Results 1 to 5 of 5

Thread: Stokes formula and the curvilinear integral of the second kind

  1. #1
    Senior Member DeMath's Avatar
    Joined
    Nov 2008
    From
    Moscow
    Posts
    474
    Thanks
    5

    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: $\displaystyle I = \int\limits_{OA} {yzdx + 3xzdy + 2xydz,}$
    where $\displaystyle OA = OB \cup BC \cup CA$, curve $\displaystyle x = t\cos t,{\text{ }}y = t\sin t,{\text{ }}z = {t^2},{\text{ }}0 \leqslant t \leqslant 2\pi ,{\text{ }}O = \left( {0,0,0} \right),{\text{ }}A = \left( {2\pi ,0,4{\pi ^2}} \right).$

    I have two carefully (I thought) addressed this problem , but my answer is clearly not the same as the right one.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,470
    Thanks
    83
    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: $\displaystyle I = \int\limits_{OA} {yzdx + 3xzdy + 2xydz,}$
    where $\displaystyle OA = OB \cup BC \cup CA$, curve $\displaystyle x = t\cos t,{\text{ }}y = t\sin t,{\text{ }}z = {t^2},{\text{ }}0 \leqslant t \leqslant 2\pi ,{\text{ }}O = \left( {0,0,0} \right),{\text{ }}A = \left( {2\pi ,0,4{\pi ^2}} \right).$

    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 $\displaystyle x = t \cos t,\; y = t \sin t,\; z = t^2$ is given parametrically as

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

    where $\displaystyle 0 \le s \le 1$ (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

    $\displaystyle
    I = \frac{32 \pi^5}{5}
    $

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

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

    so that $\displaystyle
    \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}
    $
    Attached Thumbnails Attached Thumbnails Stokes formula and the curvilinear integral of the second kind-ramp-surface.jpg  
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Senior Member DeMath's Avatar
    Joined
    Nov 2008
    From
    Moscow
    Posts
    474
    Thanks
    5
    If I understand correctly, the non-closed curve $\displaystyle OA = OB \cup BC \cup CA$ lies on the surface of paraboloid $\displaystyle z = {x^2}+{y^2}$.
    I started to solve this problem by adding a curve of the integration to a closed contour $\displaystyle G$ of the curve $\displaystyle z = {x^2}$, lying in the plane $\displaystyle XZ$. But I was mistaken in the computation, so my answer is very different from the correct answer $\displaystyle \frac{{32}}{5}{\pi ^5}$.

    Many thanks, Prof. Danny!
    Last edited by DeMath; Jul 16th 2009 at 11:37 PM.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member DeMath's Avatar
    Joined
    Nov 2008
    From
    Moscow
    Posts
    474
    Thanks
    5
    If I understood correctly (?), then this task can be interpreted graphically as follows.

    Last edited by DeMath; Jul 17th 2009 at 01:22 AM.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Jester's Avatar
    Joined
    Dec 2008
    From
    Conway AR
    Posts
    2,470
    Thanks
    83
    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

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

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

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

    and $\displaystyle \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.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Curvilinear integral doubt
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Nov 13th 2010, 03:24 PM
  2. Integral using Stokes Theorem
    Posted in the Calculus Forum
    Replies: 0
    Last Post: Jul 18th 2010, 02:13 PM
  3. Kind of tough integral
    Posted in the Calculus Forum
    Replies: 1
    Last Post: Nov 14th 2009, 01:31 PM
  4. Integral curvilinear
    Posted in the Calculus Forum
    Replies: 10
    Last Post: Apr 26th 2009, 12:07 PM
  5. What kind of integral, and how do i start?
    Posted in the Calculus Forum
    Replies: 12
    Last Post: Mar 9th 2009, 02:23 AM

Search Tags


/mathhelpforum @mathhelpforum