How to find a centroid of a quarter circle?

Question

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:

Answer 1

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 $r = \sqrt {10}$).

Answer 2

Clearly $\bar{x}$ coordinate is $0$ (by symmetry). The $\bar{y}$ coordinate is:

$$\bar{y} = \frac{ \int\limits_{y=0}^{\sqrt{5}} y \cdot 2 y\ dy + \int\limits_{\sqrt{5}}^{\sqrt{10}} y \cdot 2 \sqrt{10 - y^2}\ dy}{\pi 10/4} = \frac{8 \sqrt{5}}{3 \pi }$$

The total mass (in the denominator) is just the density times the area of a quarter circular disk of radius $\sqrt{10}$

Answer 3

Another (related$ ^* $) method for locating the centroid of a region with constant density is to apply Pappus' (Second) Centroid Theorem, which states that the volume of a solid of revolution is equal to the area of the region revolved about the rotation axis times the circumference of the circle swept out by the centroid.

Revolving the quarter-circle of radius $ \ \sqrt{10} \ $ about one of its straight edges (radii) generates a hemisphere of that radius. We then have

$$ \ V_{hemi} \ = \ \frac{2 \pi}{3} · ( \sqrt{10} )^3 \ \ = \ \ A_{qtr} \ · \ 2 \pi \bar{x} \ \ = \ \ \frac{\pi}{4} · (\sqrt{10})^2 \ · \ 2 \pi \bar{x} $$

$$ \Rightarrow \ \ \bar{x} \ \ = \ \ \frac{4}{3 \pi} \ · \ \sqrt{10} \ \ . $$

With the "center" of the quarter-circle at the origin and the edges along the coordinate axes, the centroid is thus found to be at $ \ \left( \ \frac{4\sqrt{10}}{3 \pi} \ , \ \frac{4\sqrt{10}}{3 \pi} \ \right) \ $ , by the symmetry of the region.

To connect this to the other answers given by Math Lover and David G. Stork, the distance of the centroid from the vertex of the quarter-circle is

$$ \frac{4}{3 \pi} \ · \ \sqrt{10} \ · \ \sqrt{2} \ \ = \ \ \frac{4\sqrt{20}}{3 \pi} \ \ = \ \ \frac{8 \sqrt{5}}{3 \pi} \ \ . $$

For the orientation of the quarter-circle indicated in the problem statement, the coordinates of the centroid are then $ \ \left( \ 0 \ , \ \frac{8 \sqrt{5}}{3 \pi} \ \right) \ \ . $

$$ \ $$

$^*$ This is a related method because Pappus' Centroid Theorems describe the results found from the moment integrals for centroids of surfaces and solids of revolution, used in the other answers.