Setting up exponential function in Cartesian coordinates

asked 2014-07-24 14:52:49 -0600

Christina
807 ●18 ●35 https://sites.google.com/...

updated 2014-07-24 19:20:27 -0600

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.

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answered 2014-07-24 17:00:30 -0600

Gear Junky
278 ●2 ●7 ●13

Christina,

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?

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answered 2014-07-25 06:48:55 -0600

Anonymous
740 ●2 ●27 ●49

updated 2014-07-25 09:07:16 -0600

Gear 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! :D

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Setting up exponential function in Cartesian coordinates

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