# area enclosed by

• February 11th 2010, 07:15 PM
yoman360
area enclosed by
r=1+2sin(6θ)

here's a graph of the equation: http://www.wolframalpha.com/input/?i=polar+1%2B2sin(6x)

I don't know how to set up the integral
• February 11th 2010, 07:49 PM
yoman360
$\int$
• February 12th 2010, 02:40 AM
HallsofIvy
Quote:

Originally Posted by yoman360
$\int$

I don't think that was quite his question!(Rofl)

The area of a circular sector of radius r and angle $\theta$ is [tex]\frac{1}{2}r^2\theta[/itex]. For an "infinitesmal" angle, we can think of the radius of any curve as constant and so the "differential of area", in polar coordinates, is [tex]\frac{1}{2}r^2 d\theta[/itex].

You want $\frac{1}{2}\int_{-\pi}^\pi r^2 d\theta$.