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Stokes Thm issues

  1. Jan 4, 2008 #1

    I am having some issues with this problem...

    F=( x+y, y+z, z+x) bounded by the plane with vertices at {2,0,0},{0,2,0},{0,0,2}
    I need to do both sides of stokes thm and im running into problems when I try and set up the intergral because the (curl F) comes to be (-1i - 1j) and I cant seem to get beyond that.
    Any one that can help, plaese do. Thank you very much.
  2. jcsd
  3. Jan 4, 2008 #2


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    First of all, you have "curl F" ([itex]\nabla\times F[/itex]) wrong: it is [itex]-\vec{i}-\vec{j}-\vec{k}[/itex]. (It should be clear from the symmetry of F that it couldn't be what you have.)

    Stokes Theorem says:
    [tex]\int_S\int [(\nabla \times \vec{v})\cdot \vec{n}]d\sigma= \oint_C\vec{r}\cdot dr[/tex]
    Here, the Surface S is x+ y+ z= 2, which is the same as z= 2- x- y. Using x and y as parameters, we can write its "vector" form as [itex]\vec{r}= x\vec{i}+ y\vec{j}+ (2- x- y)\vec{k}[/itex]. The "fundamental vector product" is [itex]\vec{r}_x\times\ve{r}_y[/itex] which is [itex]\vec{i}+ \vec{j}+\vec{k}[/itex]. If we want it oriented upward (to do that surface integral, we must specify an orientation and you don't mention it), that would be [itex]\vec{i}+ \vec{j}+ \vec{k}[/itex]. It is a vector normal to the surface whose length is the differential of area: itex]d\sigma[/itex] is the length of that times dxdy.

    Actually, I prefer to write [itex]\vec{v}\cdot\vec{n} d\sigma[/itex] as [itex]\vec{v}\cdot d\vec{\sigma}[/itex] where [itex]d\vec{\sigma}[/itex] is the vector normal to the surface having [itex]d\sigma[/itex] as length. That is simply the "fundamental vector product" times the differentials of the parameters. Here that is simply [itex](\vec{i}+ \vec{j}+ \vec{k})dxdy[/itex]/ Since [itex]\nabla\times F= \vec{i}+ \vec{j}+ \vec{k}[/itex], [itex]\nabla\times F \cdot d\vec{\sigma}= -3dxdy[/itex].

    Now S projects down to the xy-plane as the triangle with vertices at (0,0), (2, 0), and (0,2). The sides of that are x= 0, y= 0, and x+ y= 2 or y= 2- x. To integrate over that, taking x as the "outside" variable, we must take x from 0 to 2 and, for each x, y from 0 to 2- x:
    [tex]\int_{x=0}^2\int_{y=0}^{2-x} 3 dydx[/itex]
    That is, of course, 3 times the area of the triangle so you really don't need to do any integration at all.

    To do the path integral, break the boundary into three parts, x= 0, y= 0 and y= 2- x. Can you do that?
  4. Jan 4, 2008 #3
    I think I understand it now.

    I dont think I know how to set it into 3 parts. That was my next problem.
  5. Jan 4, 2008 #4
    any more help on this?? i cant seem to connect all the dots.

    how do you find the normal vector?
    I think it is x+y+z? (i+j+k)?

    is the tangent vector the same??

    do I need to change parameters?
    I dont think that i do for such a simple problem.
  6. Jan 5, 2008 #5


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    The tangent vector to the plane is not the same as the normal vector to it. Clearly they are perpendicular to one another. And what do you mean set it into 3 parts? And if by "change parameters" you mean change from Cartesian coordinates to cylindrical, spherical or some other fancy coordinate system, then the answer is no. Doing so only complicates the matter a lot more.

    To find the normal vector you need to first be able to see that a line drawn along the direction of i+j+k from the origin through the surface will clearly be normal to it. Can you picture it?

    As Halls stated, you have to first specify the orientation of the surface. Look at the question. It'll help if you state the question here explicitly. In fact for all surface integrals involving stokes theorem, you always have to specify the orientation of the surface. Here's how to see it intuitively:

    Imagine you are walking along the closed boundary of the surface, which is usually a closed curve such as an ellipse, circle, triangle etc. If you are walking in a anticlockwise direction, the normal vector to the surface will point in the same direction as your head.
  7. Jan 6, 2008 #6


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    Your figure is a triangle with vertices at {2,0,0} , {0,2,0}, and {0,0,2}. Since the boundary is three lines and is not "smooth" we separate it into the three individual lines.

