I want to compute the area of a right trangle with base 2 and 3, but I want to test the truth of integration of the polar coordinate, so i decide to use this integration formula: $$\int_{0}^{\arctan \frac{3}{2}} \int_0^{\frac{2}{\cos \theta}} r^2 r\;\mathrm{d}r\;\mathrm{d}\theta$$ but with Wolfram Alpha's answer, I get a weird result which differs by 0.06... That make me suspect the truth of the polar coordinate.
please help with a mathematical recreational problem
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multivariable-calculus
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2Try it without that $r^2$ in the integrand. – 2011-12-12
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0How about "testing polar integration" on $\int \sqrt{\tan (x)}$?? – 2011-12-12
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2Seems like a bit of an accusatory stance to take, to assume that your teachers (or the web, whatever the case may be) taught you something false. Perhaps you should phrase your question a bit more open-mindedly to the possibility that you made a mistake in your setup or computations. – 2011-12-12
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Corrected: The area element doesn't have the $r^2$ in it, so the integral should be $\int_{0}^{\arctan \frac{3}{2}} \int_0^{\frac{2}{\cos \theta}} r\;dr\;d\theta$. Alpha says this is $3$.
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0If $\theta$ is the angle subtended at $(0,0)$ by $(3,0)$ and $(3,2)$, then surely $\tan\theta=2/3$. I think the triangle goes $(0,0)$ to $(2,0)$ to $(2,3)$, and the limits of integration were correct in the original. – 2011-12-12
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0@GerryMyerson: you are right. Either set of limits gives the same answer (as it should-it is just repositioning the triangle). I have updated the limits to match OP. – 2011-12-12