# Set up an iterated integral in the polar coordinates for the double integral, and then find the value of I.

Helping a student out with a Calculus 3 problem and it's been a long time since I took Calc 3. Here's the statement of the problem:

Let $$I=\int_0^1 \int_0^{2-2x}f(x,y)dydx$$ be a double integral of a function $$f$$ over a region $$D$$. Let $$f(x,y)=\frac{y+2x}{x^2+y^2}$$. Set up an iterated integral in the polar coordinates for the double integral and then find the value of $$I$$.

What I can recall how to do is conver the function into polar coordinates which I did here:

$$f(x,y)=f(r,\theta)=\frac{r\sin\theta+2r\cos\theta}{(r\cos\theta)^2+(r\sin\theta)^2}=\frac{\sin\theta+2\cos\theta}{r}$$

Then, $$I=\int\int f(r,\theta)\space r\space drd\theta=\int \int \frac{\sin\theta+2\cos\theta}{r}\space r\space drd\theta=\int\int \sin\theta +2\cos\theta\space drd\theta$$

Now, obviously I need to figure out the bounds in terms of polar coordinates, but that's where I'm lost and could use some help. Also, please let me know if I've made errors elsewhere.

The region $$D$$ is simply a triangle with vertices $$(0,0)\\(1,0)\\(0,2)$$so, its representation in polar coordinates would be $$D=\left\{(r,\phi)\ \ \ |\ \ \ 0\le \phi\le {\pi\over 2}\quad,\quad 0\le r\le {2\over \sin\phi+2\cos\phi}\right\}$$