    Assuming that the plane was oriented upwards, "counterclockwise" around the triangle would be 1) from {2, 0, 0} to (0, 2, 0); 2) from (0, 2, 0) to (0, 0, 2); 3) from (0, 0, 2) to (2, 0, 0).

    1) From (2, 0, 0) to (0, 2, 0). We can write that line as x= 2- t, y= t, z= 0 with t ranging from 0 to 2 (x(0)= 2, x(2)= 0, y(0)= 0, y(2)= 2) so that dx= -dt, dy= 2dt, dz= 0. The integral of a vector function [itex]\vec{F}(x,y,z)= f(x,y,z)\vec{i}+ g(x,y,z)\vec{j}+ h(x,y,z)[/itex] on that line would be
    [tex]\int_0^2 [f(2-t, 2t, 0)+ g(2-t),t,0)+h(2-t,t,0)]dt[/tex]

    In your particular problem, f= x+ y so f(x,y,z)= 2-t+ t= t, g= y+ z so g(x,y,z)= t+ 0, and h= z+ x so h(x,y,z)= 0+ 2- t. The integral, from (2, 0, 0) to (0, 2, 0) would be
    [tex]\int_0^2 2t dt= 4[/itex]

    For the line from (0, 2, 0) to (0, 0, 2), x= 0, y= 2- t, z= t with t ranging from 0 to 2.

    For the line from (0, 0, 2) to (2, 0, 0), x= t, y= 0, z= 2- t with t ranging from 0 to 2.
  8. Jan 6, 2008 #7
    due at 7am. please help!!!

    how can you just change the sign like this??

    just to double check myself when i complete the intergral above i get... zero..
    but for the path intergral you get 4?? how? am i doing something wrong??
    Last edited: Jan 6, 2008
  9. Jan 6, 2008 #8

    So f(x,y,z)=2-t+t ....which equals 2...right??

    thus making the intergal...

    [tex]\int_0^2 2 dt= ????[/itex]
    Last edited by a moderator: Jan 7, 2008
  10. Jan 7, 2008 #9


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    Yes, of course, sorry about that.

    I did make a mess of that!
    f(x,y,z)= 2 and dx= 2dt, so f(x,y,z)= 4dt; g(x,y,z)= 2t and dy= 2 dt so g(x,y,z)dy= 4dt, h(x,y,z)= 2- 2t and dz= -2dt so h(x,y,z)= (2-t)(-2dt)= (2t-4)dt, thus making the integral
    [tex]\int_0^2 4dt+ 4tdt+ (2t- 4)dt= \int_0^2 (4+ 4+ 2t-4)dt= \int_0^2 (4+ 2t) dt[/tex].
    Of course, that's just the integral from (2, 0, 0) to (0, 2, 0).
    Last edited: Jan 7, 2008
  11. Jan 7, 2008 #10


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    Every surface has two "orientations", depending upon which direction you take the normal vector. For example, for the plane z= 0, you take the unit normal vector at a point to be either [itex]\vec{k}[/itex] or [itex]-\vec{k}[/itex]: "oriented with upward normals" or "oriented with downward normals". The paraboloid [itex]z= x^2+ y^2[/itex] can be "oriented with inward normals", normals pointing inside the paraboloid, or "oriented with outward normals", normals pointing outside the paraboloid.

    An orientation of a surface implies an orientation of a close path in the surface by the "right hand rule". If the thumb of your right hand points in the direction of the normal, you fingers will curl in the direction of the orientation on the path.

    In some problems, the orientation is given. Here, an example of Stoke's theorem, it doesn't matter as long as you use the same orientation for both sides of the equation. If you take the normal vector to be [itex]\vec{i}+ \vec{j}+ \vec{k}[/itex], the surface is "oriented upward". Curling the fingers of your right hand from (2, 0, 0) to (0, 2, 0) to (0, 0, 2) to (2, 0, 0), your thumb would be pointing "upward" so you should integrate around the triangle in that direction.

    If we used instead [itex]-\vec{i}-\vec{j}-\vec{k}[/itex] for the surface integral, then we must use (2, 0, 0) to (0, 0, 2) to (0, 2, 0) to (2, 0, 0) for the path integral. Changing orientations just changes the sign on both sides of Stokes' theorem so to show that Stokes' theorem works we can choose either orientation.

    I did have a typo in my original post: the integral should be
    [tex]\int_{x=0}^2\int_{y=0}^{2-x} 2 dydx[/itex]
    not "3 dy dx". Of course, that integral is just 2 times the area of the triangle: 2(1/2)(2)(2)= 4.
    Last edited: Jan 7, 2008
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