Simple Surface Integral - Heat Flow on Surface of Star

[wadawalnut]

Homework Statement

I have this problem in an online assignment. Someone told me the answer, so I already got it right, but I don't know why my logic leads me to the wrong answer. The problem:
The temperature u of a star of conductivity 1 is defined by u = \frac{1}{sqrt(x^2+y^2+z^2)}. If the star is a sphere of radius 5, find the rate of heat flow outward across the surface of the star.

Homework Equations

Surface integral: \int \int_S f(x,y,z) * dS
Surface area of sphere = 4\pi r^2
The correct answer is 4\pi

The Attempt at a Solution

I tried this two ways and got the same answer both ways, unfortunately this answer isn't correct.
First method was flat out doing the integral. Firstly, it is clear that u = \frac{1}{| \vec{r} |}. However, the radius on the surface of the sphere is always 5, so u = \frac{1}{5}. If we use the parametrization \vec{r} = (\rho \cos(\theta)\sin(\phi), \rho \sin(\theta) \sin(\phi), \rho \cos(\phi)) And dS is \rho^2 \sin(\phi) d\phi d\theta. Once again, \rho = 5 for every point on the surface of the star. Now we can fill out the surface integral:
\int_0^{\pi} \int_0^{2\pi} \frac{1}{5} * 25\sin(\phi)d\theta d\phi \\<br /> = 5 \int_0^{\pi} \int_0^{2\pi} \sin(\phi) d\theta d\phi \\<br /> = 10\pi \int_0^{\pi} \sin(\phi) d\phi \\<br /> = 10\pi (-\cos(\phi))|_0^{\pi} = 20\pi
I knew that wasn't correct, so I thought of an alternative: At each point on the surface, the value of the function is just \frac{1}{r} = \frac{1}{5}. So shouldn't the surface integral evaluate to the surface area of the sphere multiplied by the function u, and thus \frac{4\pi 5^2}{5} = 20\pi?
It seems like the value doesn't depend on the radius.
Can someone explain this to me?

 

[Ray Vickson]

Homework Statement

I have this problem in an online assignment. Someone told me the answer, so I already got it right, but I don't know why my logic leads me to the wrong answer. The problem:
The temperature u of a star of conductivity 1 is defined by u = \frac{1}{sqrt(x^2+y^2+z^2)}. If the star is a sphere of radius 5, find the rate of heat flow outward across the surface of the star.

Homework Equations

Surface integral: \int \int_S f(x,y,z) * dS
Surface area of sphere = 4\pi r^2
The correct answer is 4\pi

The Attempt at a Solution

I tried this two ways and got the same answer both ways, unfortunately this answer isn't correct.
First method was flat out doing the integral. Firstly, it is clear that u = \frac{1}{| \vec{r} |}. However, the radius on the surface of the sphere is always 5, so u = \frac{1}{5}. If we use the parametrization \vec{r} = (\rho \cos(\theta)\sin(\phi), \rho \sin(\theta) \sin(\phi), \rho \cos(\phi)) And dS is \rho^2 \sin(\phi) d\phi d\theta. Once again, \rho = 5 for every point on the surface of the star. Now we can fill out the surface integral:
\int_0^{\pi} \int_0^{2\pi} \frac{1}{5} * 25\sin(\phi)d\theta d\phi \\<br /> = 5 \int_0^{\pi} \int_0^{2\pi} \sin(\phi) d\theta d\phi \\<br /> = 10\pi \int_0^{\pi} \sin(\phi) d\phi \\<br /> = 10\pi (-\cos(\phi))|_0^{\pi} = 20\pi
I knew that wasn't correct, so I thought of an alternative: At each point on the surface, the value of the function is just \frac{1}{r} = \frac{1}{5}. So shouldn't the surface integral evaluate to the surface area of the sphere multiplied by the function u, and thus \frac{4\pi 5^2}{5} = 20\pi?
It seems like the value doesn't depend on the radius.
Can someone explain this to me?

Your $u = u(x,y,z)$ is the temperature at the point $(x,y,z)$ in the star. How is the rate of heat flow across the star's surface related to the function $u$? In other words, how is your function $f(x,y,z)$ obtained from the function $u(x,y,z)$?

 

[STEMucator]

In general:

$$\iint_S f(x, y, z) \space dS = \iint_D f(\vec r(u, v)) |\vec r_u \times \vec r_v| \space dA$$

For this particular problem we can adapt the notation:

$$\iint_S u(x, y, z) \space dS = \iint_D u(\vec r(\theta, \phi)) |\vec r_{\theta} \times \vec r_{\phi}| \space dA$$

You need to find a suitable $\vec r(\theta, \phi)$, and project the sphere onto the x-y plane to obtain the region $D$. Take $\rho = 5$ for the parameterization.

 

[Ray Vickson]

In general:

$$\iint_S f(x, y, z) \space dS = \iint_D f(\vec r(u, v)) |\vec r_u \times \vec r_v| \space dA$$

For this particular problem we can adapt the notation:

$$\iint_S u(x, y, z) \space dS = \iint_D u(\vec r(\theta, \phi)) |\vec r_{\theta} \times \vec r_{\phi}| \space dA$$

You need to find a suitable $\vec r(\theta, \phi)$, and project the sphere onto the x-y plane to obtain the region $D$. Take $\rho = 5$ for the parameterization.

Do we know that $f = u$? I asked the OP that, but have received no response.