# Centroid of area in the first quadrant bounded by circle and ellipse

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Centroid of area in the first quadrant bounded by circle and ellipse

Find the centroid of the area in the first quadrant bounded by the ellipse x^2/a^2+y^2/b^2=1 and the circle x^2+y^2=a^2.

Jhun Vert

Subscript c is for circle and subscript e is for ellipse:
$A_c = \frac{1}{4}\pi a^2$

$A_e = \frac{1}{4}\pi ab$

$A = A_c - A_e = \frac{1}{4}\pi a(a - b)$

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

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

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

$AX_G = \Sigma A_n x_n$

$\frac{1}{4}\pi a(a - b)X_G = \frac{1}{4}\pi a^2 \left( \dfrac{4a}{3\pi} \right) - \frac{1}{4}\pi ab \left( \dfrac{4a}{3\pi} \right)$

$\frac{1}{4}\pi a(a - b)X_G = \dfrac{a^3}{3} - \dfrac{a^2 b}{3}$

$\frac{1}{4}\pi a(a - b)X_G = \frac{1}{3}a^2(a - b)$

$\frac{1}{4}\pi X_G = \frac{1}{3}a$

$X_G = \dfrac{4a}{3\pi}$           answer

$AY_G = \Sigma A_n y_n$

$\frac{1}{4}\pi a(a - b)Y_G = \frac{1}{4}\pi a^2 \left( \dfrac{4a}{3\pi} \right) - \frac{1}{4}\pi ab \left( \dfrac{4b}{3\pi} \right)$

$\frac{1}{4}\pi a(a - b)Y_G = \dfrac{a^3}{3} - \dfrac{ab^2}{3}$

$\frac{1}{4}\pi a(a - b)Y_G = \frac{1}{3}a(a^2 - b^2)$

$\frac{1}{4}\pi a(a - b)Y_G = \frac{1}{3}a(a - b)(a + b)$

$\frac{1}{4}\pi Y_G = \frac{1}{3}(a + b)$

$Y_G = \dfrac{4(a + b)}{3\pi}$           answer

Jhun Vert

Solution by detailed integration:
$x^2 + y^2 = a^2$

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

$y = \sqrt{a^2 - x^2}$   ←   upper-end of the strip

$y_U = \sqrt{a^2 - x^2}$

$\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$

$\dfrac{y^2}{b^2} = 1 - \dfrac{x^2}{a^2}$

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

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

$y = \dfrac{b}{a}\sqrt{a^2 - x^2}$   ←   lower-end of the strip

$y_L = \dfrac{b}{a}\sqrt{a^2 - x^2}$

$dA = y \, dx = (y_U - y_L) \, dx = \left( \sqrt{a^2 - x^2} - \dfrac{b}{a}\sqrt{a^2 - x^2} \right) \, dx$

$\displaystyle A = \int_0^a \left( \sqrt{a^2 - x^2} - \dfrac{b}{a}\sqrt{a^2 - x^2} \right) \, dx$

$\displaystyle A = \int_0^a \sqrt{a^2 - x^2} \left( 1 - \dfrac{b}{a} \right) \, dx$

$\displaystyle A = \int_0^a \sqrt{a^2 - x^2} \left( \dfrac{a - b}{a} \right) \, dx$

$\displaystyle A = \dfrac{a - b}{a} \int_0^a \sqrt{a^2 - x^2} \, dx$

Let
x = a sin θ
dx = a cos θ dθ

When x = 0, θ = 0
When x = a, θ = π /2

$\displaystyle A = \dfrac{a - b}{a} \int_0^{\pi/2} \sqrt{a^2 - a^2 \sin^2 \theta} \, (a \cos \theta \, d\theta)$

$\displaystyle A = (a - b) \int_0^{\pi/2} \sqrt{a^2 - a^2 \sin^2 \theta} \, (\cos \theta \, d\theta)$

