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If i have this integral in rectangular coordinates and i want to transform in polar coordinates $$\int_{0}^2\int_0^{x} f(x,y) \,dy\, d x$$

The limits in $\displaystyle 0\le\theta\le\frac\pi4 $ but in $r$,the limits are $0$ to ? $ 2\sec[\theta]=r$ because $x=2 $ is equal to $2\sec[\theta]=r$.

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  • $\begingroup$ Try graphically shading in the region you are integrating over. Does this help? $\endgroup$ – alphacapture May 10 '16 at 1:09
  • $\begingroup$ You CANNOT convert coordinates without drawing an image. There is no rule or set of rules for converting coordinate bounds. $\endgroup$ – The Great Duck May 10 '16 at 2:21
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You are integrating over a triangle with vertices $(0,0)$, $(2,0)$ and $(2,2)$. This is bounded on the right by $x = 2$; in polarese that is $r\cos(\theta) = 2$ so your integral becomes $$\int_0^{\pi/4} \int_0^{2\sec(\theta)} f(r\cos(\theta), r\sin(\theta) r dr\,d\theta.$$

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  • $\begingroup$ That inner integral should be $\displaystyle\int_0^{2\sec \theta}$, and the $\sec\theta$ should not be part of the integrand. $\endgroup$ – Christopher Carl Heckman May 10 '16 at 1:10
  • $\begingroup$ why you add $Sec[\theta]$ in the integral? for the $rCos[\theta]=2$? $\endgroup$ – Daniel ORTIZ May 10 '16 at 1:11
  • $\begingroup$ @DanielORTIZ : I already covered this. $\endgroup$ – Christopher Carl Heckman May 10 '16 at 1:11
  • $\begingroup$ is $o=<r<=2Sec[\theta]$ frist limit? $\endgroup$ – Daniel ORTIZ May 10 '16 at 1:12
  • $\begingroup$ @DanielORTIZ : Yes. $\endgroup$ – Christopher Carl Heckman May 10 '16 at 1:14

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