Solve Double Integral with Polar System

Question

Evaluate $\displaystyle\iint\limits_{\Omega}{(x+y)dx dy}$ where the area $\Omega$ is bounded by a curve $x^2+y^2=x+y$. Note: Evaluate by converting to polar system.

My Solve: $$\iint\limits_{\Omega}{(x+y)dx dy}=\int_0^{2\pi}d\theta \int_0^{\frac{1}{\sqrt 2}} r \,(r\sin\theta + r\cos\theta)dr=$$

$$=\int_0^{2\pi}(\sin\theta+\cos\theta)d\theta*\left.{\frac{r^3}{3}}\right\rvert_0^{\frac{1}{\sqrt2}}=?$$

My Question: The answer in the book is $\frac{\pi}{2}$. I don't understand how they got it. Maybe when I switched to the polar coordinates I did something wrong? Sorry for asking help for simple question.

Answer 1

The area is bounded by a circle around $(1/2,1/2)$ with radius $1/\sqrt{2}\,:$ $$ 0=x^2-x+y^2-y=(x-\tfrac12)^2+(y-\tfrac12)^2-\tfrac12\,. $$ Expressing the integral in $(r,\theta)$ has to take into account where the center of the circle is:

$$\iint\limits_\Omega(x+y)\,dx\,dy=\int_0^{2\pi}\int_0^\frac1{\sqrt{2}}\Big\{(r\cos\theta+\tfrac12)+(r\sin\theta+\tfrac12)\Big\}r\,d\theta\,dr=\frac\pi 2\,.$$