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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.

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    $\begingroup$ 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 $\pi/2$ $\endgroup$ – JoseSquare Feb 19 '19 at 23:12
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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 $\varphi\in[0,\frac{\pi}{2}]$.

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  • $\begingroup$ You're right, I made a silly mistake there. Thanks. $\endgroup$ – Zelda_CompSci Feb 19 '19 at 23:18

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