Question

Find the area enclosed by the curve r(θ)=1+2sinθ and the rays θ=0 and θ=π3

Answer 1

area of the curve f is :

$\intop_0^{\frac{\pi}{3}} \frac{1}{2}{(f(\theta))}^2d\theta$

here, $f=1+2sin\theta$

$=\intop_0^{\frac{\pi}{3}} \frac{1}{2}{(1+2sin\theta)}^2d\theta$

$=\intop_0^{\frac{\pi}{3}} \frac{1}{2}{(1+4sin^2\theta+4sin\theta)}d\theta$

$=\frac{1}{2} [\frac{\pi}{3}-4cos\theta|_0^{\frac{\pi}{3}}+4 |\frac{\theta}{2}-\frac{sin2\theta}{4}|_0^{\frac{\pi}{3}}]$

because $\int sin^2 x dx=\frac{x}{2}-\frac{sin2x}{4}$

$= \frac{1}{2}[\frac{\pi}{3}+4.\frac{1}{2}+4(\frac{\pi}{6}-\frac{\sqrt3}{2})]$

$=\frac{\pi}{2}+1-\frac{\sqrt3}{4}$