Green's theorem: region inside 8 leaved petal

In summary, the area of one petal of an 8-leafed rose can be calculated using Green's Theorem. If you're having trouble remembering this theorem, you can use the polar formula. Finally, if you're trying to determine the limits of a curve, you would use the polar formula.
  • #1
Pete_01
51
0

Homework Statement


Use Green's theorem to compute the area of one petal of the 8-leafed rose defined by r=9sin(4theta)
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)int xdy-ydx.

Homework Equations


A =(1/2)int xdy-ydx


The Attempt at a Solution


I graphed the region on my calculator, but I am not sure what to use for x and y. Would the best way be to change r into rectangular coordinates? If so what would my limit be? 0 to pi/4? perhaps?

Thanks in advance!
 
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  • #2
Pete_01 said:

Homework Statement


Use Green's theorem to compute the area of one petal of the 8-leafed rose defined by r=9sin(4theta)
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)int xdy-ydx.

Homework Equations


A =(1/2)int xdy-ydx

The Attempt at a Solution


I graphed the region on my calculator, but I am not sure what to use for x and y. Would the best way be to change r into rectangular coordinates? If so what would my limit be? 0 to pi/4? perhaps?

Thanks in advance!

EDIT: I'm stupid, lol.
 
  • #3
Okay, so you're right with the 0 to pi/4. Now you just need to parameterize the curve. Think about finding the x and y coordinates of each of the r-vectors... Then you just integrate x(t)y'(t)dt and y(t)x'(t)dt. Hope that helps.
 
  • #4
Ok, so using the fact that x=rcostheta and y=rsintheta, would the parameter of x=9sin(4theta)cos(theta) and y=9sin(4theta)sin(theta) ?
 
  • #5
Pete_01 said:
Ok, so using the fact that x=rcostheta and y=rsintheta, would the parameter of x=9sin(4theta)cos(theta) and y=9sin(4theta)sin(theta) ?

That's correct. Now you just have to figure out your limits...
 
  • #6
So limits are 0 to pi/4 correct? When I write out the integral, its huge. Is there any trick to combining the terms together neatly?
 
  • #7
Pete_01 said:
So limits are 0 to pi/4 correct? When I write out the integral, its huge. Is there any trick to combining the terms together neatly?

Haha, yeah it's a pretty big integral. I guess this is why you wouldn't normally use Green's Theorem to find the area of a rose petal. You'd just use the polar formula...
 
  • #8
This should be your integral:

[tex]\frac{1}{2}(\int_{0}^{\pi/4}[{9sin(4t)cos(t)[9sin(4t)cos(t)+36cos(4t)sin(t)]]dt}-\int_{0}^{\pi/4}[{9sin(4t)sin(t)[-9sin(4t)sin(t)+36cos(4t)cos(t)]]dt})[/tex]

EDIT: Realistically, most people would use Maple, Mathematica, or a calculator to evaluate that monster. It doesn't look friendly even if there's something that can be done to shrink it down.
 
  • #9
Awesome got it! Thanks for all the help!
 
  • #10
nickmai123 said:
This should be your integral:

[tex]\frac{1}{2}(\int_{0}^{\pi/4}[{9sin(4t)cos(t)[9sin(4t)cos(t)+36cos(4t)sin(t)]]dt}-\int_{0}^{\pi/4}[{9sin(4t)sin(t)[-9sin(4t)sin(t)+36cos(4t)cos(t)]]dt})[/tex]

EDIT: Realistically, most people would use Maple, Mathematica, or a calculator to evaluate that monster. It doesn't look friendly even if there's something that can be done to shrink it down.

FTW:

This actually SIMPLIFIES to something much easier if you go through the distribution.
 
  • #11
keltix said:
FTW:

This actually SIMPLIFIES to something much easier if you go through the distribution.

Yeah I realized that after a few minutes but I didn't edit my post.

Also, lol @ the response 3 months later... :-)
 
  • #12
yea i know but i just spent like an hour going through this and i felt like i needed to share it with future WEBWORK kids.
 

1. What is Green's theorem?

Green's theorem is a mathematical theorem that relates the line integral of a vector field over a closed curve to a double integral over the region enclosed by the curve. It is named after the mathematician George Green.

2. How is Green's theorem applied to a region inside an 8 leafed petal?

In the case of a region inside an 8 leafed petal, Green's theorem can be used to calculate the area of the region by evaluating the double integral of the partial derivatives of the vector field over the region.

3. What is the significance of the 8 leafed petal in Green's theorem?

The 8 leafed petal is a specific shape used in mathematical examples to demonstrate the application of Green's theorem. It is chosen because it is a simple, closed, and symmetric shape that allows for easy calculation of the double integral.

4. Can Green's theorem be applied to other shapes besides the 8 leafed petal?

Yes, Green's theorem can be applied to any closed shape, as long as the region is well-defined and the vector field is continuous and differentiable over the region.

5. What are some practical applications of Green's theorem?

Green's theorem has many practical applications in physics, engineering, and other fields. It can be used to calculate work done by a force, fluid flow, and electric fields. It is also used in the study of electromagnetic fields and in the solution of differential equations.

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