The interpretation of is to give the rate of change of the amount in the account (B) over time. Hopefully, this is an increasing function. The other partial derivative, tells you how much the amount of money in the bank will increase as you adjust the principal amount invested.
At the point we have the normal vector as .
Thus, the parameterization satisfies both equations and must therefore represent a curve on each surface. Since the equations represent two different planes, and the curve that is parameterized is a straight line, this line must represent the intersection of the two planes.
This is a difficult integral to do, but the problem is simplified by using the Fundamental Theorem of Calculus for Line Integrals. Note that the vector field above is a gradient field ( ) with a potential function given by . This means that
we now take partial derivatives with respect to y and compare with the second component of the vector field.
This means that we now have . Taking the partial derivative of this with respect to z and comparing with the thrid component of the vector field we find
And so we find that the potential function is constant.
Since these partial derivatives are not equal, we know that is not a gradient field and therefore has no potential.
Now we compute the partial derivatives. For simplicity, I show only the first term, .
Similar terms arise from the other derivatives. In the end, you will find that div , as long as .
Since the vector field is pointing radially outward, it will be easy to calculate the flux through a sphere centered at the origin. For this reason, we break the cube up into two parts. The first part is the sphere, centered at the origin, with radius a/2. The second part of the region will be the cube minus the sphere. Thus, together, these regions add up to the original cube. The benefit of this approach will soon be apparent.
Through the second region (cube minus sphere) we know that the divergence of the vector field is zero. Also, since we have removed the origin (it's part of the sphere) we can use the divergence theorem. This means that the flux through the surface of region two is zero.
Through the first region, we will have to compute the flux integral. This is easy, because the vector field and the normal to the sphere are in the same direction. Furthermore, on the surface of the sphere, r = a/2, so we know . Thus,