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Homework Help: Flux of a vector

  1. Apr 21, 2012 #1
    1. The problem statement, all variables and given/known data
    The velocity of a fluid is given, in cartesian coordinates (x, y, z), by [itex]\vec{v}[/itex] = [itex]v_{0}x\hat{z}[/itex] being a constant with velocity dimensions.
    a) Calculate the flux of this vector through the closed surface composed by z-x²-y²=0 and by the plan z=8 limited by the circle with radius a=[itex]\sqrt{8}[/itex] (x²+y²=8).

    2. Relevant equations
    [itex]\int\int{\vec{v}}\cdot\vec{n}\cdot dS[/itex]

    3. The attempt at a solution
    Well. I've calculated this and it gave me 0 as a result... But I don't know if I did it right.
    I've started with

    [itex]\int\int{\vec{v}}\cdot\vec{n}\cdot dS[/itex] = [itex]\int\int{v_{0}x}\cdot dS[/itex]

    So I've transformed into polar coordinates and it gave me
    [itex]\int\int{v_{0}rcosθ}\cdot rdrdθ[/itex] = [itex]\int\int{v_{0}rcosθ}\cdot rdrdθ[/itex]
    [itex]= v_{0} \int^{2\pi}_{0}\int^{8}_{0}r²cosθdrdθ[/itex]

    And developing this, it gave me 0 as a result. Is this correct?
    Last edited: Apr 21, 2012
  2. jcsd
  3. Apr 21, 2012 #2


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    hi cristina89! :smile:
    no, [itex]\int\int{v_{0}x}(\hat{z}\cdot\hat{n})dS[/itex]
  4. Apr 21, 2012 #3
    ahh ok. so [itex]\int\int{v_{0}x}(\hat{z}\cdot\hat{n})dS = \int\int v_{0}x(\hat{z})\frac{v_{0}x\hat{z}}{\sqrt{v_{0}^{2}x^{2}}}[/itex] right?
    What should I do after that? I'm now confused if there is this normal vector in this integral. Is this correct?
    Last edited: Apr 21, 2012
  5. Apr 22, 2012 #4


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    i don't understand that :confused:

    (and that fraction is just [itex]\hat{z}[/itex])

    [itex]\hat{z}\cdot\hat{n}[/itex] is cos of the slope of the normal (= sin of the slope of the surface)
  6. Apr 22, 2012 #5
    Can't I think that [itex]\hat{n}[/itex] = [itex]\frac{v}{|v|} = \frac{v_{0}x\hat{z}}{\sqrt{v_{0}^{2}x^{2}}} = \hat{z}[/itex]?
    And then [itex]\int\int{v_{0}x\hat{z}\hat{z}dS}[/itex]?

    Can you explain to me this "[itex]\hat{z}\cdot\hat{n}[/itex] is cos of the slope of the normal (= sin of the slope of the surface)"? :S I can't understand that...
    Last edited: Apr 22, 2012
  7. Apr 22, 2012 #6


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    There are two surfaces involved here: the paraboloid [itex]z= x^2+ y^2[/itex] and the plane z= 8. Obviously (I hope it is obvious!) the outward normal to the plane z= 8 is [itex]\vec{k}[/itex] and the inward normal is [itex]-\vec{k}[/itex].

    The paraboloid [itex]z= x^2+ y^2/[/itex] can be written in vector form, using x and y as parameters, as [itex]\vec{r}(x,y)= x\vec{i}+ y\vec{j}+ z\vec{k}= x\vec{i}+ y\vec{j}+ (x^2+ y^2)\vec{l}[/itex]. The two derivative vectors, with respect to x and y are
    [itex]\vec{r}_x= \vec{i}+2x\vec{k}[/itex] and [itex]\vec{r}_y= \vec{j}+ 2y\vec{k}[/itex] are tangent vectors to the surface at each point. Their cross product,
    [tex]\left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\ 1 & 0 & 2x \\ 0 & 1 & 2y\end{array}\right|= -2x\vec{i}- 2y\vec{j}+\vec{k}[/tex]
    is normal to the surface and [itex]\vec{n}dS= (-2x\vec{i}- 2y\vec{j}+ \vec{k})dxdy[/itex]
    That, with x and y components negative, pointing inward, is the "inward normal". Multiplying by -1 to get [itex]2x\vec{i}+ 2y\vec{j}- \vec{k}[/itex] gives the "outward normal". For this problem, where you are calculating the "flux", the flux all the way through, it doesn't matter whether you use "inward" or "outward" normals as long as you are consistent, using the same for both paraboloid and plane.
  8. Apr 22, 2012 #7
    Thank you!
    Is it right to bound the integrals like this [itex]\int^{2∏}_{0}\int^{8}_{0}[/itex]?
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