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Evaluate the surface integral double integral of $S$ of $F * dS$ for the given vector field $F$ and the oriented surface $S$. In other words, find the flux of $F$ across $S$. For closed surfaces, use the positive (outward) orientation.

$$\vec{F}(x,y,z)= y\mathbf{j} - z\mathbf{k}$$ $S$ consists of the paraboloid $y = x^2 + z^2, 0 \leq y \leq 1,$ and the disk $x^2 + z^2 \leq 1, y=1$

I did $x = u\cos t, y = u^2, z =u\sin t$

So to use double integral of $$\vec{F} * dS$$

For parametrization of the paraboloid how do you find $$\vec{F} $$ to be $<0,u^2,-u\sin t>$?

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  • $\begingroup$ Do you really need to integrate over the surface, or are you allowed to use the Divergence Theorem? $\endgroup$ May 13, 2013 at 6:40
  • $\begingroup$ Would they mean the same thing? I am not exactly surface what integrating over the surface means. Does that mean finding the area of the surface? $\endgroup$
    – Walter
    May 13, 2013 at 7:25
  • $\begingroup$ if the density, or weight, was 1, then it would be area. Instead, you have a nonuniform density over the area, so the surface integral is not necessarily area. $\endgroup$ May 13, 2013 at 10:15
  • $\begingroup$ "Integrate over the surface" is exactly what you are trying to do. The point is that results such as the Divergence Theorem can often be used to avoid messy calculations. $\endgroup$ May 13, 2013 at 14:59

1 Answer 1

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Simply note that $F=(0,y,-z)$, $y=u^2$ and $z=-u\sin(t)$.

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