Find the centroid in the first quadrant which is between $x^2+y^2=4$ and $x^2+y^2=9$

Question

Let $R$ be the region in the first quadrant which is between $x^2+y^2=4$ and $x^2+y^2=9$. Find the centroid of $R$.

Answer 1

We can describe the region in polar coordinates with $r \in [2,3]$ and $\theta \in [0,\frac{\pi}{2}]$

The centroid is the center of mass of our object assuming constant $\delta=c$ density.

$$\frac{M}{A}=c$$

$$M=cA$$

$$\bar x=\frac{1}{M} \iint_{D} \delta xdA$$

$$\bar x=\frac{1}{cA}\int_{0}^{\frac{\pi}{2}} \int_{2}^{3} c r\cos(\theta) rdr d\theta$$

$$=\frac{4}{5\pi} \int_{2}^{3} r^2 dr \int_{0}^{\frac{\pi}{2}} cos(\theta) d\theta$$

$$=\frac{4}{5\pi} \frac{19}{3} =\frac{76}{5\pi}$$

$$\bar y=\frac{1}{M} \iint_{D} \delta ydA$$

$$\bar y=\frac{1}{cA}\int_{0}^{\frac{\pi}{2}} \int_{2}^{3} c r\sin(\theta) rdrd\theta$$

The mass is what you get when you add up all the densities times a small change in area,

$$M=\iint_{D} \delta dA$$

In this case we have $\delta=k$ where $k$ is a constant or $c$ if you wish, then we have:

$$M=\iint_{D} k dA=\int_{0}^{\frac{\pi}{2}} \int_{2}^{3} k rdrd\theta=\frac{5\pi}{4}k$$

In fact because we have $\frac{M}{A}=k$ then we have, $M=kA=k\frac{1}{4}(\pi(3^2)-\pi(2^2))$. Also note that,

$$\iint_{D} dA=\text{Area}(D)$$

(We're adding small areas of D to make the whole area of D).

So multiplying by $k$ gives,

$$\iint_{D} k dA=k\text{Area}(D)$$