Given $\iint\limits_{R}e ^{-x^2-y^2}dxdy$ for $R=\{(x,y):x^2+y^2 \le 9\}$. Is it $ \int\limits_{0}^{2\pi}\int\limits_{0}^{3}e ^{-r^2} r dr d \theta$
double integral $e^{-x^2-y^2}$
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calculus
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0@coffeemath But does the inner integral equal $(1-e^{-9})$ or $\frac{1}{2}(1-e^{-9})$? – 2012-11-01
1 Answers
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I assume you're asking whether $ \int \text{d}x\int \text{d}y\ e^{-x^2-y^2} = \int_0^{2\pi}\text{d}\phi\int_0^3\text{d}r\ e^{-r^2}\ , $ where the first integral is restricted to $x^2+y^2\leq9$. The answer is 'no', as $\text{d}x\text{d}y=r\text{d}r\text{d}\phi$, so you need a further factor $r$ on the right-hand-side.
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0yes i forgot to add r, but I get some weird answer like pi-(pi/e^9) – 2012-11-01