Green's Theorem to find Area help

In summary, the problem involves finding the area of the right leaf of the Lemniscate of Gerono parametrized by r(t) = <sin(t), sin(t)cos(t)> from 0 to pi. The solution involves using Green's theorem and finding the bounds of integration. After performing a dot product and integration, the final answer for the area is found to be 2/3. Alternatively, the bounds of integration can be changed to go from 0 to pi/2 and the final answer can be multiplied by 2 to get the same result of 2/3 for the area.
  • #1
RJLiberator
Gold Member
1,095
63

Homework Statement


Find the area of the right leaf of the Lemniscate of Gerono (the ∞ sign, see figure below) parametrized by
r(t)= <sin(t), sin(t)cos(t)>
from 0=<t=<pi

Picture is uploaded.

Homework Equations


Green's theorem: integral of fdx+gdy = double integral (over the region) of (gx-fy) dA
Green's theorem used to compute the area of R = double integral (over the region R) of 1dA

The Attempt at a Solution


1. First, I found the curl, I let g=x and f=0 to make the curl (gx-fy) =1.
So now this is in Green's Theorem area form.
2. Now I need to find the bounds of integration.
I am stuck here and not entirely sure what to do. Any guidance on how to proceed?
 

Attachments

  • Screen Shot 2015-04-22 at 7.13.46 AM.png
    Screen Shot 2015-04-22 at 7.13.46 AM.png
    44.9 KB · Views: 485
Physics news on Phys.org
  • #2
I've been working on this problem this morning.

Is the answer for area simply 0? It seems logical as the postive y area and the negative y area would cancel each other out.

What I did was use:
F = <0, sin(t)>
r'(t) = <cos(t), -sin^2(t)+cos^2(t)>

I then performed a dot product between them and took the integral with t going from 0 to pi as the problem stated.

I ended up after integration with
-(2/3)cos^3(t)+cos(t) from t=0 to t=pi to equal -2/3
The answer then would just be 2/3 = area?

Correct?
 
Last edited:
  • #3
Could I just go from t =0 to t=pi/2
then multiply the answer by 2, to get 2/3 as the area?
 
Last edited:

Related to Green's Theorem to find Area help

What is Green's Theorem?

Green's Theorem is a mathematical tool that relates the 2-dimensional area of a region to the line integral along the boundary of that region. It is a special case of the more general Stokes' Theorem.

How does Green's Theorem help find area?

Green's Theorem allows us to find the area of a region by evaluating a line integral along the boundary of the region. This is useful because sometimes it is easier to calculate the line integral rather than the area directly.

What is the formula for Green's Theorem?

The formula for Green's Theorem is: ∫∫(∂Q/∂x - ∂P/∂y) dA = ∫C Pdx + Qdy, where P and Q are functions of x and y, and C is the boundary of the region.

What are the requirements for using Green's Theorem?

To use Green's Theorem, the region must be simply connected (meaning it has no holes or self-intersections) and the boundary must be piecewise smooth (meaning it can be broken into a finite number of smooth curves).

What are some real-life applications of Green's Theorem?

Green's Theorem has many applications in physics, engineering and other fields. For example, it can be used to calculate the area of a 2-dimensional surface, the work done by a conservative force in a closed loop, or the flux of a vector field through a region in 2-dimensional space.

Similar threads

  • Calculus and Beyond Homework Help
Replies
3
Views
641
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
4
Views
960
  • Calculus and Beyond Homework Help
Replies
3
Views
1K
  • Calculus and Beyond Homework Help
Replies
12
Views
1K
  • Calculus and Beyond Homework Help
Replies
1
Views
2K
  • Calculus and Beyond Homework Help
Replies
8
Views
3K
  • Calculus and Beyond Homework Help
Replies
1
Views
996
  • Calculus and Beyond Homework Help
Replies
6
Views
2K
  • Calculus and Beyond Homework Help
Replies
2
Views
2K
Back
Top