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).
$dA = \frac{1}{2}r^2 \, d\theta$
$A = \frac{1}{2}{\displaystyle \int_{\theta_1}^{\theta_2}} r^2 \, d\theta$
$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
Solving for XG
$\frac{3}{2}\pi a^2 X_G = {\displaystyle \int_{\theta_1}^{\theta_2}} \frac{2}{3}r \cos \theta \left( \frac{1}{2}r^2 \, d\theta \right)$
$\frac{3}{2}\pi a^2 X_G = \frac{1}{3} {\displaystyle \int_{\theta_1}^{\theta_2}} r^3 \cos \theta \, d\theta$
$\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
Thank you sir for the help
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$dA = \frac{1}{2}r^2 \, d\theta$
$A = \frac{1}{2}{\displaystyle \int_{\theta_1}^{\theta_2}} r^2 \, d\theta$
$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
Solving for XG
$\frac{3}{2}\pi a^2 X_G = {\displaystyle \int_{\theta_1}^{\theta_2}} \frac{2}{3}r \cos \theta \left( \frac{1}{2}r^2 \, d\theta \right)$
$\frac{3}{2}\pi a^2 X_G = \frac{1}{3} {\displaystyle \int_{\theta_1}^{\theta_2}} r^3 \cos \theta \, d\theta$
$\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
Thank you sir for the help
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