Flux integral over a parabolic cylinder

  • #1
Homework Statement
Evaluate ##\int\int_S \textbf{F}\cdot\textbf{n} dS ## where ##\textbf{F}=(z^2-x)\textbf{i}-xy\textbf{j}+3z\textbf{k}## and S is the surface region bounded by ##z = 4-y^2, x=0, x=3## and the x-y plane with ##\textbf{n}## directed outward to S.

The attempt at a solution

I've worked out the correct answer but can't seem to fully understand why that is. I tried splitting up the flux integral into 3 seperate surfaces: 1 for the parabola at x=3, another for the parabola at x=0, and lastly a parametric surface between them. At each parabola I just evaluated the flux integral in cartesian coordinates, which were ##\int_{-2}^2 \int_0^{4-y^2}(2y\textbf{j}+\textbf{k})\cdot((z^2-x)\textbf{i}-xy\textbf{j}+3z\textbf{k})##, which worked out to be 256/5 and 0, at x=0 and x=3 respectively.

I parameterized the parabolic cylinder as ##\mu\textbf{i}+\lambda\textbf{j}+(4-\lambda^2)\textbf{k}##, so the flux integral for this was ## \int_0^2 \int_0^3(2\lambda\textbf{j}+\textbf{k})\cdot((z^2-x)\textbf{i}-\mu\lambda\textbf{j}+(12-3\lambda^2)\textbf{k})d\mu d\lambda## which works out to be 48, which is the correct answer. This might seem like a dumb question but I've been staring at it for hours and can't understand why the value of the flux integral at the x=0 parabola is ignored. I considered the possibility that it excludes the surfaces at x=0 and x=3 but similarly worded questions did not do this. Any help/tips will be appreciated, thanks!
 
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  • #2
vela
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Homework Statement
Evaluate ##\int\int_S \textbf{F}\cdot\textbf{n} dS ## where ##\textbf{F}=(z^2-x)\textbf{i}-xy\textbf{j}+3z\textbf{k}## and S is the surface region bounded by ##z = 4-y^2, x=0, x=3## and the x-y plane with ##\textbf{n}## directed outward to S.

The attempt at a solution

I've worked out the correct answer but can't seem to fully understand why that is. I tried splitting up the flux integral into 3 seperate surfaces: 1 for the parabola at x=3, another for the parabola at x=0, and lastly a parametric surface between them. At each parabola I just evaluated the flux integral in cartesian coordinates, which were ##\int_{-2}^2 \int_0^{4-y^2}(2y\textbf{j}+\textbf{k})\cdot((z-x^2)\textbf{i}-xy\textbf{j}+3z\textbf{k})##, which worked out to be 256/5 and 0, at x=0 and x=3 respectively.
First, is the x-component of the vector field ##z^2-x## or ##z-x^2##? You used the wrong normal for evaluating the surface integrals on the x=0 and x=3 planes.

I parameterized the parabolic cylinder as ##\int_0^2\int_0^3\mu\textbf{i}+\lambda\textbf{j}+(4-\lambda^2)\textbf{k}##, so the flux integral for this was ##(2\lambda\textbf{j}+\textbf{k})\cdot((z-x^2)\textbf{i}-\mu\lambda\textbf{j}+(12-3\lambda^2)\textbf{k})d\mu d\lambda## which works out to be 48, which is the correct answer. This might seem like a dumb question but I've been staring at it for hours and can't understand why the value of the flux integral at the x=0 parabola is ignored. I considered the possibility that it excludes the surfaces at x=0 and x=3 but similarly worded questions did not do this. Any help/tips will be appreciated, thanks!
There seems to be numerous errors or typos in what you've written here. Could you please clean it up?
 
  • #3
First, is the x-component of the vector field ##z^2-x## or ##z-x^2##? You used the wrong normal for evaluating the surface integrals on the x=0 and x=3 planes.


There seems to be numerous errors or typos in what you've written here. Could you please clean it up?
I fixed up most of the errors I think (left i component in terms of z and x since it goes to 0 anyway). The x component was supposed to be ##z^2-x##. Would the correct normals in the x=0 and x=3 planes be the unit vector i at x=3 and -i at x=0?
 
  • #4
vela
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I think you meant for the limits for ##y## on the last integral to be ##-2## and ##2##. Yes, the normals are ##\pm \hat i##. I get the same result you do integrating over just the curved part of the surface. With the three flat surface included, I get 16 for the total flux.
 
  • #5
I think you meant for the limits for ##y## on the last integral to be ##-2## and ##2##. Yes, the normals are ##\pm \hat i##. I get the same result you do integrating over just the curved part of the surface. With the three flat surface included, I get 16 for the total flux.
Guess the given answer must be off then, thanks for clearing that up.
 

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