Q. 74 - Calculus

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.

Step-by-Step Solution

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 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 $a<b.$

x coordinate 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 $a\le r\le b \& 0\le \theta \le 2\mathrm{\pi }.$

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

Step 3. y coordinate of the centroid

y coordinate 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.