707 Centroid of quarter ellipse by integration

Equation of ellipse in y as a function of x
$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$

$b^2x^2 + a^2y^2 = a^2b^2$

$a^2y^2 = a^2b^2 - b^2x^2$

$a^2y^2 = b^2(a^2 - x^2)$

$y^2 = \dfrac{b^2}{a^2}(a^2 - x^2)$

$y = \dfrac{b}{a}\sqrt{a^2 - x^2}$
 

Differential area
$dA = y \, dx$

$dA = \dfrac{b}{a}\sqrt{a^2 - x^2} \, dx$
 

Area of quarter ellipse
$A = \frac{1}{4}\pi ab$
 

 

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

$\displaystyle \frac{1}{4}\pi ab \, \bar{x} = \int_0^a x \left( \dfrac{b}{a}\sqrt{a^2 - x^2} \, dx \right)$

$\displaystyle \frac{1}{4}\pi ab \, \bar{x} = \dfrac{b}{a}\int_0^a x \sqrt{a^2 - x^2} \, dx$

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

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

$\frac{1}{4}\pi ab \, \bar{x} = -\dfrac{b}{3a} \left[ (a^2 - a^2)^{3/2} - (a^2 - 0^2)^{3/2} \right]$

$\frac{1}{4}\pi ab \, \bar{x} = -\dfrac{b}{3a} \left[ -a^3 \right]$

$\frac{1}{4}\pi ab \, \bar{x} = \frac{1}{3}a^2b$

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

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

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

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

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

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

$\frac{1}{4}\pi ab \, \bar{y} = \dfrac{b^2}{2a^2}\left[ a^2x - \frac{1}{3}x^3 \right]_0^a$

$\frac{1}{4}\pi ab \, \bar{y} = \dfrac{b^2}{2a^2}\left[ (a^3 - \frac{1}{3}a^3) - (0^3 - \frac{1}{3}\cdot 0^3) \right]$

$\frac{1}{4}\pi ab \, \bar{y} = \dfrac{b^2}{2a^2}\left[ \frac{2}{3}a^3 \right]$

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

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