Changing limits of integration to polar coordinates

Question

So, I have to calculate the integral $$\int_{B}{3\sqrt{x^2+y^2}\,dx\,dy}$$ where $$B=\{(x,y)|~ x^2+y^2 \leq 6x\}$$ I am having some trouble with setting up limits of integration in polar coordinates. Instead of $x^2 + y^2$ I wrote $r^2$ and instead of $x$ I wrote $r\cos{\varphi}$. From this I got $$0 \leq r \leq 6\cos{\varphi}.$$ Now, I need to find the limits of $\varphi$ and I think they are $$0 \leq \varphi \leq 2\pi.$$ However, when I looked up the solution, it said $$-\pi/2 \leq \varphi \leq \pi/2.$$ Could anyone explain how to get correct limits for $\varphi$? I hope you understood me since English is not my first language.

Answer 1

The region $B$ is the closed disk centered at $(3,0)$ with radius $3$. Since $(0,0)$ is located at the boundary of this dis, no you don't take $0\leqslant\varphi\leqslant2\pi$. The solution that you mentioned is correct: you do take $-\frac\pi2\leqslant\varphi\leqslant\frac\pi2$, since that corresponds to only having points $(x,y)$ with $x\geqslant0$, which is the case here. So, your integral is equal to$$3\int_{-\pi/2}^{\pi/2}\int_0^{6\cos\varphi}r^2\,\mathrm dr\,\mathrm d\varphi=216\int_{-\pi/2}^{\pi/2}\cos^3(\varphi)\,\mathrm d\varphi=288.$$

Answer 2

Note $x^2+y^2\le 6x$ is the circle region with center $(3,0)$ and radius $3$, as shown in the graph,

the radius $r$ is from $0$ to $6\cos\varphi$, and the angle $\varphi$ is from $-\frac\pi2$ to $\frac\pi2$