# Centroid of area bounded by parabola and line

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light_angel
Centroid of area bounded by parabola and line

Patulong po....
Find the centroid of the area bounded by the curves y=2x and y^2 =4ax using polar coordinates.

Jhun Vert

Solution by Integration
Area:
$A = {\displaystyle \int_{y_1}^{y_2}} (x_R - x_L) \, dy$

$A = {\displaystyle \int_0^{2a}} \left( \dfrac{y}{2} - \dfrac{y^2}{4a} \right) \, dy$

$A = \left[ \dfrac{y^2}{4} - \dfrac{y^3}{12a} \right]_0^{2a}$

$A = \dfrac{4a^2}{4} - \dfrac{8a^3}{12a}$

$A = a^2 - \frac{2}{3}a^2$

$A = \frac{1}{3}a^2$

$A \, \bar{x} = {\displaystyle \int_{y_1}^{y_2}} x_c (x_R - x_L) \, dy$

$A \, \bar{x} = {\displaystyle \int_{y_1}^{y_2}} \frac{1}{2}(x_R + x_L)(x_R - x_L) \, dy$

$A \, \bar{x} = \frac{1}{2} {\displaystyle \int_{y_1}^{y_2}} ({x_R}^2 - {x_L}^2) \, dy$

$\frac{1}{3}a^2 \, \bar{x} = \frac{1}{2} {\displaystyle \int_0^{2a}} \left[ \left( \dfrac{y}{2} \right)^2 - \left( \dfrac{y^2}{4a} \right)^2 \right] \, dy$

$\bar{x} = \dfrac{3}{2a^2} {\displaystyle \int_0^{2a}} \left[ \dfrac{y^2}{4} - \dfrac{y^4}{16a^2} \right] \, dy$

$\bar{x} = \dfrac{3}{2a^2} \left[ \dfrac{y^3}{12} - \dfrac{y^5}{80a^2} \right]_0^{2a}$

$\bar{x} = \dfrac{3}{2a^2} \left[ \dfrac{8a^3}{12} - \dfrac{32a^5}{80a^2} \right]$

$\bar{x} = \dfrac{3}{2a^2} \left[ \dfrac{2a^3}{3} - \dfrac{2a^3}{5} \right]$

$\bar{x} = \dfrac{3}{2a^2} \left[ \dfrac{4a^3}{15} \right]$

$\bar{x} = \frac{2}{5}a$

$A \, \bar{y} = {\displaystyle \int_{y_1}^{y_2}} y_c (x_R - x_L) \, dy$

$\frac{1}{3}a^2 \, \bar{y} = {\displaystyle \int_0^{2a}} y \left( \dfrac{y}{2} - \dfrac{y^2}{4a} \right) \, dy$

$\bar{y} = \dfrac{3}{a^2} {\displaystyle \int_0^{2a}} \left( \dfrac{y^2}{2} - \dfrac{y^3}{4a} \right) \, dy$

$\bar{y} = \dfrac{3}{a^2} \left[ \dfrac{y^3}{6} - \dfrac{y^4}{16a} \right]_0^{2a}$

$\bar{y} = \dfrac{3}{a^2} \left[ \dfrac{8a^3}{6} - \dfrac{16a^4}{16a} \right]$

$\bar{y} = \dfrac{3}{a^2} \left[ \dfrac{4a^3}{3} - a^3 \right]$

$\bar{y} = \dfrac{3}{a^2} \left[ \dfrac{a^3}{3} \right]$

$\bar{y} = a$

Centroid in rectangular coordinates = (0.4a, a)

Centroid In polar coordinates:

$r = \sqrt{{\bar{x}}^2 + {\bar{y}}^2} = \sqrt{(0.4a)^2 + a^2}$

$r = \frac{\sqrt{29}}{5}a = 1.077a$

$\theta = \arctan \left( \dfrac{\bar{y}}{\bar{x}} \right) = \arctan \left( \dfrac{a}{0.4a} \right)$

$\theta = 68.198^\circ$

Centroid = (1.077a, 68.198°)

