Solving for Flux. Basic Calculus question.

In summary, the conversation discusses the rate of heat transfer per unit of system area and how to solve for the flux in this case. It is explained that when integrating, the differential element is the entire argument of the integration and finding the derivative of the integral with respect to the same variable yields the desired function. This is demonstrated through the example of F(x) = \int f(x) dx and dF/dx = f(x). The conversation concludes with an understanding that dividing by dx produces the desired result.
  • #1
kdhutchi
4
0

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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  • #2
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]
 
  • #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?
 
  • #4
Right.
 
  • #5
Thank you for your help Gneill!
 

1. What is flux in calculus?

Flux in calculus refers to the flow of a vector field through a surface. It can be calculated using the dot product of the vector field and the unit normal vector to the surface.

2. How do you solve for flux using basic calculus?

To solve for flux using basic calculus, you first need to find the dot product of the vector field and the unit normal vector to the surface. Then, integrate this dot product over the surface to find the total flux.

3. What is the unit of flux in calculus?

The unit of flux in calculus depends on the units of the vector field and the surface. Generally, it is measured in units of volume per time, such as cubic meters per second.

4. Can flux be negative in calculus?

Yes, flux can be negative in calculus. This indicates that the vector field is flowing in the opposite direction of the normal vector to the surface.

5. What is the physical significance of solving for flux in calculus?

Solving for flux in calculus has many practical applications, such as determining the rate of fluid flow through a surface or calculating the electric or magnetic field passing through a given area. It is also used in many engineering and science fields, including fluid dynamics, electromagnetism, and thermodynamics.

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