Surface integral: Calculate the heat flow from a cylinder

In summary, the conversation discusses the calculation of heat flow across the boundary of a solid cylinder with a given temperature and thermal conductivity. The heat flow across the bottom and top surfaces has been calculated, but there is confusion about the heat flow across the cylindrical side due to a discrepancy in the change of limits of integration in the solution manual. The person seeking help believes the change of limits is incorrect and is looking for clarification on the issue.
  • #1
krihamm
9
0
Homework Statement
Calculate the heat flow from a cylinder
Relevant Equations
See below
Hi,

I am trying to calculate the heat flow across the boundary of a solid cylinder. The cylinder is described by x^2 + y^2 ≤ 1, 1 ≤ z ≤ 4. The temperature at point (x,y,z) in a region containing the cylinder is T(x,y,z) = (x^2 + y^2)z. The thermal conductivity of the cylinder is 55. The gradient of the temperature is nabla(T) = 2xzi + 2yzj + (x^2 + y^2)k.

I have calculated the heat flow across the bottom and top surfaces of the cylinder using the dot product of the gradient of T and the normal vector (e.g. the cross product of the tangent vectors in the u- and v-directions of the parameterized surface). They are 55pi/2 and -55pi/2, respectively. However, I am having trouble determining the heat flow across the cylindrical side.

According to the solutions manual (see figure below), the answer is -110pi. This is plausible based on the flow across the bottom and top surfaces of the cylinder. However, I do not understand the change of limits of integration in the solutions. I want to keep the limits as they are, but that results in a very high flow across the cylindrical side. Changing the limits of integration as the solution manual suggests yields a plausible answer, but I do not understand what justifies the change..

Any help is much appreciated!

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krihamm
 
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  • #2
I have no physical background, but I believe the change of limits is incorrect. Basically they are claiming that
$$\int_1^4 2vdv = \int_0^1 2vdv$$
which is clearly wrong.
 

1. What is a surface integral?

A surface integral is a mathematical tool used to calculate the flux or flow of a quantity (such as heat) across a surface. It involves integrating a function over a two-dimensional surface.

2. How is a surface integral different from a regular integral?

A regular integral involves integrating a function over a one-dimensional interval, while a surface integral involves integrating a function over a two-dimensional surface. This allows for the calculation of flux or flow across a surface rather than just along a line.

3. What is the formula for calculating a surface integral?

The formula for calculating a surface integral is ∫∫S F(x,y,z) dS, where F(x,y,z) is the function being integrated and dS represents the infinitesimal area element on the surface S.

4. How is a surface integral used to calculate heat flow from a cylinder?

A surface integral can be used to calculate the heat flow from a cylinder by integrating the heat flux (heat energy per unit area per unit time) over the surface of the cylinder. This will give the total amount of heat flowing through the surface per unit time.

5. What are some real-world applications of surface integrals?

Surface integrals have many real-world applications, including calculating the flow of fluids (such as air or water) over a surface, determining the mass and center of mass of a three-dimensional object, and calculating the electric and magnetic fields around a charged or magnetized object.

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