Green's Thm: How am I not applying it correctly?

  • Thread starter math_maj0r
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In summary: I don't know what you're trying to say.The question did say that the curve is bounded by x = 0 and y = 0....so I don't know what you're trying to say.
  • #1
math_maj0r
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Curve consists of part of a circle x2 + y2 = 4
theta from 0 to pi/4

Vector field F = <y2-yx2 , yx2>

(let the i component be P, and the j component be Q)

I used the following formula: double integral of [(partial derivative of Q with respect to x) - (partial derivative of P with respect to y) dA]

So I got: double integral of [ (2xy) - (2y - x2) dy dx ] y from 0 to x, and x from 0 to 2

I got answer = 16/3 which is wrong.

My question is: What is wrong about the way I applied Green's thm?
 
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  • #2
Well, for one thing, your bounds of integration do not describe a circle.
 
  • #3
nicksauce said:
Well, for one thing, your bounds of integration do not describe a circle.

the curve doesn't consist of the whole circle. it is only part of the circle where theta is from 0 to pi/4
 
  • #4
Still, your bounds of integration describe a triangular region, not a circular region.
 
  • #5
math_maj0r said:
the curve doesn't consist of the whole circle. it is only part of the circle where theta is from 0 to pi/4

That's the whole point! Green's theorem says that
[tex]\int Mdx+ Ldy= \int\int \left(\frac{\partial L}{\partial x}- \frac{\partial M}{\partial y}\right)dxdy[/tex]
Where the path integral is over a closed path and the double integral is over the region bounded by that path. The arc of a circle from 0 to [itex]\pi/4[/itex] is not a closed path. And your double integral was over the triangle bounded by the lines y= 0, x= 2, and y= x which doesn't have anything to do with the given path! The lines y= 0 and y= x are okay but the curved side is NOT the straight line x= 2.

If you really want to use Green's theorem to find that integral you will have to add the line y= 0, from x=0 to 2 and the line y= x. To integrate over the area bounded by those and the arc of the circle, I would recommend changing to polar coordinates, taking [itex]\theta[/itex] from 0 to [itex]\pi/4[/itex] and r from 0 to 2. If you really want to use rectangular coordinates, you will need to do two integrals, the first with y from 0 to x and then x from 0 to [itex]\sqrt{2}[/itex], since y= x intersects [itex]x^2+ y^2= 4[/itex] at [itex](\sqrt{2}, \sqrt{2})[/itex]. The second integral would be with y from 0 to [itex]\sqrt{4- x^2}[/itex] and x from [itex]\sqrt{2}[/itex] to 2.

And, of course, the integral on that region will NOT be the path integral on the arc, it will be the path integral on the arc plus the path integrals on the two straight lines. To find the path integral on the arc, you would need to calculate the path integrals on the two straight lines separately and subtract them from the area integral.
 
  • #6
HallsofIvy said:
That's the whole point! Green's theorem says that
[tex]\int Mdx+ Ldy= \int\int \left(\frac{\partial L}{\partial x}- \frac{\partial M}{\partial y}\right)dxdy[/tex]
Where the path integral is over a closed path and the double integral is over the region bounded by that path. The arc of a circle from 0 to [itex]\pi/4[/itex] is not a closed path. And your double integral was over the triangle bounded by the lines y= 0, x= 2, and y= x which doesn't have anything to do with the given path! The lines y= 0 and y= x are okay but the curved side is NOT the straight line x= 2.

If you really want to use Green's theorem to find that integral you will have to add the line y= 0, from x=0 to 2 and the line y= x. To integrate over the area bounded by those and the arc of the circle, I would recommend changing to polar coordinates, taking [itex]\theta[/itex] from 0 to [itex]\pi/4[/itex] and r from 0 to 2. If you really want to use rectangular coordinates, you will need to do two integrals, the first with y from 0 to x and then x from 0 to [itex]\sqrt{2}[/itex], since y= x intersects [itex]x^2+ y^2= 4[/itex] at [itex](\sqrt{2}, \sqrt{2})[/itex]. The second integral would be with y from 0 to [itex]\sqrt{4- x^2}[/itex] and x from [itex]\sqrt{2}[/itex] to 2.

And, of course, the integral on that region will NOT be the path integral on the arc, it will be the path integral on the arc plus the path integrals on the two straight lines. To find the path integral on the arc, you would need to calculate the path integrals on the two straight lines separately and subtract them from the area integral.

The question did say that the curve is bounded by x = 0 and y = 0. I didn't mention it because I mistakenly thought it was obvious.

Anyway, so to convert [y2 - 2y + x2] dx dy into polar coordinates I did: x = 2cos(theta), y = 2sin(theta), then got: [r2-4sin(theta)]r dr d(theta) Is that incorrect? Because I'm not getting the right answer even though I integrated r from 0 to 2, and theta from 0 to pi/4
 
  • #7
math_maj0r said:
The question did say that the curve is bounded by x = 0 and y = 0. I didn't mention it because I mistakenly thought it was obvious.

