Question

Find the area enclosed by the curve r^2=9sin(2theta).

Answer 1

is it some kind of rose?

Answer 2

have you seen Green's Theorem?

Answer 3

The blue curve is \(r=\sqrt{9\sin2\theta}\), with \(0\le t\le\dfrac{\pi}{2}\).

Answer 4

\[A=\frac{ 1 }{ 2 } \int\limits xdy - ydx\]

Answer 5

So what I have right now is A=1/2 integral of 9sin(2theta) because we know that r is already squared.

Because we know that even multiples of theta = twice that of # of loops (a flower of 4 petals)... Then here is where I get confused because there are only two loops and not 4

I don't where from what angles to integrate

Answer 6

or use \[A = \int\limits_{0}^{\frac{ \Pi }{ 2 }} \int\limits_{0}^{\sqrt{9\sin(2 \Theta)}}rdrd \]

Answer 7

or you could use something like this instead. could be more clear

Answer 8

to find the angles , you just solve to when r=0, in this case its when sin(2theta) =0, thus at 0 and pi/2

Answer 9

If you're using the positive root for the \(r\)-integral, you'll need to double the result to account for the area of the petal in the third quadrant.

Answer 10

Ohh!!! Ok Thank you so much guys!!! That makes sense now! :)