So we will discuss the Gauss's theorem using the simple cube.

So Melody, instead of the simplicity of this shape of a cube,

what is good about cube?

>> It's symmetrical.

>> It is symmetrical, it is symmetrical in three directions right?

>> Mm-hm.

>> Meaning, if I just think about one direction, say x direction,

then we can immediately understand about y and z, right?

>> Mm-hm.

>> So here we will only thing about the surface 1 and surface 2,

which has a surface normal along the x direction, okay?

And the flux through S1 is equal to integral of C dot nda over S1.

Where because the right direction is positive, the surface mole,

which is pointing to the left, will have a minus sign.

So that's why we put minus, right?

dydz will be the area of this infinitesimally small cube, right?

Cx will be the x component of the C vector.

So in a nutshell, it will be approximated by -Cx(1)

because we're thinking of surface 1, times delta y delta z.

Now, if we think about flux through S2,

which is the adjacent surface, they have the surface.

This has in the surface null to the right, which is positive, so

we don't have minus sign in front of it.

So this will be Cx(2) delta y delta z.

Now, we are going to use a first order approximation

of our function where Cx(2) is approximated by

Cx(1) + round Cx over round x times delta x.

And if I replace this in this equation and if we sum this up,

what remains is round Cx over round x times delta x, delta y, and delta z.

And this is the volume of the cube that we are now discussing, right?

And as we have solved for 1 and 2, right?

What would happen to y and x, Melody?

>> Well, because they're symmetrical they should have a very similar appearance.

>> Exactly, so if you look on this slide, the next slide,

you can see from S1 and to S2, we just solve it.

S3 and S4, it just changes the subscript from x to y.

S5 and S6, you just change the subscript to z.

And if you sum them up you will have this form for an infinitesimal cube, right?

So the flux out of a cube = (rounded

Cx over round x + round Cy over round y +

round Cz over round z) delta x delta y delta z.

And if you remember this, this was just the formula for

divergence of a vector field.

Then you can make it neater by using the vector formula (del dot C) del V.

So looking at this formula, what can we know or

what is the meaning of the divergence of a vector field, Melody?

>> So it's the flux- >> Mm-hm.

>> Over the volume is equal to the divergent of the vector field.

>> Exactly.

So what Melody told us right now is the fact

that if you're relearning this equation,

then the flux, which is D dot nda over delta V,

which is the volume of the cube is equal to the divergence of the vector field.

So this is a point function, right?

Divergence of the vector field is a point function.

And so at each point, we can define the flux out of

this point per unit volume, per unit volume.

So it's flux per unit volume, all right?

So it means, if I integrate this over an arbitrary space,

then we can evaluate the flux out of that arbitrary volume, right?

So this is really important, so

divergence of a vector field is the flux at that point per unit volume.

So in this page, we're going to discuss one more time about the heat transport,

to see how the theory we learned, the divergent theorem,

is connected to heat conduction case.

So we'll revisit the problem of outward heat flow from a body placed in

a system where the heat is conserved.

And that was an example we just discussed before about the campfire and

fictitious boundary that encloses the campfire.

In that case, total heat flow outward through

the boundary S is equal the surface integral of h dot nda.

Which is equal to minus dQ over dt,

where the Q represents total heat inside the boundary.

Now, Melody, what would happen if we

shrank the volume in this to a very little cube?

>> Thank you.

So, if you have a smaller cube,

then from the previous slide we learned this equation.

So we can now write this is the differential form with the dels,

which is more convenient later because we don't need boundary conditions.

So we set this del h times delta V equal to this differential.

And if we think of a very small cube, then you have a uniform

density inside, or in this case a density of heat.

So you can change this into a q, which represents

the heat per unit volume and multiply it by sum delta V.

And then you get this equation right here,

which is the native differential of the heat density multiplied by the volume.

>> Very good.

So from here, you can see that you have the common factor,

delta V, on both sides, left and right.

So we can remove it and derive the conservation law in a differential form.

So this was a conservation law in an integral form,

this will be in differential form.

So now you can define at each point for the conservation of heat energy.

And as you can see, the negative of the heat testing change

as a function of time is equal to the diversions of the heat flow vector, okay?

Now we're going to use Gauss' Theorem, or Divergence Theorem,

to prove the heat conservation law in integral form from differential one.

So this will be a mental exercise to go back and

forth what we just learned it for.

So let's start with the Gauss' law which states the flux of

a vector field from and arbitrary shaped volume,

where the boundary is the closed surface denoted by S.

That is equal to the volume integral of divergence

of this vector field across the volume, or including the entire volume.

Now, from the previous slide, we learned that the energy conservation,

heat energy conservation can take a differential form

where- dq over dt = del dot h, right?

So we are going to use this inside the Gauss' law, or Gauss Theorem.

And as you can see here, the C is replaced by h.

So you will see we're just repeating the first part here.

And then we are going to replace the divergence of vector

field by this equation for energy conservation.

And if I do that, instead of the divergence, we will have dq over dt.

Then we can change the sequence of integration and differentiation.

So d over dt will come out, and then it will be the volume integral

of the charged heat density in this arbitrary shape volume.

And that will really give you the total heat, right?

Total heat, so again, you can see the total heat

flow outward through S = integration of h dot nda = dQ over dt.

Which is the integral form of heat conservation, so

that will be end of proof.

So what we do in this kind of mental practice,

we can go back and forth and

corroborate what we just learned, right?