Solving for Flux. Basic Calculus question.

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kdhutchi
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Homework Statement



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.

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?
 
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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]
 
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?
 
Thank you for your help Gneill!