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

Marks : 6

Year : 2014

The curve is shown on right for A,

$Θ = 0$ and for B, $θ = 2π$

Now

$X = a(θ – sin θ)$

$\dfrac{dx}{d\theta}=a(1-\cos\theta)$

$y=a(1+\cos\theta)$

$\dfrac{dy}{d\theta}=-a\sin\theta$

$s=\int\limits_0^{2\pi}\sqrt{\Bigg(\dfrac {dx}{d\theta}\Bigg)^2+\Bigg(\dfrac {dy}{d\theta}\Bigg)^2}.d\theta$

$=\int\limits_0^{2\pi}\sqrt{a^2(1-2\cos\theta+\cos^2\theta)+a^2\sin^2\theta d\theta}$

$=\int\limits_0^{2\pi}a\sqrt{1-2\cos\theta+\cos^2\theta +\sin^2\theta.d\theta}$

$=a\int\limits_0^{2\pi}\sqrt{2-2\cos\theta d\theta}$

$=a\int\limits_0^{2\pi}\sqrt{2\times 2\sin^2\dfrac {\theta}2 d\theta}$

$s=2a\int\limits_0^{2\pi}\sin\dfrac {\theta}2 d\theta$

$=2a\Bigg[-2\cos\dfrac {\theta}2\Bigg]_0^{2\pi}$

$=-4a[\cos\pi-\cos\theta]$

$=-4a[-1-1]$

$=8a$