# Green's Theorem and Area

• Nov 20th 2008, 03:54 PM
saxyliz
Green's Theorem and Area
Use Green's theorem to compute the area of one petal of the 32-leafed rose defined by https://webwork3.asu.edu/webwork2_fi...98d574f011.png.
It may be useful for recall that the area of a region D enclosed by a curve C can be expressed as https://webwork3.asu.edu/webwork2_fi...ef87c6d231.png.

I'm pretty sure that theta is inbetween [0, pi/16] but that's really as far as I have gotten! Extensive help please!!
• Nov 20th 2008, 04:00 PM
elizsimca
Not sure if it's just me, but I can't see your equations.
• Nov 20th 2008, 04:10 PM
saxyliz
Quote:

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 $r=13sin(16theta)$
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).
• Nov 20th 2008, 08:09 PM
elizsimca
Quote:

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 $r=13sin(16theta)$
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 $\int_0^\frac{\pi}{16}\int_0^{13sin(16\theta)}rdrd\ theta$.

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 $\theta$ step to $\frac{\pi}{256}$, $\theta$min to 0 and $\theta$max to $\frac{\pi}{16}$ 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 :)
• Nov 20th 2008, 09:06 PM
saxyliz
Quote:

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 $\int_0^\frac{\pi}{16}\int_0^{13sin(16\theta)}rdrd\ theta$.

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 $\theta$ step to $\frac{\pi}{256}$, $\theta$min to 0 and $\theta$max to $\frac{\pi}{16}$ 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! :D
• Nov 20th 2008, 10:03 PM
Chris L T521
Quote:

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 $r=13sin(16theta)$
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).

Quote:

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 $\int_0^\frac{\pi}{16}\int_0^{13sin(16\theta)}rdrd\ theta$.

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 $\theta$ step to $\frac{\pi}{256}$, $\theta$min to 0 and $\theta$max to $\frac{\pi}{16}$ 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,

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

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 $\vartheta$, I end up with the same integral by applying GT to the line integral:

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

--Chris
• Nov 20th 2008, 10:19 PM
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 $\sigma$. 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!
• Nov 20th 2008, 10:23 PM
Chris L T521
Quote:

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 $\sigma$. 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:

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

--Chris
• Nov 20th 2008, 10:25 PM
elizsimca
Quote:

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

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

--Chris

Yep, I meant curl F dot k dA :)