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A spherical, $3$-dimensional planet has center at $(0, 0, 0)$ and radius $20$. At any point of the surface of this planet, the temperature is $T(x, y, z) = (x + y)^2 + (y - z)^2$ degrees. What is the average temperature of the surface of this planet?

We will define this $3$-dimensional sphere by S=$\{(x,y,z) \in \mathbb{R}: x^2 + y^2 + z^2 = 20^{2}\}$

As we can see, $T(S)$ has a uncountable number of points. So I think I have to use the integral structure to solve this problem. For well understanding, it is possible to look at Darboux Theorem for integrability. Moreover, I have the feeling we will needed of the Spherical coordinates. Is anyone is able to give me a hint to continue this question.

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    By symmetry <x^2> = <y^2> = <z^2> = 1/3 r^2. And, of course, = = 0. So, the average is 4/3 r^2. – Count Iblis Sep 27 '15 at 19:24
  • Does the average of T is simply his integral? If so, why? –  Sep 27 '15 at 19:29
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    The average is the integral divided by the surface area. You can interpret this as saying that an area element divided by the total area is the probability that a randomly chosen point falls in the are element. This multiplied by the value of the function and integrated over the area gives you, by definition, the expectation value of the function. – Count Iblis Sep 27 '15 at 19:33
  • Or you could start by finding the average temperature of two thermometers, then add more and more in between the two, up to the limit of infinitesimal densely placed thermometers, which as you said eventually leads to an integral divided by the total surface area. ...in case you don't like to throw probabilities and expectation values at this problem. – null Sep 27 '15 at 19:40
  • Are you able to explain what you wrote with the definition of the integral (Theorem Darboux integrability for maybe)? It would help me greatly. I even believe that it is instructive for me and for you. –  Sep 27 '15 at 19:42
  • and explicite formulas if possible –  Sep 27 '15 at 19:47

2 Answers2

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There is a very basic solution that takes advantages of some of the symmetries you can impose on the problem.

$T\left(x,y,z\right)=x^2+2y^2+z^2+2xy-2yz$

Now we can simplify this to

$T\left(x,y,z\right)=400+y^2+2xy-2yz$

Since $x^2+y^2+z^2=400$

But we still have a lot of left over terms to get rid of and we can do this by taking advantage of symmetry.

By switching the order of our inputs we can get:

$T\left(x,y,z\right)=400+y^2+2xy-2yz$

$T\left(z,x,y\right)=400+x^2+2zx-2xy$

$T\left(y,z,x\right)=400+z^2+2yz-2zx$

So we have

$T\left(x,y,z\right)+T\left(z,x,y\right)+T\left(y,z,x\right)=1600$

For any $x,y,z$

This is helpful because if $(x,y,z)$ is on the sphere then we know both $(z,x,y)$ and $(y,z,x)$ are also on the sphere.

Thus, for all points on our sphere such that $x\neq y\neq z$ we can group these points into groups of $3$ such that the sum of their temperatures is $1600$. This means the average temperature on the surface of the planet is $1600/3$ degrees.

Note that we do not have to take into account the cases where $x=z$ or $y=z$ or $x=y$ since they make up such a minuscule part of the actual sphere.

Anirudh
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Start by expressing $T$ in polar co-ordinates ..

$T(r,\theta,\phi) =r^2 \left [ (\sin\theta(\sin\phi + \cos \phi))^2 + (\sin\theta\cos \phi - \cos \theta)^2 \right ]$

$ =r^2 \left [ \sin^2\theta(\sin^2\phi + 2 \sin\phi\cos\phi+ \cos ^2\phi) + (\sin^2\theta\cos^2 \phi - 2\sin^2\theta\cos \phi \cos\theta+\cos ^2\theta) \right ]$

$ =r^2 \left [ \sin^2\theta(1 + 2 \sin\phi\cos\phi) + (\sin^2\theta(\cos^2 \phi - 2\cos \phi \cos\theta)+\cos ^2\theta) \right ]$

$ =r^2 \left [ 1+\sin^2\theta( 2 \sin\phi\cos\phi + \cos^2 \phi - 2\cos \phi \cos\theta)\right ]$

Average Temperature = $\frac{I}{4 \pi r^2}$ where ...

$$I= r^2\int_0^{2 \pi} \int_0^{ \pi} \left [ 1+\sin^2\theta( 2 \sin\phi\cos\phi + \cos^2 \phi - 2\cos \phi \cos\theta)\right ] d\theta d\phi $$

I suggest you split this into four double integrals and evaluate each one separately

WW1
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