Question

Prove that the centroid of a circle is the center of the circle.

Short answer

the centroid of a circle is the center of the circle.

The centroid of a circle$x^2+y^2=a^2$is at$(\overline{x},\overline{y})=(0,0)$and the center is also at$(0,0).$

Step-by-step solution

Step 1. Given information.  

The given statement is the centroid of a circle is the center of the circle.

Step 2. x coordinate of the centroid

Consider a circle $x^2+y^2=a^2$of constant density with a center at $(0,0).$

xcoordinate of the centroid of the circle is following.

$$\begin{gathered} \overline{x}=\frac{\iint x\rho (x,y)dxdy}{\iint \rho (x,y)dxdy} \\ \overline{x}=\frac{{\int }_{x=-a}^a{\int }_{y=-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}xdxdy}{{\int }_{x=-a}^a{\int }_{y=-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}{\displaystyle d}{\displaystyle x}{\displaystyle d}{\displaystyle y}} \end{gathered}$$

change the system into a polar system by substituting $x=r\cos \theta ,y=r\sin \theta \&dxdy=rdrd\theta .$

localid="1650439208659" $$\begin{gathered} \overline{x}=\frac{{\int }_{\theta =0}^{2\pi }{\int }_{\theta =0}^a(r\cos \theta )rdrd\theta }{{\int }_{\theta =0}^{2\pi }{\int }_{r=0}^{a}rdrd\theta } \\ \overline{x}=\frac{{\displaystyle \left({\int }_{r=0}^a{r}^{2}dr\right)}\times \left({\int }_{\theta =0}^{2\pi }\cos \theta d\theta \right)}{\left({\int }_{r=0}^a{r}^{2}dr\right)\times \left({\int }_{\theta =0}^{2\pi }d\theta \right)} \\ \overline{x}=\frac{\left(\frac{a^3}{3}\right)\times \left(0\right)}{\left(\frac{a^3}{3}\right)\times \left(2\pi \right)} \\ \overline{x}=0 \end{gathered}$$

Step 3. y coordinate of the centroid

ycoordinate of the centroid of the circle is following.

$$\begin{gathered} \overline{y}=\frac{\iint y\rho (x,y)dxdy}{\iint \rho (x,y)dxdy} \\ \overline{y}=\frac{{\int }_{x=-a}^a{\int }_{y=-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}ydxdy}{{\int }_{x=-a}^a{\int }_{y=-\sqrt{a^2-x^2}}^{\sqrt{a^2-x^2}}{\displaystyle }{\displaystyle d}{\displaystyle x}{\displaystyle d}{\displaystyle y}} \end{gathered}$$

change the system into a polar system by substituting$x=r\cos \theta ,y=r\sin \theta \&dxdy=rdrd\theta$

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

the centroid of the circle is $p(\overline{x},\overline{y})=(0,0).$

So the centroid of a circle is the center of the circle.