Green's theorem: region inside 8 leaved petal

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Homework Help Overview

The problem involves using Green's theorem to compute the area of one petal of an 8-leafed rose defined by the polar equation r=9sin(4θ). Participants discuss the conversion of polar coordinates to rectangular coordinates and the appropriate limits for integration.

Discussion Character

  • Exploratory, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants consider how to express x and y in terms of r and θ, and whether to convert to rectangular coordinates. There is discussion about the limits of integration, with suggestions of 0 to π/4. Some participants express uncertainty about the complexity of the resulting integral.

Discussion Status

There is ongoing exploration of the parameterization of the curve and the formulation of the integral. Some participants have provided guidance on the setup of the integral, while others note the challenges posed by its complexity. Multiple interpretations of the approach are being discussed.

Contextual Notes

Participants mention the use of computational tools like Maple or Mathematica for evaluating the integral, indicating a recognition of the integral's complexity. There is also a note of a shared experience in working through the problem, suggesting a collaborative learning environment.

Pete_01
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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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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.
 
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.
 
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) ?
 
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...
 
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?
 
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...
 
This should be your integral:

\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})

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.
 
Awesome got it! Thanks for all the help!
 
  • #10
nickmai123 said:
This should be your integral:

\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})

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.
 

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