Convert to Polar $\int_0^6\int_0^y x\; \mathrm dx \,\mathrm dy$

Question

How to calculate the integral

$$\int_0^6\int_0^y x\; \mathrm dx \,\mathrm dy$$

using polar coordinates?$$$$I know that $x=R\cos \theta$ and $y=R\sin\theta$ and that the Jacobian is $R$.

Answer 1

You have

$$\int_0^6\int_0^y x\;dx dy$$

The region that's covered is the upper left half of a square, with lower left corner at the origin and side $6$ units. This is an isoceles right triangle.

The angular part goes from $\pi/4$ to $\pi/2$.

At $\pi/4$, the integration of $r$ goes from $0 \to 6\sqrt{2}$.

At $\pi/2$, the integration of $r$ goes from $0 \to 6$.

In between, the integration goes from $0 \to 6/\sin \theta$.

So in polar coordinates the integral should be

$$\int_{\pi/4}^{\pi/2} \int_0^{6/\sin \theta} r \cos \theta\; r\;dr\;d\theta.$$

Carrying this through gives the same answer as the Cartesian integral.