Double integrals using polar coordinates help

Question

I need some help in identifying the boundaries for this equation and transforming the given double integral into a polar integral.

The problem that is given to me is: $$\int_0 ^2\int _0 ^{\sqrt{2x-x^2}} 7\sqrt{x^2 + y^2}\,dydx.$$

The only step that I'm sure of is that the function within the double integral could promptly be transformed into $7r$ and $dydx$ could be turned into $rdrd\theta$. However, I'm not really sure how to find the boundary of integration for this function.

Thanks in advance.

Answer 1

Hint. The domain of integration is $$\{(x,y): x\in [0,2],\, 0\leq y \leq \sqrt{1-(x-1)^2}\}$$ that is the upper part of the disc $$(x-1)^2+y^2\leq 1$$ which can be written in polar coordinates as $$(r\cos(\theta)-1)^2+r^2\sin(\theta)^2\leq 1\Leftrightarrow r\leq 2\cos(\theta).$$ Can you take it from here?

Answer 2

Looking at the integrand that is $\sqrt{x^2+y^2}=\rho$ the integration bounds are the following

$$\begin{cases} 0<x<2\\ y>0\\ 0<y^2<2x-x^2 \end{cases}$$

$$\begin{cases} 0<\rho cos\theta<2\\ \rho sin\theta>0\\ 0<\rho^2 sin^2\theta<2\rho cos\theta-\rho^2 cos^2\theta \end{cases}$$

The last double inequality becomes

$$\rho(\rho-2cos \theta)<0 \rightarrow \rho<2cos\theta$$

this implies that

$$\begin{cases} cos\theta>0 \\ sin\theta>0 \end{cases}\rightarrow 0<\theta<\frac{\pi}{2}$$

Thus your integral becomes

$$\int_{0}^{\frac{\pi}{2}}d\theta\int_{0}^{2cos\theta}7\rho^2d\rho $$