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A Determine the flux of the vector field trough the surface

  1. Aug 6, 2017 #1
    From my drawings it seems to be half of hemisphere. Am I right? How can I solve this task?

    Determine the flux of the vector field $$ f=(x,(z+y)e^x,-xz^2)^T$$ through the surface $Q(u,w)$, which is defined in the follwoing way:

    1) the two boundaries are given by $$\delta Q_1=\{(x,y,z):x^2+y^2=1,z=0,y\ge0\}$$ and $$\delta Q_2=\{(x,y,z):x^2+z^2=1,y=0,z\ge0\}$$

    2) the points on the two arcs $$\delta Q_1$$ and $$\delta Q_2$$ are connected by straight lines, which are parallel to the plane $$x=0$$
     
  2. jcsd
  3. Aug 6, 2017 #2

    andrewkirk

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    It's not a hemisphere. There are lines on the surface that are straight, and one cannot draw a straight line on a hemisphere.

    Also, the surface lies in only two octants, whereas a hemisphere lies in four of the eight octants. It is a surface that lies over half of the plane ##z=0##.

    To solve this, you calculate
    $$\int_{0}^1 \int_{-1}^1
    \vec f(x,y,g(x,y)) \cdot \vec n(x,y)
    \,dx\,dy$$
    where ##g(x,y)## is the ##z## coordinate of the point on the surface with ##x,y## coordinates ##(x,y)##; and ##\vec n(x,y)## is the normal to the surface at that point.

    Note that, because of the last statement in the OP, ##\vec n(x,y)## is constant over ##y## and can be obtained from the gradient of the line connecting
    • the point on the ring ##\delta Q_2## with specified ##x## coordinate and ##z\geq 0##, to
    • the point on the ring ##\delta Q_1## with the same ##x## coordinate and ##y\geq 0##
    The equation of that line can also be used to calculate the ##z## coordinate ##g(x,y)##.
     
    Last edited: Aug 7, 2017
  4. Aug 7, 2017 #3
    I don't understand, why the limit for y is -1 to 1 and not 0 to 1.
     
  5. Aug 7, 2017 #4

    andrewkirk

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    That was an error. I have corrected it.
     
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