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

Marks : 6

Year : DEC 2013

Area of one loop of temniscate = 2x Area in first quadrant

$Area=2\times\int\limits_{-\pi/4}^{\pi/4} x^2 d\theta \\ =2\times \dfrac 12\int\limits^{\pi/4}_0\cos2\theta d\theta \\ =a^2\int\limits^{\pi/4}_0\cos 2\theta d\theta\\ =a^2\Bigg[\dfrac {\sin 2\theta}2\Bigg]_0^{\pi/4} \\ =\dfrac {a^2}2\Bigg[\sin\dfrac \pi2-\sin \theta\Bigg] \\ =\dfrac {a^2}2$