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Math Help - Stokes' Theorem problem

  1. #1
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    Stokes' Theorem problem

    I have a question about a problem involving Stokes' theorem

    The question goes as follows:

    Use Stokes' theorem to evaluate:

    Double integral of the curl of F dot n ds

    Where S is the part of the cylinder z = 1 - x^2 for 0<=x<=1, -2<=y<=2 and F(x, y, z) = xyi + yzj +xzk

    So far I have the Curl of F is equal to -yi - zj - xk. I am now completely lost. Any help would be appreciated, thank you.
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  2. #2
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    Quote Originally Posted by allhalf425 View Post
    I have a question about a problem involving Stokes' theorem

    The question goes as follows:

    Use Stokes' theorem to evaluate:

    Double integral of the curl of F dot n ds

    Where S is the part of the cylinder z = 1 - x^2 for 0<=x<=1, -2<=y<=2 and F(x, y, z) = xyi + yzj +xzk

    So far I have the Curl of F is equal to -yi - zj - xk. I am now completely lost. Any help would be appreciated, thank you.
    If you want to calculate \iint \limits_S \nabla \times {\bf F} \cdot {\bf n}\, dS directly then your normal and dS are given by

     <br />
{\bf n} = \frac{<2x,0,1>}{\sqrt{1+4x^2}}<br />
and dS = \sqrt{1 + 4x^2}dA so \nabla \times {\bf F} \cdot {\bf n}\, dS = (- 2xy - x)\,dA

    and

     <br />
\int_0^1 \int_{-2}^2 \left(-2xy - x\right)\, dy\,dx = -2<br />

    To use Stokes Thm then you want to evaluate

     <br />
\int \limits_C xydx + yz dy + xz dz<br />

    There are four parts to the boundary of the surface (see the attached picture)

     <br />
C_1: x = 1, z = 0, y: -2 \to 2<br />
so  <br />
\int \limits_{C_1} xydx + yz dy + xz dz = 0<br />

    C_2: y = 2, \;\; \text{along}\; z = 1-x^2 \; \text{from}\; x = 1 \; \text{to}\; 0 so  <br />
\int \limits_{C_2} xydx + yz dy + xz dz = \int_1^0 2 x\,dx + x (1-x^2)\,(-2x)\,dx = - \frac{11}{15}<br />

     <br />
C_3: x = 0, z = 1,\; y: 2 \to -2<br />
so  <br />
\int \limits_{C_3} xydx + yz dy + xz dz = 0

     <br />
C_4: y = -2\;\; \text{along}\; z = 1-x^2 \; \text{from}\; x = 0 \; \text{to}\;1,<br />
so  <br />
\int \limits_{C_4} xydx + yz dy + xz dz = \int_0^1 -2x\,dx + x (1-x^2)\,(-2x)\,dx = - \frac{19}{15}<br />

    giving that

     <br />
\int \limits_{C} xydx + yz dy + xz dz = 0 - \frac{11}{15} + 0 - \frac{19}{15} = -2
    Attached Thumbnails Attached Thumbnails Stokes' Theorem problem-pic.jpg  
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    Thank you!
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