Green's Theorem and Line Integral

In summary, the conversation discusses how to evaluate the integral \ointxydx+x^2dy using two methods: directly and using Green's theorem. The direct method involves breaking the integral into four parts and evaluating each separately, while the Green's theorem method involves using the double integral of x with respect to y and x over the given region. The conversation also clarifies the concept of positive orientation in Green's theorem and how it affects the direction of the contour.
  • #1
EV33
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Homework Statement


[tex]\oint[/tex]xydx+x^2dy

C is the rectangle with vertices (0,0),(0,1),(3,0), and (3,1)

Evaluate the integral by two methods: (a) directly and (b) using green's theorem.

Homework Equations

The Attempt at a Solution


Evaluating the integral directly:

c1: y=0,x=t,dx=dt,dy=o {0[tex]\leq[/tex]t[tex]\leq[/tex]3}
c2: x=3, y=t, dx=0, dy=dt {0[tex]\leq[/tex]t[tex]\leq[/tex]1}
c3: y=1, dy=0, x=t, dx=dt {0[tex]\leq[/tex]t[tex]\leq[/tex]3}
c4: x=0, dx=0, y=t, dy=dt {0[tex]\leq[/tex]t[tex]\leq[/tex]1}

So I got c1 and c4 being the integral of zero which is just zero.

Then for c2 and c3...

[tex]\int[/tex]9dt {t:0[tex]\leq[/tex]t[tex]\leq[/tex]1} +[tex]\int[/tex]tdt {t:0[tex]\leq[/tex]t[tex]\leq[/tex]3}=27/2Then trying to use green's theorem:
[tex]\int[/tex][tex]\int[/tex]xdydx=[tex]\int[/tex]xdx=9/2

{y:0[tex]\leq[/tex]y[tex]\leq[/tex]1}
{x:0[tex]\leq[/tex]x[tex]\leq[/tex]3}

I am not sure where I messed up but I know I did because both of my answers should be the same. If someone could point me in the right direction, it would be much appreciated. Thank you for your time.
 
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  • #2
Sorry if it looks a little confusing. I didn't intend on my function being where my top limit of integration would be.
 
  • #3
You need to take into account the direction of the contour. You want to enclose the area going in the counter-clockwise direction.
 
  • #4
I don't really understand the positive orientation part of green's theorem. How do you even know what direction it is oriented?
 
  • #5
If you have a region enclosed by a contour, the convention is that the positive orientation is when the contour encloses the region in the counter-clockwise direction. It's similar to the convention we use that a positive angle is measured going in the counter-clockwise direction from the x-axis. There's nothing inherently positive about it. It's just that everyone agrees that that's what they mean by "positive."

In this problem, if you start at the origin, the positive sense of the contour would be going from (0,0) to (3,0) to (3,1) to (0,1) and finally back to the origin. So when you calculate C3, you have to go from (3,1) to (0,1), otherwise your answer for that leg comes out with the opposite sign and you don't get the cancellation you need.
 
  • #6
Ok. That makes perfect sense now. Thank you so much.
 

1. What is Green's Theorem and why is it important?

Green's Theorem is a fundamental concept in multivariable calculus that relates a line integral around a simple closed curve to a double integral over the region enclosed by the curve. It is important because it provides a way to evaluate the area of a region in a more efficient way than traditional methods, and it has many applications in physics, engineering, and other scientific fields.

2. How is Green's Theorem used to calculate line integrals?

Green's Theorem states that the line integral of a vector field around a simple closed curve is equal to the double integral of the curl of the vector field over the region enclosed by the curve. This means that instead of calculating a line integral directly, we can convert it into a double integral and use the properties of double integrals to evaluate it.

3. What is the relationship between Green's Theorem and Stokes' Theorem?

Green's Theorem is a special case of Stokes' Theorem, which is a more general theorem that relates line integrals to surface integrals. Green's Theorem applies specifically to 2D vector fields, while Stokes' Theorem applies to 3D vector fields. In essence, Green's Theorem can be thought of as a 2D version of Stokes' Theorem.

4. Can Green's Theorem be used for non-simple closed curves?

No, Green's Theorem only applies to simple closed curves, which are curves that do not intersect themselves and enclose a single, connected region. If the curve is not simple, then the theorem cannot be applied. In this case, other methods such as line integrals or the generalized version of Green's Theorem may be used.

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

Green's Theorem has many practical applications in physics, engineering, and other scientific fields. It can be used to calculate the work done by a conservative force, to find the center of mass of a 2D object, and to evaluate flux through a closed curve, among other things. It is also used in fluid mechanics to calculate the circulation of a fluid around a closed curve.

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