# Thread: area of a lemniscate

1. ## area of a lemniscate

i'm having trouble with these type of problems. my professor didn't really go over them in class, so i'm kinda confused here. any help would be appreciated.

find the area of the region enclosed by r^2 = 9*sin(2t)

i'm guessing the equation would be set up as: 9 integral(sin(2t))dt

but what would my limits be?

2. Hello, tomobson!

Find the area of the region enclosed by: . $r^2 \:=\: 9\sin(2\theta)$

i'm guessing the equation would be set up as: . $9\int \sin(2\theta)\,d\theta$ . . . . um, not quite

but what would my limits be?
Did you sketch the curve? . . . It always helps.

It is a figure-8, centered at the origin, tipped 45° clockwise.
. . Note: There are two loops to consider.

The limits occur when $r = 0.$,
. . when the curve leaves and returns to the origin.

We have: . $r^2 \:=\:9\sin(2\theta) \:=\:0 \quad\Rightarrow\quad \sin(2\theta) \:=\:0$

. . Then: . $2\theta \:=\:0,\:\pi,\:2\pi,\:3\pi,\hdots \quad\Rightarrow\quad \theta \:=\:0,\:\tfrac{\pi}{2},\:\pi,\:\tfrac{3\pi}{2},\: \hdots$

So one loop is formed for: . $\theta = 0\,\text{ to }\,\theta = \tfrac{\pi}{2}$

The formula for polar area is: . $A \;=\;{\color{red}\frac{1}{2}}\!\int^{\beta}_{\alph a}\!\!\! r^2\,d\theta$

The area of one loop is: . $A \;=\;\frac{9}{2}\int^{\frac{\pi}{2}}_0 \sin(2\theta)\,d\theta$

For the total area of the lemiscate, multiply that result by 2.

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# area of lemniscate of bernoulli

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