Centroid of Polar Curve by Integration

Submitted by light_angel on

Patulong po for this problem. I really need to know the solution of this.
find the centroid of the area enclosed by the cardioid r=a(1+cos theta).

Tags
Centroid of Area
Centroid by Integration
Centroid of Polar Curve

Jhun Vert Tue, 03/08/2016 - 06:58

$dA = \frac{1}{2}r^2 \, d\theta$

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

integral_009-centroid-cardioid.gif

 

$A = 2 \left[ \frac{1}{2} {\displaystyle \int_0^{\pi}} a^2(1 + \cos \theta)^2 \, d\theta \right] $

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

$A = a^2 \left( \frac{3}{2}\pi \right)$

$A = \frac{3}{2}\pi a^2$
 

By symmetry

$Y_G = 0$

 

Solving for XG

$AX_G = {\displaystyle \int_a^b} \frac{2}{3}r \cos \theta \, dA$

$\frac{3}{2}\pi a^2 X_G = {\displaystyle \int_{\theta_1}^{\theta_2}} \frac{2}{3}r \cos \theta \left( \frac{1}{2}r^2 \, d\theta \right)$

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

$\frac{3}{2}\pi a^2 X_G = \frac{1}{3} \left[ {\displaystyle 2 \int_0^{\pi}} a^3 (1 + \cos \theta)^3 \cos \theta \, d\theta \right] $

$\frac{3}{2}\pi a^2 X_G = \frac{2}{3}a^3 {\displaystyle \int_0^{\pi}} (1 + \cos \theta)^3 \cos \theta \, d\theta$

$\frac{3}{2}\pi a^2 X_G = \frac{2}{3}a^3 \left( \dfrac{15\pi}{8} \right)$

$\frac{3}{2}\pi a^2 X_G = \frac{5}{4}\pi a^3$

$X_G = \frac{5}{6}a$

 

Centroid is at (5a/6, 0)           answer