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Flux problem

  1. Apr 27, 2008 #1
    1. The problem statement, all variables and given/known data

    Evaluate the surface integral of F*n (the flux) where n is the upward-pointing unit normal vector to the plane z = 3x+2 that lies within the cylinder x^2 + y^2 = 4. F = <0, 2y, 3z>

    2. The attempt at a solution

    The normal vector is <3,0,1> so that F*n = 3z. dS equals sqrt(10). I transform to polar coordinates and integrate 3*sqrt(10)*(3r^2*cos(theta) + 2r) over the cylinder. The answer I get is 24sqrt(10)*pi, while the correct answer is 24*pi. Does this mean that dS=1? Why?
     
  2. jcsd
  3. Apr 27, 2008 #2

    Dick

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    You said n is a unit vector. Then you said n=<3,0,1>. That's not a unit vector. <3,0,1>/sqrt(10) is.
     
  4. Apr 28, 2008 #3

    HallsofIvy

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    Time for me to rush to battle again! I fight constantly against the very notation "[itex]\vec{f}\cdot\vec{n}d\sigma[/itex]" because if you follow that literally, computing [itex]\vec{n}[/itex] and [itex]d\sigma[/itex] separately you wind up calculating the length of a vector twice and then watch them cancel out! (Unless you forget one of them!)

    In this problem, we can write the position vector of a point on the plane as [itex]\vec{r}(x,y)= x\vec{i}+ y\vec{j}+ (3x+ 2)\vec{k}[/itex].
    Differentiating: [itex]\vec{r}_x= \vec{i}+ 3\vec{k}[/itex] and [itex]\vec{r}_y= \vec{j}[/itex]. The cross product of those two vectors, [itex]-3\vec{i}+ \vec{k}[/itex], is the "fundamental vector product" for the plane and the vector differential of area, [itex]d\vec{\sigma}[/itex] is [itex](-3\vec{i}+ \vec{k})dxdy[/itex].

    Now [itex]\int\int \vec{F}\cdot\vec{d\sigma}[/itex] [itex]= \int\int (2y\vec{j}+ 3z\vec{k})\cdot(-3\vec{i}+ \vec{k})dxdy[/itex] [itex]= \int\int 3z dxdy= 3\int\int (3x+ 2)dxdy[/itex].

    Integrate that over the circle of radius 2.

    Obviously the way to do that is to switch to polar coordinates. We could do that right at the beginning:
    write [itex]\vec{r}(r,\theta)= rcos(\theta)\vec{i}+ rsin(\theta)\vec{j}+ (3rcos(\theta)+ 2)\vec{k}[/itex], differentiate with respect to r and [itex]\theta[/itex] and the "r" you need for rdrd[itex]\theta[/itex] pops out automatically!

    [Here endeth the sermon]
     
  5. Apr 28, 2008 #4

    Dick

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    Amen. It is clearer to keep the volume and the direction information in the same package.
     
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