# How to determine the interval of $\varphi$ for integration?

I have $$\int_{D_2}(x^3+xy^2+y) d(x,y)$$ with $$D_2 := \{(x,y)^T\in{\mathbb R^2} : 4 \leq x^2+y^2 \leq 9, x \geq 0, y \geq 0\}$$.

Obviously, I want to transform this into polar coordinates and determine the intervals of $$r$$ and $$\varphi$$ to calculate the definite integrals.

I know that $$r^2 = x^2+y^2$$ and $$r = \sqrt {x^2+y^2}$$. Thus, $$4 \leq r^2 \leq 9 <=> 2 \leq r \leq 3$$.

Now, I have successfully found the interval of $$r$$.

My polar coordinates are $$x = r \cdot cos(\varphi)$$ and $$y = r \cdot sin(\varphi)$$.

How do I now calculate the interval of $$\varphi$$? I know the solution is $$\varphi \in [0, \frac{\pi}{2}]$$, but I don't understand how this solution was obtained.

My first thought was to look at the quadrants of the unit circle.

• The best for this is drawing what is $D_2$. You will notice is a quarter of "Donut" in the first Quadrant, where the angle that $\varphi$ moves bewteen $0$ and $\pi/2$ – JoseSquare Feb 19 '19 at 23:12

Consider the point in the $$(x,y)$$. The variable $$\varphi$$ is angle that its radius vector forms with the positive part of the $$x$$-axis. When $$x\geq0,y\geq0$$, the point $$(x,y)$$ lies in the first quadrant and thus, $$\varphi\in[0,\frac{\pi}{2}]$$.
Your mistake lies in the fact that $$x\geq0 \implies \varphi \in [0,\frac{\pi}{2}] \cup[\frac{3\pi}{2},2\pi]$$ $$y\geq0 \implies \varphi \in [0,\pi]$$ which differs from what you suggested.
The intersection of these sets indeed does give $$\varphi\in[0,\frac{\pi}{2}]$$.