MAN: Verify the divergence theorem for vector field F of x comma y comma z equal to angle bracket x plus y plus z comma y comma 2x minus y close angle bracket and surface S given by the cylinder x squared plus y squared equals 1, with 0 less than or equal to z less than or equal to 3 plus the circular top and bottom of the cylinder. Assume that S is positively oriented. The divergence theorem set states that if S is a surface that encloses a solid E, the triple integral over E of the divergence of F is equal to the double integral over S of F dot ds.

In order to verify this, we're going to calculate both integrals separately and compare are the results. Let's begin with the triple integral over E of the divergence of F. To do this, we need to note that the divergence of F, which is p sub x plus q sub y plus r sub z, is equal to 1 plus 1 plus 0. So the divergence of F equal to 2, which means the triple integral over E of the divergence of F. dv is equal to 2 times the volume of E.

Well, since the volume of this solid, the cylinder, after all is found by pi times 1 squared times 3. To times that volume would be equal to 6 pi. To verify the divergence there means that the second integral needs to be equal to 6 pi, so let's go ahead and follow that through.

Now, notice when we find the surface integral, the triple integral over S of F dotted with ds, there are three parts to this integral. We need to find the double integral over surface 1, which I'm going to define as the sides or the side of the cylinder of F dotted with ds 1, surface 1, plus the double integral over surface 2 of F dotted with d surface 2 plus double integral over S3, or surface 3, of F dotted with ds 3, surface 3 again.

So we'll find these three separate things. Well, the first, I said surface 1 is the side of the cylinder. Surface 2, I'm going to call the bottom of the cylinder, and surface 3 is the top.

As it's going to turn out, the bottom and the top are going to have very similar parameterizations, and in fact the side will have a different one, which is why we have to break this up into three integrals. So let's begin with surface 1. Surface 1, the parameterization will be r of u comma v equals angle bracket cosine u comma sine u coma v end angle bracket, where 0 is less than or equal to u is less than or equal to 2 pi and 0 is less than or equal to v is less than or equal to 3.

Think about this for a moment. A cylinder is made up an infinite number of stacked circles. So cosine u sine u, where u is between 0 and 2 pi, represents each of those circles. Then the z component ranges from 0 to 3 because it has a height of 3. Well, z is actually from 0 to 3.

All right. From here, we need to find t sub u, which will be equal to angle bracket negative sine u comma cosine of u comma 0 close angle bracket. And t sub v is equal to open angle bracket negative sine u comma cosine u-- oh, I'm rewriting the same one. t sub v equals open angle bracket 0 comma 0 comma 1, which means that t sub u crossed with t sub v will be equal to-- with some scratch work on your part and on mine-- open angle bracket cosine u comma sine u comma 0 end angle bracket.

Now, to calculate this integral, we need to find F dotted with parentheses t sub u cross t sub v close parentheses. So let's use our parameterization. Replace that in for the function for-- replace that in the function of the vector field F of x, y, z with that parameterization.

That means that this will be equal to open angle bracket cosine u plus sine u plus v comma sine u comma 2 cosine u minus sine u end angle bracket dotted with open angle bracket cosine u comma sine u comma 0 end angle bracket. So F with the values from r of u comma v replaced in there, and then our t sub you cross t sub v replaced the second vector.

Now, if we find that dot product, that will be equal to-- well, since sine squared plus cosine squared equals 1, this will be equal to 1 plus sine u times cosine u plus v cosine u, so that the surface integral on surface 1 of F dotted with d surface 1 becomes the integral from 0 to 3 with respect to v of the integral from 0 to 2 pi with respect to u of 1 plus sine u times cosine u plus v cosine u.

And because the inner integral has a value of 2 pi, this becomes the integral from 0 to 3 with respect to v of 2 pi, which is equal to 6 pi. So we're on our way to showing these two are equivalent or equal. We just need to find the two other surfaces, the bottom and the top of our cylinder.

Now, in just a moment, I'm going to point out specifically what I mean. But surface 2 and surface 3 have very similar parameterization, so in reality we're only going to find the bottom surface, surface 2. And then I'll point out why the two are actually the same. We don't need to calculate that second one.

So let's find surface 2. All right. For surface 2, we're going to use the-- well, we're ultimately going to find the double integral over surface 2 of F dotted with ds sub 2, but I'll come back to that. So let's see.

r of u comma v is equal to angle bracket u cosine v comma u sine v comma 0, where 0 is less than or equal to u is less than or equal to 1 and 0 is less than or equal to v is less than or equal to 2 pi. This requires a different parameterization because in the case of the side of the cylinder we have a constant radius, whereas when we're finding the side we're actually beginning with a radius of 0, the center of that circle, and working our way out. So we have a range on u, which is representing our radius, but we're still working in 2 pi radians all the way around the circle. So we still have the range on v.

All right. So this means t sub u will be equal to cosine-- or angle bracket cosine-- I left something off. t sub u-- no, that's right-- equals angle bracket cosine v comma sine v comma 0 end angle bracket. T sub v is angle bracket negative u sine v comma u cosine v comma 0 end angle bracket. And t sub u crossed with t sub v will be the vector open bracket 0 comma 0 comma u end angle bracket.

Now we'll find F dotted with open parentheses t sub u cross t sub v close parentheses. That'll be equal to angle bracket u cosine v plus u sine v plus 0 comma u sine v comma 2u cosine v minus u sine v end angle bracket dotted with open angle bracket 0 comma 0 comma u end angle bracket. Calculating that dot product, that would be equal to 2u squared cosine v minus u squared sine v.

So that integral on surface 2 of F dotted with d surface 2 is the integral from 0 to 1 with respect to u of the integral from 0 to 2 pi with respect to v of the function 2u squared cosine v minus u squared sine v. The inner integral has a value of 0, so this equals the integral from 0 to 1 with respect to u of 0, which is equal to 0. Now let's think through what surface 3 would look like.

Our parameterization would be the same as surface 2, except for the z component because in that case there is a fixed z component of 3 at the top of the cylinder. However, changing that 0 to a 3 does not change t sub u or t sub v. It also doesn't change the vector field, so in fact nothing else hinges on that. So it would also be 0, therefore the double integral of surface 1 actually represents the entire surface integral, all three terms. So that sum is equal to 6 pi, which verifies that the divergence theorem holds for the specific case of this vector field with this surface.