Jhun Vert

Solution by Composite Area
Parabolic segment

$A_1 = \frac{2}{3}a(2a) = \frac{4}{3}a^2$

$x_1 = \frac{3}{5}a$

$y_1 = \frac{3}{8}(2a) = \frac{3}{4}a$

Triangle

$A_2 = \frac{1}{2}a(2a) = a^2$

$x_2 = \frac{2}{3}a$

$y_2 = \frac{1}{3}(2a) = \frac{2}{3}a$

Required area

$A = A_1 - A_2 = \frac{4}{3}a^2 - a^2 = \frac{1}{3}a^2$

$A \, \bar{x} = A_1 x_1 - A_2 x_2$

$\frac{1}{3}a^2 \, \bar{x} = \frac{4}{3}a^2 (\frac{3}{5}a) - a^2(\frac{2}{3}a)$

$\frac{1}{3}a^2 \, \bar{x} = \frac{2}{15}a^3$

$\bar{x} = \frac{2}{5}a$

$A \, \bar{y} = A_1 y_1 - A_2 y_2$

$\frac{1}{3}a^2 \, \bar{y} = \frac{4}{3}a^2 (\frac{3}{4}a) - a^2 (\frac{2}{3}a)$

$\frac{1}{3}a^2 \, \bar{y} = \frac{1}{3}a^3$

$\bar{y} = a$

Centroid in rectangular coordinates = (0.4a, a)

Calculator Technique
The above calculations can be done in calculator with linear regression capability in Statistics. I am using <strong>CASIO fx-991ES PLUS</strong> for the following operations:

Centroid in rectangular coordinates = (0.4a, a)

Centroid In polar coordinates:

$r = \sqrt{{\bar{x}}^2 + {\bar{y}}^2} = \sqrt{(0.4a)^2 + a^2}$

$r = \frac{\sqrt{29}}{5}a = 1.077a$

$\theta = \arctan \left( \dfrac{\bar{y}}{\bar{x}} \right) = \arctan \left( \dfrac{a}{0.4a} \right)$

$\theta = 68.198^\circ$

Centroid = (1.077a, 68.198°)

Jhun Vert

Solution by Integration in Polar Form (pahirapan ang self method... lol)
$y^2 = 4ax$

$(r \sin \theta)^2 = 4ar \cos \theta$

$r \sin^2 \theta = 4a \cos \theta$

$r = \dfrac{4a \cos \theta}{\sin^2 \theta}$

$r = 4a \cdot \dfrac{\cos \theta}{\sin \theta} \cdot \dfrac{1}{\sin \theta}$

$r = 4a \cot \theta \csc \theta$

Point of intersection (a, 2a) in polar form
$r = \sqrt{a^2 + (2a)^2} = a\sqrt{5}$

$\theta = \arctan \left( \dfrac{2a}{a} \right) = \arctan (2)$

$A = \frac{1}{2} {\displaystyle \int_{\theta_1}^{\theta_2}} r^2 \, d\theta$

$A = \frac{1}{2} {\displaystyle \int_{\arctan (2)}^{\pi / 2}} (4a \cot \theta \csc \theta)^2 \, d\theta$

$A = 8a^2 {\displaystyle \int_{\arctan (2)}^{\pi / 2}} \cot^2 \theta \csc^2 \theta \, d\theta$

$A = -8a^2 {\displaystyle \int_{\arctan (2)}^{\pi / 2}} \cot^2 \theta (-\csc^2 \theta \, d\theta)$

$A = -8a^2 \left[ -\dfrac{\cot^3 \theta}{3} \right]_{\arctan (2)}^{\pi / 2}$

$A = -\frac{8}{3}a^2 \left[ \dfrac{1}{\tan^3 \theta} \right]_{\arctan (2)}^{\pi / 2}$

$A = -\frac{8}{3}a^2 \left[ \dfrac{1}{\tan^3 (\pi / 2)} - \dfrac{1}{\tan^3 (\arctan 2)} \right]$

$A = -\frac{8}{3}a^2 \left[ \dfrac{1}{\infty^3} - \dfrac{1}{2^3} \right]$

$A = -\frac{8}{3}a^2 \left[ 0 - \dfrac{1}{8} \right]$

$A = \frac{1}{3}a^2$

$A \, \bar{x} = {\displaystyle \int_{\theta_1}^{\theta_2}} (\frac{2}{3}r \cos \theta)(\frac{1}{2} r^2 \, d\theta)$

$\frac{1}{3}a^2 \, \bar{x} = \frac{1}{3} {\displaystyle \int_{\theta_1}^{\theta_2}} r^3 \cos \theta \, d\theta$

$\bar{x} = \dfrac{1}{a^2} {\displaystyle \int_{\arctan (2)}^{\pi / 2}} (4a \cot \theta \csc \theta)^3 \cos \theta \, d\theta$

$\bar{x} = 64a {\displaystyle \int_{\arctan (2)}^{\pi / 2}} \cot^3 \theta \csc^3 \theta \cos \theta \, d\theta$

$\bar{x} = 64a {\displaystyle \int_{\arctan (2)}^{\pi / 2}} \dfrac{\cos^3 \theta}{\sin^3 \theta} \cdot \dfrac{1}{\sin^3 \theta} \cdot \cos \theta \, d\theta$

