Green's Theorem on a Polar Rose

[wonderinglady]

The problem is this:
Use Green's Theorem to express the area inside one petal (the one in the first quadrant) of the polar rose r= sin(3 Theta) as a line integral, then evaluate to compute the area.

I set it up as (with t's here for the thetas for brevity)
1/2 the integral from 0 to pi/3 of [(-sin(3t)sin(t))*(cos(2t)+2cos(4t))+(sin(3t)cos(t))* (sin(2t)-2sin(4t))] dt

I thought I did it correctly, but I ran it though Mathmatica and got 0, which both makes sense and doesn't. It seems wrong to have the area be 0, but if it's graphed in rectangular coordinates half the area is above the x-axis and half is below.

So the question is, did I mess something up?

 

[Deleted member 4993]

The problem is this:
Use Green's Theorem to express the area inside one petal (the one in the first quadrant) of the polar rose r= sin(3 Theta) as a line integral, then evaluate to compute the area.

I set it up as (with t's here for the thetas for brevity)
1/2 the integral from 0 to pi/3 of [(-sin(3t)sin(t))*(cos(2t)+2cos(4t))+(sin(3t)cos(t))*(sin(2t)-2sin(4t))] dt

I thought I did it correctly, but I ran it though Mathmatica and got 0, which both makes sense and doesn't. It seems wrong to have the area be 0, but if it's graphed in rectangular coordinates half the area is above the x-axis and half is below.

So the question is, did I mess something up?

The polar plot should look like above - not "half the area is above the x-axis and half is below."

 

[wonderinglady]

Ah, right, I'm sorry. I was thinking about the sin function by itself, trying to figure out if 0 was a reasonable answer, but you're right that that doesn't actually apply here.

That just takes me back to square one though. 0 doesn't seem like an acceptable answer for area, so I obviously went wrong in my set up.

 

[galactus]

Area using Green's Theorem is given by:

\(\displaystyle \displaystyle \frac{1}{2}\oint_{C}-ydx+xdy\)

One petal is swept out at \(\displaystyle \theta=\frac{\pi}{3}\)

\(\displaystyle \displaystyle y=\cos(3t)\cos(t), \;\ x=\cos(3t)\sin(t)\)

\(\displaystyle \displaystyle dx=\cos(t)\cos(3t)-3\sin(t)\sin(3t)\)

\(\displaystyle \displaystyle dy=-\sin(t)\cos(3t)-3\cos(t)\sin(3t)\)

So, \(\displaystyle \displaystyle -ydx+xdy=-(\cos(3t))^{2}\)

\(\displaystyle \displaystyle \frac{1}{2}\int_{0}^{\frac{\pi}{3}}\cos^{2}(3t)dt=\frac{\pi}{12}\)

The equations above give a rose that is rotated, but essentially the same as \(\displaystyle r=\sin(3\theta)\).

If this is done in polar, then the area should be the same:

\(\displaystyle \displaystyle \frac{1}{2}\int_{0}^{\frac{\pi}{3}} \sin^{2}(3\theta)d\theta=\frac{\pi}{12}\)

It's been a while since I messed with this particular stuff. Check me out.