Finding $\iint_{B} x \,dx\,dy$ where $B$ is the circle $x^2+y^2 -x \leq 0$

Question

$\displaystyle\iint_{B} x \,dx\,dy$ where $B$ is the circle $x^2+y^2 -x \leq 0$

I am having trouble with this integral.

I already know that $B$ is a circle with center $(1/2,0)$ and radius $1/2$.

I think it would be better to transform in polar coordinates, but what is the variation of $r$ then?

Answer 1

hint

Your disk can be parametrized as $$x=\frac 12 +r\cos(t)$$ $$y=r\sin(t)$$ with $$0\le t\le 2\pi$$ and $$0\le r\le \frac 12$$

the integral becomes

$$\int_0^{\frac 12}\int_0^{2\pi}(\frac 12+r\cos(t)rdrdt$$ $$=\frac{\pi}{8}$$

because $$\int_0^{2\pi}\cos(t)dt=0$$

Answer 2

Recenter the circle with $x-\frac12=u$ and $y=v$. Then, $u^2+v^2\leqslant\frac14=r^2$ and the integral is

$$\begin{align}\iint_{B} x\, dx\,dy&=\iint_{u^2+v^2\le r^2} (u+\frac12)\,du\,dv\\[1ex]&= \frac12 \iint_{u^2+v^2\le r^2}\,du\,dv =\frac12 (\pi r^2) = \frac\pi8\end{align}$$

where the integral on $u$ vanishes due to symmetry and what remains is a plain circle area.