Surface integrals and heat flow

[csnsc14320]

Homework Statement

The temperature u in a star of conductivity 6 is inversely proportional to the distance from the center:

u = \frac{3}{\sqrt{x^{2} + y^{2} + z^{2}}}

If the star is a sphere of radius 3, find the rate of heat flow outward across the surface of the star.

Homework Equations

The Attempt at a Solution

I really don't know what to do here - I have not learned about "conductivity" in terms of taking surface integrals and every example problem my teacher has done has involved a force function, yet this seems to only give a scalar temperature function.

So, I am thoroughly confused as to where to start?

 

[keltix]

im doing the exact same one , it is due in 7 hrs
im lost too but

you know z^2+y^2+x^2 = r^2 so plug that into the eq. and solve for u.

after that I am lost

 

[HallsofIvy]

That looks to me like you need Gauss' theorem:
\int\int\int \nabla \cdot \vec{f} dV= \int\int \vec{f}\cdot \vec{n}dS
where the first integral is over a bounded region and the second over the surface of that area. The right side is the flow across the surface and so the integrand is the rate of heat flow across the surface.

 

[keltix]

you mean divergence theorem? but what is the vector field?

 

[lanedance]

can you make a vector heat field from your scalar tempertaure field?

any differential conduction equations that might hep?

 

[keltix]

the gradient.

then what?

use DIV theorem and spherical coordinates?

 

[lanedance]

right so you get the heat field from something like

\vec{q} = -k \nabla u

and
u(r) = \frac{3}{r}

now try the left hand integral given by Halls in Gauss's theorem, i think this is effectively summing up all the heat generated by the star, which in steady state will also be the amount leaving the surface of the star (by the way does the form of the field look familiar? maybe you can think of the temperature as the potential)

knowing the surface area of the star & due to the spherical symmetry of the heat field, you should be able to work back to the integrand on the right