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Consider the paraboloid $B := \{(x, y, z) ∈ R^ 3 : x^ 2 + y^ 2 = z, z < 1\}$ The flux $\int_ B F · ν dS$(from the inside out) of the vector field $F : R ^3 → R ^3 , F(x, y, z) := (xy, xz, −zy)$

I am not sure if I am correct in doing this, but I use z=0, however I get $\iiint \text{div}\; \vec F . dV=F_1+F_2$ both $F_1, F_2$ are equal to 0. Should I use z=1? I am a bit lost here. Thanks for any given advice!

For the outer normal vector I get $(\frac{2x} {\sqrt{4x^2+4y^2+1}}, \frac{2y} {\sqrt{4x^2+4y^2+1}}, \frac{-1} {\sqrt{4x^2+4y^2+1}})$

ugjumb
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  • $z$ is not equal to $0$. $z$ is equal to $x^2+y^2$. – coudy Mar 12 '23 at 11:36
  • So should I do 3 integrations? dzdydx? And the domain of z would be [0,1]? of y[$-\sqrt {-x^2}, \sqrt {-x^2}$] and of x [0,1]? – ugjumb Mar 12 '23 at 11:37
  • Hint: how many integrations you do over a volume? over a surface? what $x,y,z$ are follows from the definition of $B$. $y$ is certainly not in the interval $[-\sqrt{\color{red}{-}x^2},\sqrt{\color{red}{-}x^2}],.$ – Kurt G. Mar 12 '23 at 11:43
  • @KurtG. How should I get the interval of y? Also is x [0,1] is correct? – ugjumb Mar 12 '23 at 12:08
  • Which $x,y,z$ satisfy $x^2+y^2=z,z<1,?$ You are permitted to change this to $z\le 1,.$ Now, for every fixed $z\in[0,1]$ can you tell me the ranges of $x$ and $y$? Hint: start fixing $x\in[0,z],.$ Why? $z$ is your radius. – Kurt G. Mar 12 '23 at 12:14
  • $y\in [-\sqrt{1-x^2}, \sqrt{1-x^2}]$? – ugjumb Mar 12 '23 at 12:16
  • No. $y\in [-\sqrt{\color{red}{z}-x^2},\sqrt{\color{red}{z}-x^2}],.$ and $x\in[0,\sqrt{z}]$ (correction to previous comment: $\sqrt{z}$ is the radius). – Kurt G. Mar 12 '23 at 12:29
  • Before we do that cumbersome $3d$ integral in Cartesian coordinates. Can you first tell me the divergence of $F$ ? – Kurt G. Mar 12 '23 at 12:35
  • @KurtG. if I take y as what you suggested I am getting integral of dy is =0 – ugjumb Mar 12 '23 at 14:52
  • All of the substitutions are with y^2 so it will be 0 , what do I do? – ugjumb Mar 12 '23 at 15:01
  • The divergence of $F(x, y, z) := (xy, xz, −zy)$ is $y+0-y=0$ which leads to $\iiint{\rm div F},dV=0,.$ What do we do now? Nothing. That was it. – Kurt G. Mar 12 '23 at 15:19
  • Really? it could be that the flux of the vector field is simply 0 ? I was searching for a mistake.... Well in any case, thank you so much!!!! – ugjumb Mar 12 '23 at 15:27
  • You are welcome. For an example that is divergence free except in one point (and has non zero flux) please see this answer. – Kurt G. Mar 12 '23 at 15:30

1 Answers1

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You can parameterize your paraboloid by $x$ and $y$, in which case $z$ is given by $z = x^2+y^2$. Integrate with respect to $x$ and $y$ replacing $z$ by $x^2+y^2$ everywhere and you should be fine. The paraboloid is a surface, it is two-dimensional, you are computing an integral depending on two variables.

coudy
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  • Is the outer normal vector correct, then or should it be (x, y, x^2+y^2)? – ugjumb Mar 12 '23 at 11:45
  • Yes it is correct – coudy Mar 12 '23 at 11:47
  • Which one? $(\frac{2x} {\sqrt{4x^2+4y^2+1}}, \frac{2y} {\sqrt{4x^2+4y^2+1}}, \frac{-1} {\sqrt{4x^2+4y^2+1}})$? – ugjumb Mar 12 '23 at 11:50
  • Yes, that one is the correct one. – coudy Mar 12 '23 at 11:51
  • I want to ask, why is the ${\sqrt{4x^2+4y^2+1}$ part come from squaring 2x+2y-1 individually and then taking the root of the whole? Will it always be like this? or how it is determined? – ugjumb Mar 12 '23 at 12:05
  • You want a unit vector. A vector of norm one. So you need to divide by the norm of the gradient. In the flux formula, this is the unit normal that appears. – coudy Mar 12 '23 at 13:06
  • Do you perhaps know what should be the domain of y? because I am getting 0 – ugjumb Mar 12 '23 at 15:12
  • $x$ and $y$ belong to the unit disc, $D = {(x,y) \in {\bf R}^2 \mid x^2+y^2 \leq 1}$. The final integral is probably easiest to compute in polar coordinates. – coudy Mar 12 '23 at 17:24