# Integral limits of polar coordinate transformation

In an integration exercise, I need to find the volume of the region bounded by the paraboloid $$z=x^2+y^2$$ and below by the triangle enclosed by the lines $$y=x$$, $$x=0$$, and $$x+y=2$$ in the $$xy$$-plane.

Here is the illustration of $$xy$$-plane and the standard solution: However, I got some troubles when I try to use the method of polar coordinate transformation to solve it.

$$\left\{ \begin{array}{ll} x=r\cdot cos\theta\\ y=r\cdot sin\theta \end{array} \right.$$ , where the Jacobian is $$J=r$$.

The integral should be

$$\iint_R[(r\cdot cos\theta)^2+(r\cdot sin\theta)^2]\cdot|J|~drd\theta=\iint_R r^3~drd\theta$$

I think the limits of $$\theta$$ should be $$\frac{\pi}{4}\sim\frac{\pi}{2}$$, but I'm not sure what are the inner limits. Thanks for helps.

For the limits, as you said, $$\theta$$ goes from $$\dfrac{\pi}{4}$$ and $$\dfrac{\pi}{2}$$. For $$r$$, some drawing helps: Here $$\alpha=\theta-\dfrac{\pi}{4}$$.
The limits of $$r$$ is from $$0$$ to some length that changes according to $$\theta$$ (or $$\alpha$$). For the highest value, you see in the drawing that $$\cos\alpha=\dfrac{\sqrt{2}}{r}$$ so $$r=\dfrac{\sqrt{2}}{\cos\alpha}$$, and you have $$\alpha=\theta-\dfrac{\pi}{4}$$...