Polar Curves stuff

**leinadwerdna** · April 27th 2009, 04:37 PM

find the area of the region inside one loop of the lemniscate r^2=4sin2theta

**Soroban** · April 27th 2009, 05:05 PM

Hello, leinadwerdna!

Find the area of the region inside one loop of the lemniscate: $r^2\:=\:4\sin2\theta$

We really have very little do . . .

Formula: . $A \;=\;\tfrac{1}{2}\int^{\beta}_{\alpha} r^2\,d\theta$

If you recognize the curve or can make a sketch, you see that
. . it is a figure-8 tipped so the loops are in Quadrants 1 and 3.

We already have $r^2$ . . . all we need are the limits.

Set $r = 0\!:\quad 4\sin2\theta \:=\:0\quad\Rightarrow\quad\sin2\theta \:=\:0 \quad\Rightarrow\quad 2\theta \:=\:0,\:\pi \quad\Rightarrow\quad \theta \:=\:0,\:\tfrac{\pi}{2}$

There! . . . $A \;=\;\tfrac{1}{2}\int^{\frac{\pi}{2}}_0 4\sin2\theta\,d\theta$

Got it?

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