Heat flow surface integral problem

Question

I have a Rate of heat flow problem,

There is a temperature at a point in a substance,

$$u(x,y,z) = 4y^2 + 4z^2 $$ $$K = 5 $$

K is the conductivity constant.

I have to find the rate of heat flow inward across a cylindrical surface $$y^2 + z^2 = 324$$ $$ -10 \le x \le 3 $$

Here is a formula to compute this

$$ \iint_S F \cdot dS = -K \iint_S \nabla u \cdot dS$$

I would say the integrand is $8y+8z$ per the gradient but the limits of integration are something I cannot find.

Answer 1

$$u=4r^2$$ where r is in cylindrical coordinates With the symmetries you have, the gradient of $u$ in cylindrical coordinates is $\frac{\partial u}{\partial r} \hat{\rho}=8r \hat{\rho} $

So the integral on right hand side of your equation is $$-K \int\int_S \nabla u \cdot dS = -K \times 8R \times A$$ where $A$ is the surface area of the cylinder of radius $R=18$ And length$ h= 13$, i.e. $2\pi R h$.

So the heat flux you need is $$-K \times 8R \times 2\pi Rh$$.

Alternatively, use the divergence theorem to compute the integral over the volume. $$ -K \int\int_{S+C} \nabla u \cdot dS = -K\int \int\int_V \operatorname{div} (\nabla u) dV $$

Where $C$ denotes the two ends of that cylinder in question. Noting the following 3 things:

1. The divergence is 16

2. The flux through the end caps is zero (so all of it is through the surface S.

3. Since the divergence is constant, the integral is simply the product of the volume and $\pi R^2 h$ and the divergence 16.

Again, we get that the flux is $$-K \times 16 \times \pi R^2 h$$.

Answer 2

I know you picked a best answer already, but I was the in the middle of writing this and didn't feel like deleting, so I'll keep it here in case someone else happens to have the same question.

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Using cylindrical coordinates $$ y = r\cos \phi. \ z = r\sin\phi,\ x = x $$

We can parametrize the surface as $$ r = 18, \ -10 \le x \le 3 $$

From here you can either do the computations using standard Cartesian vectors (common in math) or curvilinear unit vectors aligned with the new coordinates (common in physics). I'll detail both so you can pick one depending on what you know.

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First method: Cartesian vectors

Using the parametrization $$ x = x, \ y = 18\cos\phi, \ z = 18\sin\phi $$ We have $$ \nabla u = 8y \ \hat{\boldsymbol{\imath}} + 8z \ \hat{\boldsymbol{\jmath}} = 8\cdot 18(\cos\phi \ \hat{\boldsymbol{\imath}} + \sin\phi \ \hat{\boldsymbol{\jmath}} ) $$

and the surface is $$ \boldsymbol{S}(x,\phi) = x\ \hat{\boldsymbol{\imath}} + 18\cos \phi \ \hat{\boldsymbol{\jmath}} + 18\sin \phi \ \hat{\boldsymbol{k}} $$

The normal vector is $$ \begin{align} d\boldsymbol{S} &= \frac{\partial \boldsymbol{u}}{\partial x}\ dx \times \frac{\partial \boldsymbol{u}}{\partial \phi} \ d\phi \\ &= \hat{\boldsymbol{\imath}} \ dx \times (-18\sin\phi \ \hat{\boldsymbol{\jmath}} +18\cos\phi \ \hat{\boldsymbol{k}}) \ d\phi \\ &= -18(\cos\phi \ \hat{\boldsymbol{\jmath}} + \sin\phi \ \hat{\boldsymbol{k}} ) \ dx \ d\phi \end{align} $$

The integrand simplifies to $$ \nabla u \cdot d\boldsymbol{S} = -8\cdot 18^2 (\cos^2 \phi + \sin^2 \phi) \ dx \ d\phi = -8\cdot 18^2 \ dx \ d\phi $$

The limits are the the same for $x$ and one full revolution for $\phi$, so the integral looks something like

$$ \iint_S = -8\cdot 18^2 \int_0^{2\pi}\int_{-10}^3 \ dx \ d\phi $$

which is a constant

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Second method: Curvilinear vectors

In cylindrical coordinates, the unit vectors have the form $$ \hat{\boldsymbol{r}} = \cos\phi \ \hat{\boldsymbol{\jmath}} + \sin\phi \ \hat{\boldsymbol{k}}, \ \hat{\boldsymbol{\phi}} = -\sin\phi \ \hat{\boldsymbol{\jmath}} + \cos\phi \ \hat{\boldsymbol{k}}, \ \hat{\boldsymbol{x}} = \hat{\boldsymbol{\imath}} $$

with $$ u(x,r,\phi) = 4r^2 $$

The gradient is expressed as $$ \nabla u = \frac{\partial u}{\partial x} \ \hat{\boldsymbol{x}} + \frac{\partial u}{\partial r} \ \hat{\boldsymbol{r}} + \frac{1}{\phi} \frac{\partial u}{\partial \phi} \ \hat{\boldsymbol{\phi}} = 8r \ \hat{\boldsymbol{r}} $$

$S$ is one of the coordinate surfaces, so the normal is simply $$ d\boldsymbol{S} = - r \ \hat{\boldsymbol{r}} \ dx \ d\phi $$

with the negative sign indicating the inward direction

The integral simplifies to $$ \iint_S = \left( -8 r^2 \left.\int_0^{2\pi}\int_{-10}^3 \ dx \ d\phi \right) \right|_{r = 18} $$

which, again, is a constant. Intuitively this makes sense as we have a radially symmetric heatflow over a radially symmetric surface.