A plane region $R$ is determined by the inequalities $y\ge0$, $y\ge-x$, $x^2+y^2\le3\sqrt{x^2+y^2}-3x$. Sketch the region and find it's area.
I have foregone sketching the area and tried to use algebra instead. A bit of substitution with the polar coordinates $x=r\cos{\theta}$, $y=r\sin{\theta}$ yields $$r\le3(1-\cos{\theta})$$ which should give me $0\le r\le6$, and after inspection $0\le\theta\ \le2 \pi$. Right?
So the integral should be $$\int_{0}^{2\pi}\int_{0}^{6}rdrd\theta,$$ correct? The book doesn't provide answers so I just want to know if I got the bounds right, I can do the actual computation myself.
Thanks!
I assume you intended to do the problem for the following areas, unless, as @Andrea noted we have nothing to do.
According to the plot, we have two evaluate two areas:
$$A_1:~~~\theta|_{7\pi/4}^0, r|_{3(1-\cos(\theta))}^0$$
$$A_2:~~~\theta|_0^{3\pi/4}, r|_0^{3(1-\cos(\theta))}$$