$\bar{x} = 64a {\displaystyle \int_{\arctan (2)}^{\pi / 2}} \dfrac{\cos^4 \theta}{\sin^6 \theta} \, d\theta$

$\bar{x} = 64a {\displaystyle \int_{\arctan (2)}^{\pi / 2}} \dfrac{\cos^4 \theta}{\sin^4 \theta} \cdot \dfrac{1}{\sin^2 \theta} \, d\theta$

$\bar{x} = 64a {\displaystyle \int_{\arctan (2)}^{\pi / 2}} \cot^4 \theta \csc^2 \theta \, d\theta$

$\bar{x} = -64a {\displaystyle \int_{\arctan (2)}^{\pi / 2}} \cot^4 \theta (-\csc^2 \theta \, d\theta)$

$\bar{x} = -64a \left[ \dfrac{\cot^5 \theta}{5} \right]_{\arctan (2)}^{\pi / 2}$

$\bar{x} = -\frac{64}{5}a \left[ \dfrac{1}{\tan^5 \theta} \right]_{\arctan (2)}^{\pi / 2}$

$\bar{x} = -\frac{64}{5}a \left[ \dfrac{1}{\tan^5 (\pi / 2)} - \dfrac{1}{\tan^5 (\arctan 2)} \right]$

$\bar{x} = -\frac{64}{5}a \left[ \dfrac{1}{\infty^5} - \dfrac{1}{2^5} \right]$

$\bar{x} = -\frac{64}{5}a \left[ 0 - \dfrac{1}{32} \right]$

$\bar{x} = 0.4a$

$A \, \bar{y} = {\displaystyle \int_{\theta_1}^{\theta_2}} (\frac{2}{3}r \sin \theta)(\frac{1}{2} r^2 \, d\theta)$

$\frac{1}{3}a^2 \, \bar{y} = \frac{1}{3} {\displaystyle \int_{\theta_1}^{\theta_2}} r^3 \sin \theta \, d\theta$

$\bar{y} = \dfrac{1}{a^2} {\displaystyle \int_{\arctan (2)}^{\pi / 2}} (4a \cot \theta \csc \theta)^3 \sin \theta \, d\theta$

$\bar{y} = 64a {\displaystyle \int_{\arctan (2)}^{\pi / 2}} \cot^3 \theta \csc^3 \theta \sin \theta \, d\theta$

$\bar{y} = 64a {\displaystyle \int_{\arctan (2)}^{\pi / 2}} \dfrac{\cos^3 \theta}{\sin^3 \theta} \cdot \dfrac{1}{\sin^3 \theta} \cdot \sin \theta \, d\theta$

$\bar{y} = 64a {\displaystyle \int_{\arctan (2)}^{\pi / 2}} \dfrac{\cos^3 \theta}{\sin^3 \theta} \cdot \dfrac{1}{\sin^2 \theta} \, d\theta$

$\bar{y} = 64a {\displaystyle \int_{\arctan (2)}^{\pi / 2}} \cot^3 \theta \csc^2 \theta \, d\theta$

$\bar{y} = -64a {\displaystyle \int_{\arctan (2)}^{\pi / 2}} \cot^3 \theta (-\csc^2 \theta) \, d\theta$

$\bar{y} = -64a \left[ \dfrac{\tan^4 \theta}{4} \right]_{\arctan (2)}^{\pi / 2}$

$\bar{y} = -16a \left[ \dfrac{1}{\tan^4 \theta} \right]_{\arctan (2)}^{\pi / 2}$

$\bar{y} = -16a \left[ \dfrac{1}{\tan^4 (\pi / 2 )} - \dfrac{1}{\tan^4 (\arctan 2 )} \right]$

$\bar{y} = -16a \left[ \dfrac{1}{\infty^4} - \dfrac{1}{\tan^4 (\arctan 2)} \right]$

$\bar{y} = -16a \left[ 0 - \dfrac{1}{16} \right]$

$\bar{y} = a$

Centroid in rectangular coordinates = (0.4a, a)

Centroid In polar coordinates:

$r = \sqrt{{\bar{x}}^2 + {\bar{y}}^2} = \sqrt{(0.4a)^2 + a^2}$

$r = \frac{\sqrt{29}}{5}a = 1.077a$

$\theta = \arctan \left( \dfrac{\bar{y}}{\bar{x}} \right) = \arctan \left( \dfrac{a}{0.4a} \right)$

$\theta = 68.198^\circ$

Centroid = (1.077a, 68.198°)

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