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Math Help - Stokes formula and the curvilinear integral of the second kind

  1. #1
    Senior Member DeMath's Avatar
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    Stokes formula and the curvilinear integral of the second kind

    Hello all!
    I can not solve a difficult task from the vector analysis:
    Using Stokes formula, calculate the curvilinear integral of the second kind: I = \int\limits_{OA} {yzdx + 3xzdy + 2xydz,}
    where OA = OB \cup BC \cup CA, curve x = t\cos t,{\text{ }}y = t\sin t,{\text{ }}z = {t^2},{\text{ }}0 \leqslant t \leqslant 2\pi ,{\text{ }}O = \left( {0,0,0} \right),{\text{ }}A = \left( {2\pi ,0,4{\pi ^2}} \right).

    I have two carefully (I thought) addressed this problem , but my answer is clearly not the same as the right one.
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  2. #2
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    Quote Originally Posted by DeMath View Post
    Hello all!
    I can not solve a difficult task from the vector analysis:
    Using Stokes formula, calculate the curvilinear integral of the second kind: I = \int\limits_{OA} {yzdx + 3xzdy + 2xydz,}
    where OA = OB \cup BC \cup CA, curve x = t\cos t,{\text{ }}y = t\sin t,{\text{ }}z = {t^2},{\text{ }}0 \leqslant t \leqslant 2\pi ,{\text{ }}O = \left( {0,0,0} \right),{\text{ }}A = \left( {2\pi ,0,4{\pi ^2}} \right).

    I have two carefully (I thought) addressed this problem , but my answer is clearly not the same as the right one.
    I'm not sure what your B and C is but I'm going to guess.

    The surface whose boundary is x = t \cos t,\; y = t \sin t,\; z = t^2 is given parametrically as

    \vec{r}(s,t) = <s \,t \cos t,\; s \, t \sin t,\; t^2>

    where 0 \le s \le 1 (see attached graph). The surface is blue, your boundary is in black and I think the B and C you mention are the red lines in the graph one horizontal and one vertical. Now the line integral is easy. Introducing your parametric equations directly into the line integral gives

     <br />
I = \frac{32 \pi^5}{5}<br />

    Now for the Stoke's part. If {\bf F} = <yz, 3xz, 2xy> so \nabla \times {\bf F} = <-x, -y, 2z> = < - s \,t \cos t, - s \, t \sin t, 2t^2>

    Further \vec{n} = \vec{r}_s \times \vec{r}_t = <-2 t^2 \sin t, 2 t^2 \cos t, s t^2>

    so that  <br />
\iint_s \nabla \times {\bf F} \cdot \vec{n} dS = \int_0^{2\pi} \int_0^1 2s t^4 ds dt = \frac{32 \pi^5}{5}<br />
    Attached Thumbnails Attached Thumbnails Stokes formula and the curvilinear integral of the second kind-ramp-surface.jpg  
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  3. #3
    Senior Member DeMath's Avatar
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    If I understand correctly, the non-closed curve OA = OB \cup BC \cup CA lies on the surface of paraboloid z = {x^2}+{y^2}.
    I started to solve this problem by adding a curve of the integration to a closed contour G of the curve z = {x^2}, lying in the plane XZ. But I was mistaken in the computation, so my answer is very different from the correct answer \frac{{32}}{5}{\pi ^5}.

    Many thanks, Prof. Danny!
    Last edited by DeMath; July 16th 2009 at 11:37 PM.
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  4. #4
    Senior Member DeMath's Avatar
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    If I understood correctly (?), then this task can be interpreted graphically as follows.

    Last edited by DeMath; July 17th 2009 at 01:22 AM.
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    Quote Originally Posted by DeMath View Post
    If I understood correctly (?), then this task can be interpreted graphically as follows.

    Then all we need is a different parametization of the surface. In this case

     <br />
\vec{r} = < s \cos t, s \sin t, s^2><br />
    so

     <br />
\vec{r}_u \times \vec{r}_v = <-2s^2 \cos t, - 2 s^2 \sin t, s><br />

    so \nabla \times {\bf F} \cdot \vec{n} = 4s^3

    and \iint_S \nabla \times {\bf F} \cdot \vec{n} dS = \int_0^{2 \pi} \int _0^t 4s^3 ds\,dt which gives the same answer.
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