Calculate double integral of $$\iint_\Omega \sqrt{x^2+y^2} \;\mathrm{d}x\mathrm{d}y,\;\mathrm{where} \;\Omega:(x-1)^2+y^2\leq 1, \; 0\leq y$$ I try to figure out the boundaries so $0\leq x \leq 2$ and $0 \leq y \leq \sqrt{1-(x-1)^2}$
I try to change into polar coordinates where I even substituted $\sqrt{x^2+y^2}=r$ and I got $8\pi/3$ (for $y=0 \mathrm{\;and\;} y=\pi$) but the result should be $16/9$.
On
HINT
Make the change of variables $x = 1 + r\cos(\theta)$ and $y = r\sin(\theta)$. Then the proposed integral becomes \begin{align*} \iint_{\Omega}\sqrt{x^{2} + y^{2}}\mathrm{d}x\mathrm{d}y = \int_{0}^{1}\int_{0}^{\pi}r\sqrt{1 + 2r\cos(\theta) + r^{2}}\mathrm{d}\theta\mathrm{d}r \end{align*}
On
Let $u=x-1$ and $v=y$ to rewrite the integral
$$I= \iint_\Omega \sqrt{x^2+y^2}{d}x{d}y = \int_{u^2+v^2\le 1, v>0}\sqrt{(u+1)^2+v^2}du dv $$
Then, with polar coordinates
$$I=\int_0^{\pi}\int_0^1\sqrt{r^2-2 r\cos\theta+1}\>rdr d\theta$$
Fisrt, evaluate with $t=r-\cos\theta$
\begin{align} f(\theta)& =\int_0^1\sqrt{r^2-2 r\cos\theta+1}\>rdr = \int_{-\cos\theta}^{1-\cos\theta}\sqrt{t^2 +\sin^2\theta}\> (t+\cos\theta)dt\\ &= \frac13(8\sin^3\frac\theta2-1)+\frac12\cos\theta[4\sin^3\frac\theta2+\cos\theta+\sin^2\theta\ln(1+\csc\frac\theta2)] \end{align}
Then, integrate over $\theta$ to obtain $$I=\int_0^{\pi}f(\theta)d\theta = \frac{16}9 $$
The region that we are talking about here is the upper half of the circle with center $(1,0)$ and radius $1$. So, the possible values of $\theta$ lie in the range $\left[0,\frac\pi2\right]$. For each such $\theta$, let $t=\tan(\theta)$. One must determine where $y=tx$ intersects the circle $(x-1)^2+y^2=1$ (see the picture below). So, we solve the equation$$(x-1)^2+t^2x^2=1\left(\iff(t^2+1)x^2-2x=0\right)$$and we get that $x=0$ or that $x=\frac2{t^2+1}=2\cos^2\theta$. And, when $x=2\cos^2\theta$, then$$y=2\cos^2(\theta)\tan(\theta)=2\cos(\theta)\sin(\theta),$$and therefore $r=\sqrt{x^2+y^2}=2\cos\theta$. So, compute$$\int_0^{\pi/2}\int_0^{2\cos\theta}r^2\,\mathrm dr\,\mathrm d\theta.$$