If you're seeing this message, it means we're having trouble loading external resources on our website.

If you're behind a web filter, please make sure that the domains *.kastatic.org and *.kasandbox.org are unblocked.

Main content

# Explanation of example 1

## Video transcript

in the last video we use the divergence theorem to show that the flux across this surface right now which is equal to the divergence of F along or summed up and throughout the entire region it's equal to zero now what I want to do in this video is think a little bit whenever you get an answer like zero you want to think about why is it this is saying that there's no there's no net flux across this surface right over here or if you sum up all of the divergences in this volume you are getting zero so why is that well the simple way to think about it is when we took the divergence of F this vector field F is hard to visualize but the divergence of F is fairly easy to visualize the divergence is equal to 2 times X so over here you're going to get as you go further and further in this direction as X becomes larger your divergence becomes more and more positive so you have very you have kind of a divergence of two right over here you have a divergence of one along that line and you have a divergence of zero right there and that's also true obviously as you go higher because they're just changing the Z you're not changing the X so all over here you have positive divergence over there you have positive entrance and not just along that plane but if you go in the X direction as well it's kind of this whole region of space you have positive divergence I guess you could say in the positive x side of our of our octant but then as you go in that side on the other side you have negative divergence and this this diagram is symmetric with respect to the zy plane and so those divergences cancel them out you would have had a positive flux across the surface or positive value right over here instead of calculating it for this region we had calculated it for a region that was just between X is 0 and 1 so let's just think about that region so that region that would been cutoff right over here that would have been cutoff right over there and so the back I guess you could say the back wall of that the back wall of this would have been would have been the Z Y plane now if we care about this if we care about this volume so we're essentially eliminating the rest of it so let me try to eliminate it as best as I can change the colors so if I eliminate that part of it over there and all even what we see all of that I should have deleted that first so if we eliminate the back part of it and we're just dealing with we're just dealing it with with it when X is positive then our entire solution that we did in the last video would've been the exact same except now X is going to vary between instead of negative 1 & 1 it'll vary between 0 & 1 and so our bounds of integration X is going to go between 0 & 1 0 & 1 and then in that situation our final answer this part this would be between 0 & 1 that would all be 0 and we would be left with 3 halves minus 1/2 3 halves minus 1/2 is 1 minus 1/6 which is just going to be 5 6 and so when you just think about this part of it this side of it this one that I've just drawn you had a positive flux of 5 6 on the other side you had a negative flux of 5 6 and then they cancelled it out one way to interpret that is if you think about that entire if you if we thought about the entire surface like what we saw in the last video that you had an aggregate inward flux in the last video that's why it was negative and that is completely offset by an outward flux right over here the positive or you had a negative divergence in that other region and you have a positive divergence that completely offsets it in this region right over here