706 Centroid of quarter circle by integration

Equation of the circle with center at the origin and radius r is
$x^2 + y^2 = r^2$

$y^2 = r^2 - x^2$

$y = \sqrt{r^2 -x^2}$
 

Differential area
$dA = y \, dx$

$dA = \sqrt{r^2 -x^2} \, dx$
 

Area of the quarter circle
$A = \frac{1}{4}\pi r^2$
 

 

x-coordinate of the centroid
$\displaystyle A \, \bar{x} = \int_a^b x_c \, dA$

$\frac{1}{4}\pi r^2 \, \bar{x} = \displaystyle{\int_0^r} x\sqrt{r^2 - x^2} \, dx$

$\frac{1}{4}\pi r^2 \, \bar{x} = -\frac{1}{2}\displaystyle{\int_0^r (r^2 - x^2)^{1/2}(-2x \, dx)}$

$\frac{1}{4}\pi r^2 \, \bar{x} = -\frac{1}{2} \left[ \dfrac{(r^2 - x^2)^{3/2}}{3/2} \right]_0^r$

$\frac{1}{4}\pi r^2 \, \bar{x} = -\frac{1}{3} \Big[ (r^2 - r^2)^{3/2} - (r^2 - 0^2)^{3/2} \Big]$

$\frac{1}{4}\pi r^2 \, \bar{x} = -\frac{1}{3} \Big[ -r^3 \Big]$

$\frac{1}{4}\pi r^2 \, \bar{x} = \frac{1}{3}r^3$

$\bar{x} = \dfrac{4r}{3\pi}$           answer
 

y-coordinate of the centroid
$\displaystyle A \, \bar{y} = \int_a^b y_c \, dA$

$\frac{1}{4}\pi r^2 \, \bar{y} = \displaystyle{\int_0^r} (\frac{1}{2}y)(y \, dx)$

$\frac{1}{4}\pi r^2 \, \bar{y} = \frac{1}{2}\displaystyle{\int_0^r} y^2 \, dx$

$\frac{1}{4}\pi r^2 \, \bar{y} = \frac{1}{2}\displaystyle{\int_0^r} (r^2 - x^2) \, dx$

$\frac{1}{4}\pi r^2 \, \bar{y} = \frac{1}{2}\Big[ r^2x - \frac{1}{3}x^3 \Big]_0^r$

$\frac{1}{4}\pi r^2 \, \bar{y} = \frac{1}{2}\Big[ (r^3 - \frac{1}{3}r^3) - (0 - \frac{1}{3}0^3) \Big]$

$\frac{1}{4}\pi r^2 \, \bar{y} = \frac{1}{2}\Big[ \frac{2}{3}r^3 \Big]$

$\frac{1}{4}\pi r^2 \, \bar{y} = \frac{1}{3}r^3$

$\bar{y} = \dfrac{4r}{3\pi}$           answer