greene's theorem

[moy1989]

Hello, I'm having trouble getting started with the following problem:

Use Green's theorem to compute the area of one petal of the 24-leafed rose defined by r = 3sin(12theta)

It may be useful to recall that the area of a region D enclosed by a curve C can be expressed as A = (1/2) INTEGRAL( xdy - ydx).

I tried to draw a graph of r, but I don't know what to do with it?
Is y equal to 3sin(12theta)*sin(theta)?

help please.

 

[Deleted member 4993]

Hello, I'm having trouble getting started with the following problem:

Use Green's theorem to compute the area of one petal of the 24-leafed rose defined by r = 3sin(12theta)

It may be useful to recall that the area of a region D enclosed by a curve C can be expressed as A = (1/2) INTEGRAL( xdy - ydx).

I tried to draw a graph of r, but I don't know what to do with it?
Is y equal to 3sin(12theta)*sin(theta)?

help please.

Your given equation is in polar co-ordinates.

The expression of area is in cartesian co-ordinate.

You need to transform one into the other - choose your poison.

 

[galactus]

Green's theorem would just be:

\(\displaystyle \int_{0}^{\frac{\pi}{12}}\int_{0}^{3sin(12\theta)} r \;\ drd{\theta}\)

 

[mashadar]

Green's theorem would just be:

\(\displaystyle \int_{0}^{\frac{\pi}{12}}\int_{0}^{3sin(12\theta)} r \;\ drd{\theta}\)

Why is the angle going from 0 to pi/12 ? and not doing a full circle i.e. from 0 to pi/6 ?

thanks!