could someone please explain this:
find the area bound by the curve r= a(1-sin theta) between theat = 0 and theta = 2pi.
Assuming that $\displaystyle a$ is a constant and the integral is taken with respect to $\displaystyle \theta$...
$\displaystyle \int_0^{2\pi}{a(1 - \sin{\theta})\,d\theta} = a\int_0^{2\pi}{1 - \sin{\theta}\,d\theta}$
$\displaystyle = a\left[\theta + \cos{\theta}\right]_0^{2\pi}$
$\displaystyle = a\left[\left(2\pi + \cos{2\pi}\right) - \left(0 + \cos{0}\right)\right]$
$\displaystyle = a\left(2\pi + 1 - 0 - 1\right)$
$\displaystyle = 2\pi a$.