Let the vector field v equal angle bracket, a fraction of x divided by z, end fraction, fraction of y divided by z, end fraction, 0, end angle bracket be the velocity field of a fluid. Let C be the solid cube given by 1 is less than or equal to x is less than or equal to 4, and 2 is less than or equal to y is less than or equal to 5, and 1 is less than or equal to z is less than or equal to 4, and let S be the boundary of this cube. Find the flow rate of the fluid across S.

Let's apply the divergence theorem, so that us finding-- as we find the velocity or the flow rate of the fluid across S, we're going to take the double integral-- well, ultimately, we want to find the double integral on the surface S of the vector v dotted with dS. But the divergent theorem says that will be equal to the triple integral and the solid E that is contained by S of the divergence of the vector field v, dv.

So let's go ahead and find what the divergence of v would be. Taking as we know divergence of v is equal to Px plus P sub x plus Q sub y plus R sub z. And each of those partials, respectively, are 1 over z. The second is 1 over Z. And the third is 0.

So 1 over z plus 1 over z plus 0, which is equal to 2 over z. So we need to find the integral from 1 to 4 with respect to z of the integral from 2 to 5 with respect to y of the integral from 1 to 4 with respect to x of 2 over z.

And because we have some independent integrals here-- independent functions-- I'm going to write this as this is equal to open parentheses integral from 1 to 4 with respect to z of the function 2 over z, end parentheses, multiplied by open parentheses, integral from 2 to 5 with respect to y of the function 1, end parentheses, open parentheses, integral from 1 to 4 with respect to x of the function 1.

And the second two integrals that I stated have a value of 3. The first integral has a value of 18 ln(4). So the end value of this-- actually, sorry. That first one isn't-- so the second two integrals have value of 3 and 3 respectively. The first is 2 ln(4). So 18 ln(4) is our flow rate.

Now it's actually written in the textbook, Open Stacks Calculus Volume 3 that this is also in fact equal to 9 natural log of 16. I would prefer 18 natural log of 4. And just to show those do match, those are equivalent answers.