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Math Help - help with finding dA in computing the flux of the vector field F

  1. #1
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    help with finding dA in computing the flux of the vector field F

    Compute the flux of the vector field F(x, y, z) = 2xj + yk through the surface S, where S is the part of the surface z = -y + 4 above the square 0 ≤ x ≤ 3, 0 ≤ y ≤ 1, oriented upward.

    Again, having trouble finding dA. is it not dA = (0i + j + k)dxdy?
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  2. #2
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    Re: help with finding dA in computing the flux of the vector field F

    Do you mean "dA", which is a scalar differential or " \vec{n} dA= d\vec{S}", a vector differential? Since you are talking about the flux of a vector function through the surface, you must mean the latter and you want to find \int\int \vec{F}\cdot d\vec{S}.

    You can write the surface, z= 4- y as x= x, y= y, z= 4- y or as the vector equation \vec{r}(x, y)= x\vec{i}+ y\vec{j}+ (4- y)\vec{k}. The derivatives, \vec{r}_x= \vec{i} and \vec{r}_y= \vec{j}- \vec{k} lie in the tangent plane at each point and their cross product,
    \left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ 1 & 0 & 0 \\ 0 & 1 & -1\end{array}\right|= -\vec{j}+ \vec{k}
    with "upward z component" or \vec{j}- \vec{k} with "downward z component". Which of those you use depends upon how you want the surface oriented- whether you want to think of the flux as positive or negative.

    Taking the "upward z compontent" d\vec{S}= (\vec{j}- \vec{k})dxdy so you have a sign wrong.
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  3. #3
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    Re: help with finding dA in computing the flux of the vector field F

    Ok. thanks. I used the scalar differential <0, 1, -1>dxdy and then tried to calculate the flux by
    ∫_s F dA = ∫_R <0, 2x, y> <0, 1, -1> dxdy
    = ∫(y = 0 to 1)∫(x = 0 to 3) (2x - y) dxdy

    and got 15/2 but that was wrong.
    I also tried <0, -1, 1>dxdy and got -15/2 and that was also incorrect. what did I do wrong?
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