APEX Calculus

We begin by parametrizing the planar surface \(\surfaceS\text{.}\) Using the techniques of the previous section, we can let \(x=u\) and \(y=v(2-2u)\text{,}\) where \(0\leq u\leq 1\) and \(0\leq v\leq 1\text{.}\) Solving for \(z\) in the equation of the plane, we have \(z=3-2x-y\text{,}\) hence \(z = 3-2u-v(2-2u)\text{,}\) giving the parametrization \(\vec r(u,v) = \langle u, v(2-2u), 3-2u-v(2-2u)\rangle\text{.}\)

We need \(dS=\snorm{\vec r_u\times \vec r_v}dA\text{,}\) so we need to compute \(\vec r_u\text{,}\) \(\vec r_v\) and the norm of their cross product. We leave it to the reader to confirm the following:

We need to be careful to not “simplify” \(\snorm{\vec r_u\times \vec r_v} = 2\sqrt{6}\sqrt{(u-1)^2}\) as \(2\sqrt{6}(u-1)\text{;}\) rather, it is \(2\sqrt{6}|u-1|\text{.}\) In this example, \(u\) is bounded by \(0\leq u\leq 1\text{,}\) and on this interval \(|u-1| = 1-u\text{.}\) Thus \(dS = 2\sqrt{6}(1-u)dA\text{.}\)

The density is given as a function of \(x\text{,}\) \(y\) and \(z\text{,}\) for which we'll substitute the corresponding components of \(\vec r\) (with the slight abuse of notation that we used in previous sections):

Thus the mass of the sheet is:

Using our work from the previous example, we have \(\vec n = \vec r_u\times\vec r_v = \langle 4-4u,2-2u,2-2u\rangle\text{.}\) We also need \(\vec F\big(\vec r(u,v)\big) = \langle 1, u, -v(2-2u)\rangle\text{.}\)

Thus the flux of \(\vec F\) across \(\surfaceS\) is:

To make full use of this numeric answer, we need to know the direction in which the field is passing across \(\surfaceS\text{.}\) The graph in Figure 14.6.8 helps, but we need a method that is not dependent on a graph.

Pick a point \((u,v)\) in the interior of \(R\) and consider \(\vec n(u,v)\text{.}\) For instance, choose \((1/2,1/2)\) and look at \(\vec n(1/2,1/2) = \langle 2,1,1\rangle/\sqrt{6}\text{.}\) This vector has positive \(x\text{,}\) \(y\) and \(z\) components. Generally speaking, one has some idea of what the surface \(\surfaceS\) looks like, as that surface is for some reason important. In our case, we know \(\surfaceS\) is a plane with \(z\)-intercept of \(z=3\text{.}\) Knowing \(\vec n\) and the flux measurement of positive \(5/3\text{,}\) we know that the field must be passing from “behind” \(\surfaceS\text{,}\) i.e., the side the origin is on, to the “front” of \(\surfaceS\text{.}\)

We begin by parametrizing each surface.

The boundary of the unit disk in the \(xy\)-plane is the unit circle, which can be described with \(\langle \cos u,\sin u,0\rangle\text{,}\) \(0\leq u\leq 2\pi\text{.}\) To obtain the interior of the circle as well, we can scale by \(v\text{,}\) giving

As the boundary of \(\surfaceS_2\) is also the unit circle, the \(x\) and \(y\) components of \(\vec r_2\) will be the same as those of \(\vec r_1\text{;}\) we just need a different \(z\) component. With \(z = 1-x^2-y^2\text{,}\) we have

where \(0\leq u\leq 2\pi\) and \(0\leq v\leq 1\text{.}\)

We now compute the normal vectors \(\vec n_1\) and \(\vec n_2\text{.}\)

For \(\vec n_1\text{:}\) \(\vec r_{1u}= \langle -v\sin u, v\cos u,0\rangle\text{,}\) \(\vec r_{1v} = \langle \cos u,\sin u,0\rangle\text{,}\) so

As this vector has a negative \(z\)-component, we instead use

Similarly, \(\vec n_2\text{:}\) \(\vec r_{2u}= \langle -v\sin u, v\cos u,0\rangle\text{,}\) \(\vec r_{2v} = \langle \cos u,\sin u,-2v\rangle\text{,}\) so

Again, this normal vector has a negative \(z\)-component so we use

We are now set to compute flux. Over field \(\vec F_1=\langle 0,0,1\rangle\text{:}\)

These two results are equal and positive. Each are positive because both normal vectors are pointing in the positive \(z\)-directions, as does \(\vec F_1\text{.}\) As the field passes through each surface in the direction of their normal vectors, the flux is measured as positive.

We can also intuitively understand why the results are equal. Consider \(\vec F_1\) to represent the flow of air, and let each surface represent a filter. Since \(\vec F_1\) is constant, and moving “straight up,” it makes sense that all air passing through \(\surfaceS_1\) also passes through \(\surfaceS_2\text{,}\) and vice-versa.

If we treated the surfaces as creating one piecewise-smooth surface \(\surfaceS\text{,}\) we would find the total flux across \(\surfaceS\) by finding the flux across each piece, being sure that each normal vector pointed to the outside of the closed surface. Above, \(\vec n_1\) does not point outside the surface, though \(\vec n_2\) does. We would instead want to use \(-\vec n_1\) in our computation. We would then find that the flux across \(\surfaceS_1\) is \(-\pi\text{,}\) and hence the total flux across \(\surfaceS\) is \(-\pi + \pi = 0\text{.}\) (As \(0\) is a special number, we should wonder if this answer has special significance. It does, which is briefly discussed following this example and will be more fully developed in the next section.)

We now compute the flux across each surface with \(\vec F_2=\langle 0,0,z\rangle\text{:}\)

This time the measurements of flux differ. Over \(\surfaceS_1\text{,}\) the field \(\vec F_2\) is just \(\vec 0\text{,}\) hence there is no flux. Over \(\surfaceS_2\text{,}\) the flux is again positive as \(\vec F_2\) points in the positive \(z\) direction over \(\surfaceS_2\text{,}\) as does \(\vec n_2\text{.}\)