# Need help on setting up a double integral over a given region, in polar coordinates

The task is as follows:

Find double integral (in polar coordinates) over region $$D$$ of $$f(x,y) = (x^2 + y)$$ by $$dxdy$$.

The region $$D$$ is bounded by a circle of radius $$3$$ and a circle of radius $$2$$, with upper angle $$a1$$ being $$\frac{\pi}{4}$$ and lower angle $$a2$$ being $$\frac{\pi}{6}$$

The region is as the sketch below:

I need help on setting up the double integral.

My work so far:

(1) Let $$x = r \cos \theta$$ and $$y = r \sin \theta$$

(2) The double integral in polar coordinates should be something in nature of $$\int_{\theta_1} ^ {\theta_2} \int_{r_1}^{r_2} f(r\cos\theta,r\sin\theta) r dr d{\theta}$$

(3) Base on the sketch of the region, $$r$$ should go from $$2$$ to $$3$$,

But I don't see the lower and upper bounds for $$\theta$$. Should it be from $$0$$ to $$a1 + a2$$?

Or should I consider two integrals instead? One with $$\theta$$ from $$0$$ to $$a1$$ and another integral with $$\theta$$ from $$0$$ to $$a2$$?

I just want to make sure that my original thoughts are right (or not).

Would someone please help me on this?

Thank you.

## 1 Answer

the angular range will be as given from $-\frac{\pi}{6}$ to $\frac{\pi}{4}$ assuming $a1$ is the angle above the $x-$axis not including the angle $a2$, otherwise the upper value is $\frac{\pi}{4} - \frac{\pi}{6} = \frac{\pi}{12}$