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Solving for Flux. Basic Calculus question.

  1. Oct 5, 2011 #1
    1. The problem statement, all variables and given/known data

    Basic Question. My book describes the rate of heat transfer per unit of system area as [itex]\dot{Q}[/itex]=[itex]\int[/itex][itex]\dot{q}[/itex]dA.

    3. The attempt at a solution

    I'm trying to solve for the flux. I've already discovered that the correct answer question is just the derivative of [itex]\dot{Q}[/itex] with respect to area, but this goes against other things I've read. I thought taking the derivative of an integral with respect to the same variable as the integration yields zero. For example, if I was integrating something with respect to area, and I take the derivative of the integral with respect to area, then the whole thing cancels out. Isn't this true? How would I solve for the flux in this case?
     
    Last edited: Oct 5, 2011
  2. jcsd
  3. Oct 5, 2011 #2

    gneill

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    Staff: Mentor

    When you integrate, the entire "argument" of the integration is the differential element of the desired function. The integration just sums them up. So for example,
    [tex] F(x) = \int f(x) dx [/tex]
    then the differential element for F(x) is f(x)dx. That is to say, dF = f(x)dx.

    Rearranging, you find dF/dx = f(x).

    So in your case [itex] d\dot{Q} = \dot{q}dA[/itex] and [itex] \frac{d\dot{Q}}{dA} = \dot{q}[/itex]
     
  4. Oct 5, 2011 #3
    I think I understand what you are saying. The part where you mention dF=f(x)dx comes from just taking a derivative from both sides, correct? So dividing by dx will produce the dF/dx=f(x) I sought earlier correct?
     
  5. Oct 5, 2011 #4

    gneill

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    Staff: Mentor

    Right.
     
  6. Oct 6, 2011 #5
    Thank you for your help Gneill!
     
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