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
\begin{equation*}
\vec r_1(u,v) = \langle v\cos u,v\sin u, 0\rangle, 0\leq u\leq 2\pi 0\leq v\leq 1\text{.}
\end{equation*}
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
\begin{equation*}
\vec r_2(u,v) = \langle v\cos u,v\sin u, 1-v^2\cos^2u-v^2\sin^2u\rangle = \langle v\cos u,v\sin u, 1-v^2\rangle\text{,}
\end{equation*}
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
\begin{equation*}
\vec n_1 = \vec r_{1u}\times \vec r_{1v} = \langle 0,0,-v\rangle\text{.}
\end{equation*}
As this vector has a negative \(z\)-component, we instead use
\begin{equation*}
\vec n_1 = \vec r_{1v}\times \vec r_{1u} = \langle 0,0,v\rangle\text{.}
\end{equation*}
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
\begin{equation*}
\vec n_2 = \vec r_{2u}\times \vec r_{2v} = \langle -2v^2\cos u,-2v^2\sin u,-v\rangle\text{.}
\end{equation*}
Again, this normal vector has a negative \(z\)-component so we use
\begin{equation*}
\vec n_2 = \vec r_{2v}\times \vec r_{2u} = \langle 2v^2\cos u,2v^2\sin u,v\rangle\text{.}
\end{equation*}
We are now set to compute flux. Over field \(\vec F_1=\langle 0,0,1\rangle\text{:}\)
\begin{align*}
\text{ Flux across } \surfaceS_1 \amp = \iint_{\surfaceS_1} \vec F_1\cdot \vec n_1\, dS\\
\amp = \iint_R\langle 0,0,1\rangle\cdot\langle 0,0,v\rangle\, dA\\
\amp = \int_0^1\int_0^{2\pi} (v)\, du\, dv\\
\amp = \pi\text{.}
\end{align*}
\begin{align*}
\text{ Flux across } \surfaceS_2 \amp = \iint_{\surfaceS_2} \vec F_1\cdot \vec n_2\, dS\\
\amp = \iint_R\langle 0,0,1\rangle\cdot\langle 2v^2\cos u,2v^2\sin u,v\rangle\, dA\\
\amp = \int_0^1\int_0^{2\pi} (v)\, du\, dv\\
\amp = \pi\text{.}
\end{align*}
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 \(\vec F_2=\langle 0,0,z\rangle\text{:}\)
\begin{align*}
\text{ Flux across } \surfaceS_1 \amp = \iint_{\surfaceS_1} \vec F_2\cdot \vec n_1\, dS .\\
\end{align*}
Over \(\surfaceS_1\text{,}\) \(\vec F_2 = \vec F_2\big(\vec r_2(u,v)\big) = \langle 0,0,0\rangle\text{.}\) Therefore,
\begin{align*}
\amp = \iint_R\langle 0,0,0\rangle\cdot\langle 0,0,v\rangle\, dA\\
\amp = \int_0^1\int_0^{2\pi} (0)\, du\, dv\\
\amp = 0\text{.}
\end{align*}
\begin{align*}
\text{ Flux across } \surfaceS_2 \amp = \iint_{\surfaceS_2} \vec F_2\cdot \vec n_2\, dS .\\
\end{align*}
Over \(\surfaceS_2\text{,}\) \(\vec F_2 = \vec F_2\big(\vec r_2(u,v)\big) = \langle 0,0,1-v^2\rangle\text{.}\) Therefore,
\begin{align*}
\amp = \iint_R\langle 0,0,1-v^2\rangle\cdot\langle 2v^2\cos u,2v^2\sin u,v\rangle\, dA\\
\amp = \int_0^1\int_0^{2\pi} (v^3-v)\, du\, dv\\
\amp = \pi/2\text{.}
\end{align*}
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 \(\vec n_2\text{.}\)