This question is the second part to a previous question on the same problem.
I have the following problem:
The temperature at a point in a cylinder of radius $a$ and height $h$, and made of material with conductivity $k$, is inversely proportional to the distance from the centre of the cylinder. Find the heat flow across the curved surface of the cylinder.
The solution is as follows:
Use cylindrical coordinates with the cylinder centred at the origin. Now
$T = \dfrac{\alpha}{\sqrt{x^2 + y^2 + z^2}}$ where $\alpha$ is a constant
and hence
$\mathbf{F} = \dfrac{\alpha k(x, y, z)}{(x^2 + y^2 + z^2)^{3/2}}$
$= \dfrac{\alpha k(r\cos(\theta), r\sin(\theta), z)}{(r^2 + z^2)^{3/2}}$
The solution then says that the normal vector is $\mathbf{n} = (a\cos(\theta), a\sin(\theta), 0)$.
Finally, the solution says that the heat flux on the curved surface $r = a$ is
$\int^{h/2}_{z = -h/2} \int^{2\pi}_{\theta = 0} \mathbf{F} \mid_{r = a} \cdot \mathbf{n} d\theta dz$
$= \int^{h/2}_{z = -h/2} \int^{2\pi}_{\theta = 0} \dfrac{a^2 \alpha k}{(a^2 + z^2)^{3/2}}d\theta dz$
$= \dfrac{4\pi \alpha k h}{\sqrt{4a^2 + h^2}}$ where the substitution $z = a \tan(\theta)$ has been employed.
The following are my misunderstandings:
- When doing the dot product $\mathbf{F} \mid_{r = a} \cdot \mathbf{n}$, shouldn't we be using $ a d\theta dz $ as the scaling factor for cylindrical coordinates instead of just $d\theta dz$?
- As far as I can tell, $\mathbf{F} \mid_{r = a} \cdot \mathbf{n} \not = \dfrac{a^2 \alpha k}{(a^2 + z^2)^{3/2}}$? Where did the $\cos(\theta)$ go?
I would greatly appreciate it if people could please take the time to clarify this solution.