are bound by polar coordinates

**hyzlemon** · March 5th 2010, 01:38 AM

could someone please explain this:
find the area bound by the curve r= a(1-sin theta) between theat = 0 and theta = 2pi.

**Prove It** · March 5th 2010, 01:51 AM

Originally Posted by hyzlemon

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 $a$ is a constant and the integral is taken with respect to $\theta$ ...

$\int_0^{2\pi}{a(1 - \sin{\theta})\,d\theta} = a\int_0^{2\pi}{1 - \sin{\theta}\,d\theta}$

$= a\left[\theta + \cos{\theta}\right]_0^{2\pi}$

$= a\left[\left(2\pi + \cos{2\pi}\right) - \left(0 + \cos{0}\right)\right]$

$= a\left(2\pi + 1 - 0 - 1\right)$

$= 2\pi a$ .

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