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

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- Feb 11th 2010, 07:15 PMyoman360area 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 - Feb 11th 2010, 07:49 PMyoman360
$\displaystyle \int$

- Feb 12th 2010, 02:40 AMHallsofIvy
I don't think that was quite his question!(Rofl)

The area of a circular sector of radius r and angle $\displaystyle \theta$ 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 $\displaystyle \frac{1}{2}\int_{-\pi}^\pi r^2 d\theta$.