# Rectangular coordinates transform in polar coordinates $\int_{0}^2\int_0^{x} f(x,y) \,dy\,dx$

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 $2\sec[\theta]=r$.

• Try graphically shading in the region you are integrating over. Does this help? – alphacapture May 10 '16 at 1:09
• You CANNOT convert coordinates without drawing an image. There is no rule or set of rules for converting coordinate bounds. – The Great Duck May 10 '16 at 2:21

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 $$\int_0^{\pi/4} \int_0^{2\sec(\theta)} f(r\cos(\theta), r\sin(\theta) r dr\,d\theta.$$
• That inner integral should be $\displaystyle\int_0^{2\sec \theta}$, and the $\sec\theta$ should not be part of the integrand. – Christopher Carl Heckman May 10 '16 at 1:10
• why you add $Sec[\theta]$ in the integral? for the $rCos[\theta]=2$? – Daniel ORTIZ May 10 '16 at 1:11
• is $o=<r<=2Sec[\theta]$ frist limit? – Daniel ORTIZ May 10 '16 at 1:12