Line Integral, Green's Theorem

In summary, the conversation is about verifying Green's Theorem by evaluating a double integral and a line integral. The content discusses the necessary steps to do so, including finding the limits of the integral, splitting the integral into two parts, and being careful about the direction of integration. The final answer is -32.
  • #1
Angello90
65
0
Homework Statement
[tex]\int_{C} (xy^{2}-3y)dx + x^{2}y dy[/tex]
G is finite region enclosed by:
[tex]y=x^{2}[/tex]

[tex]y=4[/tex]

C is boundary curve of G. Verify Green's Theorem by evaluating double integral and line integral.

The attempt at a solution
[tex]Q = x^{2}y[/tex]

[tex]dQ/dx = 2xy[/tex]

[tex]P = xy^{2}-3y[/tex]

[tex]dP/dy = 2xy-3y[/tex]

Limits to integral are:
[tex]from x = - \sqrt{y} to \sqrt{y}[/tex]

[tex]from y = 0 to 4[/tex]

Thus integral is:
[tex]\int_{G} dQ/dx - dP/dy dA = \int_{G} 3 dA[/tex]

Therefore Green's Theorem gives me: 32

How the hell do I do Line integral? Spend entire day looking it up, but examples on internet uses C which is given in a nice polar form. Where in here I don't have such a nice form - I think. I would assume that dy = 4 (from 0 to 4) but than I don't know what would be dx.

Simply confused! Any hints guys?

Thanks
Angello
 
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  • #2
Hmm, what I would do here is split the integral into two, the part along y=4 and the part along y=x^2, For the line y=0, the line integral would transform into dy=0 and so:

[tex]
\int_{C} (xy^{2}-3y)dx + x^{2}y dy=\int_{-2}^{2}16x-12dx
[/tex]

Likewise for the other part of the integral, use y=x^2 and hence dy=2xdx
 
  • #3
Ok. Could you explain that slower? So it's like, you find area under the rectangle x = -2,2 y= 0,4, and than the area under the paraboloid y=x^2 again from x= -2,2 and y= 0,4. Than subtract both answers?
 
Last edited:
  • #4
No, there is no "area" involved- you are integrating along a path, not over a region.

To integrate f(x,y)dx+ g(x,y)dy over a path given by x= u(t), y= v(t), [itex]a\le t\le b[/itex], You determine that dx= u' dt and dy= u' dt so that your integral becomes
[tex]\int_a^b f(u(t), v(t))u'(t)dt+ g(u(t),v(t))v'(t)dt[/tex]
I am sure that is what the examples you saw did. (You say you found them on the internet. Does your textbook not have examples? Not to mention a definition of "path integral".

The examples you saw did NOT use polar coordinates as you seem to think. I suspect that the path was a circle or part of a circle so they use the "standard" parameterization for a circle with radius R, centered at the origin, x= cos(t), y= sin(t).
 
  • #5
As HallsofIvy has said, you're integrating along a path, the path i presented to you was the path y=4, you have to find where that intersects with the other path, so you set x^2=4 which shows that the intersection points are -2 and +2. So you substitute y=4 and dy=0 in your contour integral to get the integral along that particular segment of the path.
 
  • #6
I don't have any textbook, just lectures notes, who didn't cover the material, as he was sick, but still expect us to know it.

Ok so in my case I have one path which can be split into two, the paraboloid path y=x^2 and line y=4 yes? Than I just add both?

But I get different answer. The path y=4 I get:
[tex]\int_{C} (xy^{2}-3y)dx + x^{2}y dy=\int_{-2}^{2}16x-12dx = -48[/tex]
And for y=x^2; dy=2xdx
[tex]\int_{C} (xy^{2}-3y)dx + x^{2}y dy=\int_{-2}^{2}x(x^{2})^{2}-3(x^{2})dx + x^{2}(x^{2})2xdx = \int_{-2}^{2}3x^{5} - 3x^{2} dx = -16[/tex]

I should add them but this gives me -64. Why is that? Should I get 48 rather than -48. Than I would get 32, as Green's result.
 
  • #7
Be careful about the direction of integration. Green's theorem requires that you integrate around a closed path counter-clockwise. On the parabola, that means you are integrating from x= -2 to x= 2, but on the line y= 4, to complete the path, you have to go from x= 2 to x= -2, reversing the sign of the integral.
 
  • #8
You have to integrate from 2 to -2 on your second integral as you're integrating around a path...

So the total integral should be -48+16=-32. If you integrated in the opposite direction (I did clockwise) then your answer would have become 32.
 
  • #9
That's what I was thinking. Was confused cause it is usual donated it by
[tex]\oint[/tex]
isn't it? Thanks a million guys! So it turns out to be fairly simple example!

Cheers
Angello
 
  • #10
It is, I use this notation when I write countour integrals down, and it is a more applied mathematicians approach, the pure mathematician will just write it as your teacher did.

Glad we could be of help.
 

What is a line integral?

A line integral is a mathematical concept that allows us to calculate the total value of a function along a given curve in a two-dimensional or three-dimensional space. It is represented by the symbol ∫ and can be thought of as the area under a curve.

What is Green's Theorem?

Green's Theorem is a fundamental theorem in vector calculus that relates a line integral around a simple closed curve to a double integral over the region enclosed by the curve. It states that the line integral of a two-dimensional 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.

What is the difference between a line integral and a surface integral?

A line integral is used to calculate the total value of a function along a given curve, while a surface integral is used to calculate the total value of a function over a given surface. Line integrals are one-dimensional, while surface integrals are two-dimensional.

How is Green's Theorem used in real-world applications?

Green's Theorem has many practical applications, including calculating work done by a force along a path, calculating the flow of a fluid through a region, and determining the area of a region bounded by a curve. It is also used in electromagnetic theory and in physics to calculate electric and magnetic fields.

What are some common misconceptions about Green's Theorem?

One common misconception about Green's Theorem is that it only applies to simple closed curves. In reality, it can be applied to any closed curve, regardless of its complexity. Another misconception is that Green's Theorem is only applicable in two-dimensional space. However, it can also be extended to three-dimensional space in the form of the Divergence Theorem.

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