Answer to Question #268885 in Calculus for Bhuvana

Question

Find the centroid of the region enclosed by the cardioid r = a(1 + sin θ)

Answer 1

$A=\frac{1}{2}\int\limits_0^{2\pi}r^2d\theta = \frac{a^2}{2}\int\limits_0^{2\pi}(1+\sin \theta)^2d\theta = \frac{a^2}{2}\int\limits_0^{2\pi}(1+2\sin\theta +\sin^2\theta)d\theta = \\ = \frac{a^2}{2}\int\limits_0^{2\pi}(1+2\sin\theta +\frac{1-\cos2 \theta}{2})d\theta =\frac{a^2}{2}\big( \theta -2\cos\theta +\theta /2 -\frac{\sin 2\theta}{4} \big)\big|_0^{2\pi}= \\ \frac{a^2}{2}(2\pi-2+\pi -0)-\frac{a^2}{2}(0-2+0-0)=\frac{3a^2\pi}{2}$

$\iint \limits_RxdA=\int\limits_0^{2\pi}\int\limits_0^{a(1+\sin \theta)}x\cdot r\ dr \ d\theta = \int\limits_0^{2\pi}\int\limits_0^{a(1+\sin \theta)}r^2\cos \theta \ dr \ d\theta = \\ =\int\limits_0^{2\pi}\frac{1}{3}a^3(1+\sin \theta)^3\cos\theta \ d\theta =\frac{a^3}{12}(1+\sin \theta )^4\big|_0^{2\pi}=0$

$\iint \limits_RydA=\int\limits_0^{2\pi}\int\limits_0^{a(1+\sin \theta)}y\cdot r\ dr \ d\theta = \int\limits_0^{2\pi}\int\limits_0^{a(1+\sin \theta)}r^2\sin \theta \ dr \ d\theta = \\ =\int\limits_0^{2\pi}\frac{1}{3}a^3(1+\sin \theta)^3\sin\theta \ d\theta =\frac{a^3}{3}\int\limits_0^{2\pi}(\sin \theta +3\sin^2\theta +3\sin^3\theta+\sin^4\theta)d\theta= \\= \frac{a^3}{3}\int\limits_0^{2\pi}(3\sin^2\theta+\sin^4\theta)d\theta=\frac{a^3}{3}\int\limits_0^{2\pi}(3\sin^2\theta +(1-\cos^2\theta)\sin^2\theta)d\theta= \\ =\frac{a^3}{3}\int\limits_0^{2\pi}\bigg(4\sin^2\theta -\sin^2\theta \cos^2\theta \bigg)d\theta =\frac{a^3}{3}\int\limits_0^{2\pi}\bigg(2(1-\cos2\theta) -\tfrac{1}{4}\sin^2 2\theta \bigg)d\theta= \\= \frac{a^3}{3}\int\limits_0^{2\pi}\big(2-2\cos2\theta -\tfrac{1}{8}+\tfrac{1}{8}\cos 4\theta\big)d\theta=\frac{a^3}{3}\bigg(\tfrac{15}{8}\theta -\sin 2\theta +\tfrac{1}{32}\sin 4\theta\bigg)\bigg|_0^{2\pi}=\frac{15a^3\pi}{12}$

$\overline {x}=\frac{1}{A}\iint \limits_RxdA=0$

$\overline {y}=\frac{1}{A}\iint \limits_RydA=\frac{15a^3\pi}{12}:\frac{3a^2\pi}{2}=\frac{5a}{6}$

Answers: centroid is $(0,\ \frac{5a}{6})$ .