Solving for Flux. Basic Calculus question.

AI Thread Summary
The discussion revolves around solving for heat flux using the equation \dot{Q}=\int\dot{q}dA. The user initially struggles with the concept that taking the derivative of an integral with respect to the same variable results in zero, but later clarifies that the derivative of \dot{Q} with respect to area yields the flux \dot{q}. They confirm that the relationship d\dot{Q} = \dot{q}dA leads to the conclusion that \frac{d\dot{Q}}{dA} = \dot{q}. The conversation emphasizes the correct application of calculus in understanding heat transfer. Ultimately, the user gains clarity on the relationship between integration and differentiation in this context.
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Homework Statement



Basic Question. My book describes the rate of heat transfer per unit of system area as \dot{Q}=\int\dot{q}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 \dot{Q} 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,
F(x) = \int f(x) dx
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 d\dot{Q} = \dot{q}dA and \frac{d\dot{Q}}{dA} = \dot{q}
 
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?
 
Right.
 
Thank you for your help Gneill!
 
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