Line Integral, Green's Theorem

[Angello90]

Homework Statement
$$\int_{C} (xy^{2}-3y)dx + x^{2}y dy$$
G is finite region enclosed by:
$$y=x^{2}$$

$$y=4$$

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

The attempt at a solution
$$Q = x^{2}y$$

$$dQ/dx = 2xy$$

$$P = xy^{2}-3y$$

$$dP/dy = 2xy-3y$$

Limits to integral are:
$$from x = - \sqrt{y} to \sqrt{y}$$

$$from y = 0 to 4$$

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

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

 

[hunt_mat]

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:

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

Likewise for the other part of the integral, use y=x^2 and hence dy=2xdx

 

[Angello90]

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?

 

[HallsofIvy]

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), $a\le t\le b$, You determine that dx= u' dt and dy= u' dt so that your integral becomes
$$\int_a^b f(u(t), v(t))u'(t)dt+ g(u(t),v(t))v'(t)dt$$
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).

 

[hunt_mat]

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.

 

[Angello90]

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:
$$\int_{C} (xy^{2}-3y)dx + x^{2}y dy=\int_{-2}^{2}16x-12dx = -48$$
And for y=x^2; dy=2xdx
$$\int_{C} (xy^{2}-3y)dx + x^{2}y dy=\int_{-2}^{2}x(x^{2})^{2}-3(x^{2})dx + x^{2}(x^{2})2xdx = \int_{-2}^{2}3x^{5} - 3x^{2} dx = -16$$

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.

 

[HallsofIvy]

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.

 

[hunt_mat]

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.

 

[Angello90]

That's what I was thinking. Was confused cause it is usual donated it by
$$\oint$$
isn't it? Thanks a million guys! So it turns out to be fairly simple example!

Cheers
Angello

 

[hunt_mat]

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.