I have the following integral
$$\iint\limits_D \sqrt{x^2+y^2}\,\, dA$$
where $D$ is the region bounded by the curves $y= \sqrt{2x-x^2}$ and $y=x$.
I'v been trying to solve it by change of variable to polar coordinates (centered at $(1,0)$) because the region looks like a circle and by some trigonometric rules I rewrite it like this
$$\int_\frac{\pi}{2}^\pi \int_{\sec\theta}^1 r\sqrt{(x(r,\theta))^2 + (y(r,\theta))^2} \, dr\,d\theta$$
where $$x=r\cos\theta +1$$
$$y = r\sin\theta$$
However, this just complicates the calculus.
Edit:
This is how the region looks.
$y= \sqrt{2x-x^2}$ and
Best Answer
Given the line $y = x$ passes through origin and integrand is $\sqrt{x^2+y^2}$, it is easier to use,
$x = r \cos \theta, y = r \sin \theta$
Please note that for circles with center on one of the coordinate axes and the origin being a point on the circle, it is another standard parametrization that we can use.
$y \leq \sqrt{2x-x^2} \implies x^2 + y^2 \leq 2x, y \geq 0$
So in polar coordinates, semicircle can be written as, $r^2 \leq 2 r \cos\theta \ $ i.e. $r \leq 2 \cos\theta, 0 \leq \theta \leq \frac{\pi}{2}$
Now there are two regions bound between $y = x$ and the semi-circle. The question should clarify which region it refers to. Based on your diagram, it is the shaded region to the left of $y = x$. So we have,
$y \geq x \implies \theta \geq \frac{\pi}{4} \ $ and the integral simplifies to,
$$ \int_{\pi/4}^{\pi/2} \int_0^{2\cos\theta} r^2 \ dr \ d\theta$$