Question

Find the area of the region enclosed by one loop of the curve. r = 3/2 sin(2θ)

Answer 1

we are given that
`r = (3)/(2) sin(2\theta)`

when
`r=0, \theta=0, \pi/2` which corresponds to one loop.

Therefore area of the region enclosed by this curve is

`A=(9)/(4)* (1)/(2) \int_0^ {\pi/2} sin^2 2\theta d\theta\\ \\  A=(9)/(8)\int_0^ {\pi/2} (1-cos 4\theta)/(2)  d \theta\\ \\ A=(9)/(8) [\theta-(sin 4\theta)/(2) ]_0^(\pi/2)\\ \\ A=(9)/(8)*\pi/2\\ \\ A=(9 \pi)/(16)\\`

Area of the required loop is
`(9 \pi)/(16) )` only.