Converting $ \int_{1}^{2}\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}{x\,dy\,dx} $ to polar coordinates

Question

An integral is given. $$ \int_{1}^{2}\int_{-\sqrt{4-x^2}}^{\sqrt{4-x^2}}{x\,dy\,dx} $$ It is asked to convert it into the polar coordinate system. The limit of $r$ I have calculated is $0$ to $2$. I found the following after converting into the polar form. $$ \int_{0}^{\frac{\pi}{3}}\int_{\sec{\theta}}^{2\sec{\theta}}r^2 cos{\theta}\,dr\,d{\theta} $$ Is my attempt correct? Also, the value of the double integral in the Cartesian coordinate and that of the polar coordinate are different. Kindly explain.

Answer 1

If you sketch the region on the plane, it will help you to see what is going on. We have a vertical slice of a circle of radius 2 centered at the origin.

You are correct that the maximum value of $\theta$ is $\frac{\pi}{3}$ but the circle also lies below the $x$-axis so the minimum value of $\theta$ is $-\frac{\pi}{3}$.

For a given value of $\theta=\theta^*$, the maximum value of $r$ is of course 2, but the minimum value occurs at the intersection of the line $\theta=\theta^*$ and the line $x=1$, which is $\sec\theta^*$.

So the integral should be \begin{align} \int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}\int_{\sec{\theta}}^2 r^2\cos\theta\ drd\theta. \end{align}