Mumbai University > First Year Engineering > sem 2 > Applied Maths 2

Marks : 6

Year : DEC 2014

Solving simultaneously

$a (1 + \cos θ) = a (1 – \cos θ) \\ 1 + \cos θ = 1 – \cos θ\\ \cos θ = 0\\ \theta=\dfrac \pi2 \space or \space \theta=\dfrac {3\pi}2$

Area common to the cardiode = 4 x common area in first quadrant

$$=4\times\int\limits_0^{\pi/2}\int\limits_0^{a(1-\cos\theta)}r\space dr\space d\theta $$

$=4\int\limits_0^{\pi/2}\Bigg[\dfrac {r^2}2\Bigg]_0^{a(1-\cos\theta)}d\theta \\ =\dfrac 42\int\limits_0^{\pi/2}[a^2(1-\cos\theta)^2-0]d\theta\\ =2a^2\int\limits(1-2\cos\theta +\cos^2\theta)\\ =2a^2\int\limits_0^{\pi/2}(1-2\cos\theta +\dfrac {1+\cos2\theta}2)d\theta \\ =2a^2\int\limits_0^{\pi/2}(1+\dfrac 12 -2\cos\theta +\dfrac {\cos2\theta}2)d\theta \\ =2a^2\int\limits_0^{\pi/2}(\dfrac 32-\dfrac {4\cos\theta}2+\dfrac {\cos\theta }2)d\theta \\ =a^2[3\theta-4\sin\theta+\dfrac {\sin2\theta}2]_0^{\pi/2} \\ =a^2[\dfrac {3\pi}2-4\sin\dfrac {\pi}2+ \sin\dfrac \pi2]-[0]\\ =a^2[\dfrac {3\pi}2-4]\\ =\dfrac {a^2}2(3\pi-8)$