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Calculate the flux of F=$<xy,yz,xz>$ out of $S$, ,which is the part of the paraboloid $x^2+y^2+z=4$ that lies above the square $[0,1]\times[0,1]$.

I know we need to solve $\iint_{s}F\cdot ds$ which can be written as $\iint_{D}F(u,v)\cdot (r_u\times r_v)dA$, where $r(u,v)$ is the parameterization of the surface, however I am unaware how to account for the square in the parameterization. Thanks

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  • $\begingroup$ what do you mean by "account for the square in the parameterization" $\endgroup$ – qbert Apr 18 '18 at 18:12
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You need to find a parameterization for your surface. Since $S$ is described as a graph of a function $$z=4-x^2-y^2$$ you can use $x$ and $y$ (your independent variables) as parameters.

You get $x=x$, $y=y$, and $z=4-x^2-y^2$. In other words, $${\bf r}(x,y) = \langle x,y,4-x^2-y^2 \rangle$$

The referenced square tells you what part of the paraboloid you're dealing with. In particular, it is telling you that your domain is: $0 \leq x \leq 1$ and $0\leq y \leq 1$.

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If $z = f(x,y)$ then you already have $z$ parameterized in terms of $x,y$

And since your region is a k-cell, there is no reason to change coordinates.

$r_x\times r_y= (\frac {\partial x}{\partial x},\frac {\partial y}{\partial x}, \frac {\partial z}{\partial x})\times (\frac {\partial x}{\partial y},\frac {\partial y}{\partial y}, \frac {\partial z}{\partial y}) \ dx\ dy$

Which might look a little silly.. but $\frac {\partial x}{\partial x} =1,$ and $\frac {\partial y}{\partial x} = 0$

$dS = (1,0, \frac {\partial z}{\partial x})\times (0,1, \frac {\partial z}{\partial y}) = (-\frac {\partial z}{\partial x},-\frac {\partial z}{\partial x},1)\ dx\ dy$

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