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Math Help - Surface Integrals

  1. #1
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    Surface Integrals

    Help!!!

    Consider the surface given by the equation  2{x}^{2} + 2{y}^{2} + z = 6 , oriented upward, and let S be the part of this surface that is above the plane z = 4. Given the vector field:
     \vec{F} = x\vec{i} + y\vec{j} + 3\vec{k} ,
    evaluate the surface integral
     \iint \vec{F} \cdot d\vec{S} .
    Last edited by topsquark; April 30th 2011 at 08:37 PM. Reason: Fixed up the LaTeX
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by raven17 View Post
    Help!!!

    Consider the surface given by the equation  2{x}^{2} + 2{y}^{2} + z = 6 , oriented upward, and let S be the part of this surface that is above the plane z = 4. Given the vector field:
     \vec{F} = x\vec{i} + y\vec{j} + 3\vec{k} ,
    evaluate the surface integral
     \iint \vec{F} \cdot d\vec{S} .
    \iint_S \mathbf{F}\cdot d\mathbf{S}=\iint_D\mathbf{F}\cdot \left(-\frac{\partial z}{\partial x}\mathbf{i}-\frac{\partial z}{\partial y}\mathbf{j} +\mathbf{k}  \right)dA

    The area of integration can be found by the equation.

    2x^2+2y^2+4=6 \iff x^2+y^2+1 so the domain of integration is the unit circle.

    \iint_S \mathbf{F}\cdot d\mathbf{S}=\iint_D\left( x\mathbf{i}+y\mathbf{j}+3\mathbf{k}\right)\cdot \left(4x\mathbf{i}+4y\mathbf{j} +\mathbf{k}  \right)dA=

    \iint 4(x^2+y^2)+3dA

    Now change to polar coordinates to finish.
    Last edited by TheEmptySet; April 30th 2011 at 10:21 PM. Reason: typo
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  3. #3
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    Thanks sooo much! Just a question though... where did you get the "(4xi + 4yj + k)" from?
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  4. #4
    Behold, the power of SARDINES!
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    Quote Originally Posted by raven17 View Post
    Thanks sooo much! Just a question though... where did you get the "(4xi + 4yj + k)" from?
    -\frac{\partial z}{\partial x}\mathbf{i}-\frac{\partial z}{\partial y}\mathbf{j} +\mathbf{k}

    z=6-2x^2-2y^2

    Now just take the partials and plug in.
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  5. #5
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    Quote Originally Posted by raven17 View Post
    Thanks once again! I am sorry for asking again but how did you get -dz/dx (i) - dz/dy (j) + k?
    If you solve the equation for z=f(x,y)

    we can parametrize the surface as follows

    \mathbf{r}(x,y)=x\mathbf{i}+y\mathbf{j}+f(x,y) \mathbf{k}

    So \mathbf{r}_x=\mathbf{i}+\frac{\partial f}{\partial x}\mathbf{k}

    and \mathbf{r}_y=\mathbf{j}+\frac{\partial f}{\partial y}\mathbf{k}

    Now if you cross these two vectors you will get the formula
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