Question

Recall that an annulus is the region between two concentric circles. Prove that the centroid of an annulus is the common center of the two circles.

Short answer

the centroid of an annulus is the common center of the two concentric circles.

The Center of two concentric circles $x^2+y^2=a^2\&x^2+y^2=b^2$is $(0,0)$and the centroid of the annulus is$(\overline{x},\overline{y})=(0,0).$

Step-by-step solution

Step 1. Given information.   

The given statement state that the centroid of an annulus is the common center of the two concentric circles.

Step 2. x coordinate of the centroid 

Consider two concentric circles $x^2+y^2=a^2\&x^2+y^2=b^2$of constant density with a center at $(0,0)$where $a<b.$

xcoordinate of the centroid of the annulus is following.

$\overline{x}=\frac{\iint x\rho (x,y)dxdy}{\iint \rho (x,y)dxdy}$

change the system into a polar system by substituting $x=r\cos \theta ,y=r\sin \theta \&dxdy=rdrd\theta$and limits are role="math" localid="1650440508501" $a\le r\le b\&0\le \theta \le 2\mathrm{\pi }.$

$$\begin{gathered} \overline{x}=\frac{{\int }_{\theta =0}^{2\pi }{\int }_{\theta =a}^b(r\cos \theta )rdrd\theta }{{\int }_{\theta =0}^{2\pi }{\int }_{\theta =a}^{b}rdrd\theta } \\ \overline{x}=\frac{{\displaystyle \left({\int }_{\theta =a}^b{r}^{2}dr\right)\times \left({\int }_{\theta =0}^{2\pi }\cos \theta d\theta \right)}}{\left({\int }_{\theta =a}^b{r}^{2}dr\right)\times \left({\int }_{\theta =0}^{2\pi }d\theta \right)} \\ \overline{x}=\frac{\left(\frac{b^3-a^3}{3}\right)\times \left(0\right)}{\left(\frac{a^3}{3}\right)\times \left(2\pi \right)} \\ \overline{x}=0 \end{gathered}$$

Step 3. y coordinate of the centroid 

ycoordinate of the centroid of the annulus is following.

$\overline{y}=\frac{\iint y\rho (x,y)dxdy}{\iint \rho (x,y)dxdy}$

change the system into a polar system by substituting $x=r\cos \theta ,y=r\sin \theta \&dxdy=rdrd\theta$and limits are $a\le r\le b\&0\le \theta \le 2\mathrm{\pi }.$

$\begin{array}{rl}\overline{y}& =\frac{{\int }_{\theta =0}^{2\pi }{\int }_{r=a}^b\left(r\sin \theta \right)rdrd\theta }{{\int }_{\theta =0}^{2\pi }{\int }_{r=a}^{b}rdrd\theta }\\ \overline{y}& =\frac{{\int }_{r=a}^b{r}^{2}dr\times {\int }_{\theta =0}^{2\pi }\sin \theta d\theta }{{\int }_{r=a}^b{r}^{2}dr\times {\int }_{\theta =0}^{2\pi }d\theta }\\ \overline{y}& =\frac{\left(\frac{b^3-a^3}{3}\right)\times \left(0\right)}{\left(\frac{b^3-a^3}{3}\right)\times \left(2\pi \right)}\\ \overline{y}& =0\end{array}$

The centroid of annulus is $(\overline{x},\overline{y})=(0,0).$

the centroid of an annulus is the common center of the two circles.