Find the centroid of the quarter circle $x^2+y^2 \le 10$, $y\ge |x|$ assuming the density $ {\displaystyle \delta }(x,y) = 1$. give your answer as point coordinates in the form (X,Y)
So I first convert to polar coordinates:
$$rsin(\theta)^2+rcos(\theta)^2=10$$
Then I know it's a quarter circle that's bounded by $y=|x|$
So $\theta = \frac{\pi}{4}$ to $\frac{3\pi}{4}$
So now I integrate:
$$\int _{\frac{\pi }{4}}^{\frac{3\pi }{4}}\int _0^{2.5}\:\:\:rsin\left(\theta \right)^2+rcos\left(\theta \right)^2drd\theta $$
But this is just one integral. How do I get two? What am I missing?
Thank you.
edit:
Best Answer
To do this in polar coordinates as the question asks, please consider the surface as a triangle with two sides as $r$ and the third side as $r \mathrm{d}\theta$.
Area of triangle, $dA = \frac{1}{2} r^2 \mathrm{d}\theta$. As we know, the centroid of the triangle will be at $\frac{2}{3}r$.
So,
$\overline{x} = \displaystyle \dfrac{2}{3A}\int_{\pi/4}^{3\pi/4} r\cos\theta \, \mathrm{d}A = \dfrac{1}{3A}\int_{\pi/4}^{3\pi/4} r^3 \cos\theta \, \mathrm{d} \theta$
$\overline{y} = \displaystyle \dfrac{2}{3A}\int_{\pi/4}^{3\pi/4} r\sin\theta \, \mathrm{d}A = \dfrac{1}{3A}\int_{\pi/4}^{3\pi/4} r^3 \sin\theta \, \mathrm{d} \theta$
$A = \displaystyle \dfrac{1}{2}\int_{\pi/4}^{3\pi/4}r^2 \,\mathrm{d}\theta = \frac{\pi r^2}{4}$
So, $\overline {x} =\displaystyle \frac{4r}{3\pi} [\sin \theta]_{\pi / 4}^{3 \pi / 4} = 0$ (which is obvious due to symmetry around $y$ axis).
$\overline {y} =\displaystyle \frac{4r}{3\pi} [-\cos \theta]_{\pi / 4}^{3 \pi / 4} = \frac{4 \sqrt2 \, r}{3\pi} = \frac {8 \sqrt5}{3 \pi} \,$ (for $r = \sqrt {10}$).