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### Course: Multivariable calculus>Unit 5

Lesson 7: 3D divergence theorem

# Explanation of example 1

Intuition as to why we got no net flux in the last worked example. Created by Sal Khan.

## Want to join the conversation?

• At why didn't he draw the arrows the other way?. In that case the flux (out) would be equal to the flux (in) and there would be 0 flux, right?
• I agree. Since positive divergence is defined in a direction out of a surface, a negative divergence would in this case mean that the arrows on the negative x-axis would point in the opposite direction.
• I completely understand Sal's intuitive argument explaining why the eventual result of this problem is 0. However, when Sal offers up the possibility of changing the bounds of "x" (and thus our region R) to show that the flux would not equal 0 for this new region, does it not contradict the idea that Gauss' theorem (the divergence theorem) can only be used for symmetric regions? In my calculus text, R is defined as a "symmetric elementary region." Is this new region actually "symmetric" & "elementary" and I'm actually lost in translation here?
• A Gaussian surface need not be symmetric, only closed. I guess that one often chooses a symmetrical surface because it's easier to do calculations with.
• If all of the arrows are pointing outward in the original problem, why isn't the total outward flux equal to 2*(5/6)?
• The arrows he drew were just examples of the normal direction. The direction that flux would have to travel in to be considered "outward flux".

Sal did not draw the direction that mass was traveling in. To do that he would need to draw the vector field F (the velocity of the mass), that would be difficult to do by hand for this example.
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• Would the solution change if F is not a divergent?
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• at those arrows should be pointing inwards correct? That would denote a negative divergence as particles are coming into the surface.
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• He was just drawing the direction that mass/stuff has to travel in to be considered "outward flux" (basically the normal vector, he drew the normal vector direction when z=0).

If you wanted to see the direction that mass traveled in when it crossed the surface you would need to draw the vector field F, which would be difficult to do by hand in this example.

Also remember divergence is a scalar field, not a vector field (it has no direction).
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• Would anyone know if there are videos on transport theorems ? I'm a vector calculus student trying to get some help on Reynolds transport and flux transport theorem
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• Is this condsidered as Calculus 3
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• Would it be valid to recognize right away that, because the integral with respect to x is being evaluated over an odd function (2x) and the bounds of this integration are a = -b and b=b, the integral is zero knowing that this is an inherent property of integrals of this type? In other words, once the bounds and flux are known, can all this calculation be skipped and the drawing be left un-drawn by calling this property? I feel like it would save a lot of calculus time if this is generalized.
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• You can, but you need to make sure that you are actually integrating over an odd function. If for instance, the bounds of integration of the z-integral were not the even function 1 - x^2, then the final integral might not go over an odd function.
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• Is it only because de base is symetric(one half in negative x et one half in positive x) that it's equals zero or because the entire volume is symetric? if the entire volume wasn't symétrique it wouldn't have been zero right?
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• Given a divergence of 2x, if the volume of our region is not symmetric about the yz plane, then the flux of F across the surface will be none-zero since the positive divergence on one side of the yz plane cannot completely cancel the negative divergence on the other side owing to a lack of symmetry.
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• Will you always get a flux or a divergence equal to zero when the shape is symmetric about an axis?