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Stoke's theorem on a cube

  1. Dec 2, 2005 #1
    Hi, can someone help me through the following question.

    Q. Use Stoke's Theorem to evaluate [tex]\int\limits_{}^{} {\int\limits_S^{} {curl\mathop F\limits^ \to } } \bullet d\mathop S\limits^ \to [/tex]

    Here [tex]\mathop F\limits^ \to \left( {x,y,z} \right) = xyz\mathop i\limits^ \to + xy\mathop j\limits^ \to + x^2 yz\mathop k\limits^ \to [/tex] and S is the cube which consists of all sides apart from the bottom. Ie. S is a 'square hemisphere' without the bottom.

    I have a solution to this problem but I don't understand it. I think that I would benefit greatly if someone helps me by explaining what I should be looking for. Anyway here is what I've thought about so far.

    [tex]
    \int\limits_{}^{} {\int\limits_S^{} {curl\mathop F\limits^ \to } } \bullet d\mathop S\limits^ \to = \int\limits_C^{} {\mathop F\limits^ \to } \bullet d\mathop r\limits^ \to
    [/tex]

    In the above equation F is usually F(r(something)) and dr is r'(something). I usually need to parameterise the surface but in this case I don't think that is such an easy task. The solution considers two surfaces, one being S and the other being the bottom 'missing' face of the cube. A normal vector also comes into it. I must admit, this question has got me lost for ideas. I would appreciate any pointers to get me started.
     
  2. jcsd
  3. Dec 2, 2005 #2

    Galileo

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    Correct. That's what Stokes' theorem tells you. So instead of evaluating the flux of the curl of F through S, you evaluate the line integral of F along the boundary line C of S, which is the square formed by the four edges of the bottom of the cube.

    So the line integral consists of 4 straight line segments which are easily parametrized. I assume you know how to solve a line integral if you know the function and the curve along which to integrate.

    Another way is to realize that Stokes' theorem implies that the value of the integral is independent of the surface that is bounded by the given curve C. So maybe a different surface choice S' will make the calculation even easier.
     
    Last edited: Dec 2, 2005
  4. Dec 2, 2005 #3
    Thanks for the help Galileo. I managed to get an answer of zero which is the same as the book's answer - the solution provided is far more complicated than using the parameterisation you alluded to.
     
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