# Thread: Green's Theorem and Area

1. ## Green's Theorem and Area

Use Green's theorem to compute the area of one petal of the 32-leafed rose defined by .
It may be useful for recall that the area of a region D enclosed by a curve C can be expressed as .

I'm pretty sure that theta is inbetween [0, pi/16] but that's really as far as I have gotten! Extensive help please!!

2. Not sure if it's just me, but I can't see your equations.

3. Originally Posted by elizsimca
Not sure if it's just me, but I can't see your equations.
Hmm... all of the other times I've copied and pasted like that no one has ever said anything...

I'll do it the right way this time:

Use Green's theorem to compute the area of one petal of the 32-leafed rose defined by $\displaystyle r=13sin(16theta)$
It may be useful for recall that the area of a region D enclosed by a curve C can be expressed as A=(1/2)[integral_C xdy - ydx]

I don't know if you understand it now, the word theta is supposed to be the symbol, the second equation is supposed to represent the integral over C of xdy - ydx (a variant of green's theorem).

4. Originally Posted by saxyliz
Hmm... all of the other times I've copied and pasted like that no one has ever said anything...

I'll do it the right way this time:

Use Green's theorem to compute the area of one petal of the 32-leafed rose defined by $\displaystyle r=13sin(16theta)$
It may be useful for recall that the area of a region D enclosed by a curve C can be expressed as A=(1/2)[integral_C xdy - ydx]

I don't know if you understand it now, the word theta is supposed to be the symbol, the second equation is supposed to represent the integral over C of xdy - ydx (a variant of green's theorem).
Hm...

I guess I'm just confused on how to use GT in this context. Is there a vector field given? If not do we just assume the vector field is just <0,0,0>? If so, Green's Theorem would just simplify to finding the area of a region using a double integral. If this is the case, you should evaluate the double integral in polar coordinates. So, you will have $\displaystyle \int_0^\frac{\pi}{16}\int_0^{13sin(16\theta)}rdrd\ theta$.

This is the only thing I can figure out. I got the r limits from the equation given, and the theta limits by graphing in polar coords on my calculator. If you set your $\displaystyle \theta$ step to $\displaystyle \frac{\pi}{256}$, $\displaystyle \theta$min to 0 and $\displaystyle \theta$max to $\displaystyle \frac{\pi}{16}$ then you will see exactly one petal of that function.

This seems to be the only way to do it to me. Like I said, I'm not sure why they ask you to use GT...there might be something I'm missing. Hope this helps

5. Originally Posted by elizsimca
Hm...

I guess I'm just confused on how to use GT in this context. Is there a vector field given? If not do we just assume the vector field is just <0,0,0>? If so, Green's Theorem would just simplify to finding the area of a region using a double integral. If this is the case, you should evaluate the double integral in polar coordinates. So, you will have $\displaystyle \int_0^\frac{\pi}{16}\int_0^{13sin(16\theta)}rdrd\ theta$.

This is the only thing I can figure out. I got the r limits from the equation given, and the theta limits by graphing in polar coords on my calculator. If you set your $\displaystyle \theta$ step to $\displaystyle \frac{\pi}{256}$, $\displaystyle \theta$min to 0 and $\displaystyle \theta$max to $\displaystyle \frac{\pi}{16}$ then you will see exactly one petal of that function.

This seems to be the only way to do it to me. Like I said, I'm not sure why they ask you to use GT...there might be something I'm missing. Hope this helps
You are my savior!! Thank you so much!!
I totally understood that. I went in for help today and my calc teacher did something really complex and awful to solve it, I was so confused!
You're super awesome! Thanks so much for your time and patience!

6. Originally Posted by saxyliz
Hmm... all of the other times I've copied and pasted like that no one has ever said anything...

I'll do it the right way this time:

Use Green's theorem to compute the area of one petal of the 32-leafed rose defined by $\displaystyle r=13sin(16theta)$
It may be useful for recall that the area of a region D enclosed by a curve C can be expressed as A=(1/2)[integral_C xdy - ydx]

I don't know if you understand it now, the word theta is supposed to be the symbol, the second equation is supposed to represent the integral over C of xdy - ydx (a variant of green's theorem).
Originally Posted by elizsimca
Hm...

I guess I'm just confused on how to use GT in this context. Is there a vector field given? If not do we just assume the vector field is just <0,0,0>? If so, Green's Theorem would just simplify to finding the area of a region using a double integral. If this is the case, you should evaluate the double integral in polar coordinates. So, you will have $\displaystyle \int_0^\frac{\pi}{16}\int_0^{13sin(16\theta)}rdrd\ theta$.

This is the only thing I can figure out. I got the r limits from the equation given, and the theta limits by graphing in polar coords on my calculator. If you set your $\displaystyle \theta$ step to $\displaystyle \frac{\pi}{256}$, $\displaystyle \theta$min to 0 and $\displaystyle \theta$max to $\displaystyle \frac{\pi}{16}$ then you will see exactly one petal of that function.

This seems to be the only way to do it to me. Like I said, I'm not sure why they ask you to use GT...there might be something I'm missing. Hope this helps
Green's Theorem doesn't require vector fields! By definition,

$\displaystyle \int_Cf\left(x,y\right)\,dx+g\left(x,y\right)\,dy= \iint\limits_R\left(\frac{\partial g}{\partial x}-\frac{\partial f}{\partial y}\right)\,dA$

Where C is the contour path, and R is the region enclosed by the contour path.

If I'm not mistaken, using your values for $\displaystyle \vartheta$, I end up with the same integral by applying GT to the line integral:

$\displaystyle \tfrac{1}{2}\int_C -y\,dx+ x\,dy=\int_0^{\frac{\pi}{16}}\int_0^{13\sin(16\var theta)}r\,dr\,d\vartheta$

--Chris

7. Thanks Chris, that makes sense.

The definition for GT I have just relates the work integral to the double integral over R of curl F dot n d$\displaystyle \sigma$. I guess I just didn't think about finding area over a region as an application of GT...I would have just done in straight away using a double integral in polar coords.

Thanks for expanding my understanding of the application/usefulness of GT!

8. Originally Posted by elizsimca
Thanks Chris, that makes sense.

The definition for GT I have just relates the work integral to the double integral over R of curl F dot n d$\displaystyle \sigma$. I guess I just didn't think about finding area over a region as an application of GT...I would have just done in straight away using a double integral in polar coords.

Thanks for expanding my understanding of the application/usefulness of GT!
I think your mixing Green's Theorem up with Stokes Theorem, which states:

$\displaystyle \oint_C\mathbf{F}\cdot\mathbf{T}\,ds=\iint\limits_ {\sigma}\left(\text{curl}\,\mathbf{F}\right)\cdot\ mathbf{n}\,dS$

--Chris

9. Originally Posted by Chris L T521
I think your mixing Green's Theorem up with Stokes Theorem, which states:

$\displaystyle \oint_C\mathbf{F}\cdot\mathbf{T}\,ds=\iint\limits_ {\sigma}\left(\text{curl}\,\mathbf{F}\right)\cdot\ mathbf{n}\,dS$

--Chris
Yep, I meant curl F dot k dA