Chapter 6.6, Problem 50E

To determine

To Find: The centroid of a region shown on a coordinate plane

Answer to Problem 50E

The centroid of the region is at $\left(\frac{4r}{3\pi },\frac{4r}{3\pi }\right)$ .

Explanation of Solution

Given Information: The density of the plate is $\text{ρ}=2$ .

The region is

  

Method:

• The equation of a circle of radius $r$ with center at the origin is

  $x^2+y^2=r^2$

• The coordinates of centroid of a region which is bounded by $x=a,b$ are

  $\left(\overline{x},\overline{y}\right)=\left(\frac{M_y}{A},\frac{M_x}{A}\right)$

where

  $M_y=\rho \underset{a}{\overset{b}{{\displaystyle \int }}}xf(x)dx$ , $M_x=\rho {\displaystyle \underset{a}{\overset{b}{\int }}\frac{1}{2}}f{(x)}^{2}dx$ and $A=\rho {\displaystyle \underset{a}{\overset{b}{\int }}f}(x)dx$

• The quantities . $M_x$ . and $M_y$ are called the moments of the system about the respective axes.

Calculation:Solve the equation of the circle $x^2+y^2=r^2$ for $y$ in terms of $x$ to get

  $f\left(x\right)=y=\sqrt{r^2-x^2}$

The bounds of the region on the $x$ -axis are $0$ and $r$ . So,

  $M_y=\rho \underset{0}{\overset{r}{{\displaystyle \int }}}xf\left(x\right)dx=2\underset{0}{\overset{r}{{\displaystyle \int }}}x\sqrt{r^2-x^2}dx$

  $=2\times {\frac{-1}{3}{(r^2-x^2)}^{\frac{3}{2}}]}_0^r=\frac{-2}{3}\times (0-r^3)=\frac{2r^3}{3}$

  $M_x=\frac{1}{2}\rho \underset{0}{\overset{r}{{\displaystyle \int }}}f{(x)}^{2}dx=\frac{2}{2}\times {{\displaystyle \int }}_0^r{\left(\sqrt{r^2-x^2}\right)}^{2}dx$

  $={r^{2}x-\frac{x^3}{3}]}_0^r=r^3-\frac{r^3}{3}=\frac{2r^3}{3}$

  $A=2\underset{0}{\overset{r}{{\displaystyle \int }}}\sqrt{r^2-x^2}dx=2\underset{0}{\overset{\frac{\pi }{2}}{{\displaystyle \int }}}{r}^2{\cos }^2\theta d\theta =r^2\underset{0}{\overset{\frac{\pi }{2}}{{\displaystyle \int }}}(1+\cos 2\theta )d\theta =\frac{\pi r^2}{2}$ .

Therefore, $\overline{x}=\frac{\left(\frac{2r^3}{3}\right)}{\left(\frac{\pi r^2}{2}\right)}=\frac{4r}{3\pi }$ and $\overline{y}=\frac{\left(\frac{2r^3}{3}\right)}{\left(\frac{\pi r^2}{2}\right)}=\frac{4r}{3\pi }$

So, the centroid of the region is at $\left(\frac{4r}{3\pi },\frac{4r}{3\pi }\right)$ .