Anyway, so to convert [y2 - 2y + x2] dx dy into polar coordinates I did: x = 2cos(theta), y = 2sin(theta), then got: [r2-4sin(theta)]r dr d(theta) Is that incorrect? Because I'm not getting the right answer even though I integrated r from 0 to 2, and theta from 0 to pi/4
No. [itex]x= r cos(\theta)[/itex] and [itex]y= r sin(\theta)[/itex], not "[itex]2cos(\theta)[/itex] and [itex]2sin(\theta)[/itex]. [itex]y^2- 2y+ ^2[/itex] becomes [itex]r^2- 2r sin(\theta)[/itex]. Your integral should be
[tex]\int_{\theta= 0}^{\pi/4}\int_{r= 0}^2 (r^2- 2r sin(\theta))r drd\theta[/tex]

And that double integral will NOT be the path integral over the curve because you have not taken into acount the two straight line boundaries of the region. That will be equal to the integral on the x-axis, from x= 0 to x= 2 plus the integral over the curve, plus the integral on y= x from x= [itex]\sqrt{2}[/itex] to 0.
 
  • #8
HallsofIvy said:
No. [itex]x= r cos(\theta)[/itex] and [itex]y= r sin(\theta)[/itex], not "[itex]2cos(\theta)[/itex] and [itex]2sin(\theta)[/itex]. [itex]y^2- 2y+ ^2[/itex] becomes [itex]r^2- 2r sin(\theta)[/itex]. Your integral should be
[tex]\int_{\theta= 0}^{\pi/4}\int_{r= 0}^2 (r^2- 2r sin(\theta))r drd\theta[/tex]

And that double integral will NOT be the path integral over the curve because you have not taken into acount the two straight line boundaries of the region. That will be equal to the integral on the x-axis, from x= 0 to x= 2 plus the integral over the curve, plus the integral on y= x from x= [itex]\sqrt{2}[/itex] to 0.

i can't figure out how to take into account the line segments from (0,0) to (2,0), and from (square root of 2, square root of 2) to (0, 0)

what's the most efficient way to set up the path integral using green's thm?
 
  • #9
I don't know what you mean by "set up the path integral using green's thm". You don't use Green's theorem to set up a path integral! You use it to give a hopefully simpler area integral equal to the path integral on a closed path. One difficulty I have with this is that you have never said what the original problem was!

If the problem is just to find the path integral you cite- the integral of [itex]<y^2-yx^2 , yx^2>[/itex] over the unit circle from [itex]\theta= 0[/itex] to [itex]2\pi[/itex], I see no reason to use Green's theorem at all.

If you must use Green's theorem then, like I said, you must take into account the straight line sides. On the x-axis, y= 0 so you can take x= t, y= 0, dx= dt, dy= 0. The vector function [itex]<y^2-yx^2 , yx^2>= < 0, 0>[/itex] so the integral is just 0.

On the line y= x, take x= t, y= t so dx= dy= dt. [itex]<y^2-yx^2 , yx^2>= <t^2- t^3, t^3>[/itex] so the integral is

[tex]\int_{t=\sqrt{2}}^0 (t^2- t^3)dt+ t^3dt[/tex]

[tex]= \int_{t= \sqrt{2}}^0[/tex][tex] t^2 dt[/tex]
 
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  • #10
thank you so much :)
 

1. How do I know when to use Green's Theorem?

Green's Theorem is used to evaluate the line integral around a closed curve in a two-dimensional plane. In order to use it, the curve must be simple and closed, and the region enclosed by the curve must be simply connected. In simpler terms, the curve cannot intersect itself and there cannot be any holes or gaps in the region enclosed by the curve.

2. What is the formula for Green's Theorem?

The formula for Green's Theorem is:
C(Mdx + Ndy) = ∬R ( ∂N/∂x - ∂M/∂y ) dA
where C is the closed curve, R is the region enclosed by the curve, and M and N are functions of x and y.

3. Can Green's Theorem be used for non-rectangular regions?

Yes, Green's Theorem can be used for any simply connected region, regardless of its shape. The important factor is that the curve must be closed and the region must be simply connected.

4. How do I know if I'm applying Green's Theorem correctly?

To apply Green's Theorem correctly, you must first identify the region and the closed curve. Then, you must find the functions M and N and their partial derivatives. Finally, you can plug these values into the formula and evaluate the double integral to find the line integral along the curve.

5. Can Green's Theorem be used in three-dimensional space?

No, Green's Theorem can only be used in two-dimensional space. In three-dimensional space, we use a similar theorem called Stokes' Theorem to evaluate line integrals along a closed curve.

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