$\displaystyle A = (a - b) \int_0^{\pi/2} \sqrt{a^2(1 - \sin^2 \theta)} \, (\cos \theta \, d\theta)$

$\displaystyle A = (a - b) \int_0^{\pi/2} \sqrt{a^2 \cos^2 \theta} \, (\cos \theta \, d\theta)$

$\displaystyle A = (a - b) \int_0^{\pi/2} a \cos \theta \, (\cos \theta \, d\theta)$

$\displaystyle A = a (a - b) \int_0^{\pi/2} \cos^2 \theta \, d\theta$

$\displaystyle A = a(a - b) \int_0^{\pi/2} \frac{1}{2}(1 + \cos 2\theta) \, d\theta$

$A = a(a - b) \left[ \dfrac{1}{2} \left( \theta - \dfrac{\sin 2\theta}{2} \right) \right]_0^{\pi/2}$

$A = \dfrac{a(a - b)}{2} \left[ \left( \dfrac{\pi}{2} - \dfrac{\sin \pi}{2} \right) - \left( 0 - \dfrac{\sin 0}{2} \right) \right]$

$A = \dfrac{a(a - b)}{2} \left( \dfrac{\pi}{2} \right)$

$A = \frac{1}{4}\pi a(a - b)$

$\displaystyle AX_G = \int_a^b x_c \, dA$

$\displaystyle AX_G = \int_a^b x_c (y_U - y_L) \, dx$

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

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

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

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

$\dfrac{\pi a(a - b)}{4}X_G = -\dfrac{1}{3} \left[ (a^2 - a^2)^{3/2} - (a^2 - 0^2)^{3/2} \right] + \dfrac{b}{3a} \left[ (a^2 - a^2)^{3/2} - (a^2 - 0^2)^{3/2} \right]$

$\dfrac{\pi a(a - b)}{4}X_G = -\dfrac{1}{3}(-a^3) + \dfrac{b}{3a}(-a^3)$

$\dfrac{\pi a(a - b)}{4}X_G = \dfrac{a^3}{3} - \dfrac{a^2b}{3}$

$\dfrac{\pi a(a - b)}{4}X_G = \dfrac{a^2}{3}(a - b)$

$\dfrac{\pi}{4}X_G = \dfrac{a}{3}$

$X_G = \dfrac{4a}{3\pi}$           answer

$\displaystyle AY_G = \int_a^b y_c \, dA$

$\displaystyle AY_G = \int_{x_1}^{x_2} \left( \dfrac{y_U + y_L}{2} \right) \left( y_U - y_L \right) \, dx$

$\displaystyle AY_G = \dfrac{1}{2} \int_{x_1}^{x_2} \left( {y_U}^2 - {y_L}^2 \right) \, dx$

$\displaystyle \dfrac{\pi a(a - b)}{4}Y_G = \dfrac{1}{2} \int_0^a \left[ (a^2 - x^2) - \dfrac{b^2}{a^2}(a^2 - x^2) \right] \, dx$

$\displaystyle \dfrac{\pi a(a - b)}{4}Y_G = \dfrac{1}{2} \int_0^a (a^2 - x^2)\left[ 1 - \dfrac{b^2}{a^2} \right] \, dx$

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

$\dfrac{\pi a(a - b)}{4}Y_G = \dfrac{1}{2} \left( \dfrac{a^2 - b^2}{a^2} \right) \left[ a^2x - \dfrac{x^3}{3} \right]_0^a$

$\dfrac{\pi a(a - b)}{4}Y_G = \dfrac{1}{2} \left[ \dfrac{(a - b)(a + b)}{a^2} \right] \left[ \dfrac{2a^3}{3} \right]$

$\dfrac{\pi a(a - b)}{4}Y_G = \dfrac{a(a - b)(a + b)}{3}$

$\dfrac{\pi}{4}Y_G = \dfrac{a + b}{3}$

$Y_G = \dfrac{4(a + b)}{3\pi}$           answer

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