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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: $$\int_0^{2\pi}\int_0^R e^{-r^2}r dr d\Theta$$ $$u=-r^2$$ $$du=-2rdr$$ $$\frac{du}{-2}=rdr$$ $$\int_0^{2\pi} \int_0^R e^u(-1/2) du d\Theta$$ $$\pi \int_0^R e^e du$$ $$\pi(e^u) \biggr|_0^R$$ $$\pi(e^R - e^0)$$ $$\pi e^R - \pi$$

This is definitely easy! :D

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: $$\int_0^{2\pi}\int_0^R e^{-r^2}r dr d\Theta$$ $$u=-r^2$$ $$du=-2rdr$$ $$\frac{du}{-2}=rdr$$ $$\int_0^{2\pi} \int_0^R e^u(-1/2) du d\Theta$$ $$\pi $$-\pi \int_0^R e^e du$$ $$\pi(e^u) $$-\pi(e^u) \biggr|_0^R$$ $$\pi(e^R $$-\pi(e^R - e^0)$$ $$\pi e^R - \pi$$\pi e^R$$

This is definitely easy! :D