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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.

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Try it without that $r^2$ in the integrand. – Gerry Myerson Dec 12 '11 at 2:12
How about "testing polar integration" on $\int \sqrt{\tan (x)}$?? – The Chaz 2.0 Dec 12 '11 at 2:49
Seems 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. – Dustan Levenstein Dec 12 '11 at 3:01
up vote 3 down vote accepted

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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If $\theta$ 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. – Gerry Myerson Dec 12 '11 at 3:15
@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. – Ross Millikan Dec 12 '11 at 3:33

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