! :DFri, 25 Jul 2014 06:48:55 -0500http://calc3.askbot.com/question/169/setting-up-exponential-function-in-cartesian-coordinates/?answer=177#post-id-177