Three. (Full marks 20 points) Let $y^{2}=2^{n+1} x(n=1,2$, $3, \cdots, 2000)$ be a family of parabolas. Draw a line with an inclination angle of $45^{\circ}$ through the focus of each parabola, and intersect the parabola to get the chord $A_{n} B_{n}(n=1,2, \cdots, 2000)$. By rotating each chord $90^{\circ}$ around its corresponding directrix, 2000 rotational surfaces are obtained. Try to find the sum of the areas of all these rotational surfaces.
Three, the equation of the $n$-th parabola is $y^{2}=2^{n+1} x=2 \times 2^{n} x$, hence its focus coordinates are $\left(2^{n-1}, 0\right)$. The equation of the line passing through this focus with an inclination angle of $45^{\circ}$ is
$$
y=x-2^{n-1} \text {. }
$$

By solving the system of equations
$$
\left\{\begin{array}{l}
y^{2}=2^{n+1} x, \\
y=x-2^{n-1}
\end{array}\right.
$$

we can find the chord length.
Substituting (2) into (1) and simplifying, we get
$$
x^{2}-\left(2^{n}+2^{n+1}\right) x+2^{2 n-2}=0 \text {. }
$$

Thus, $x_{1}+x_{2}=2^{n}+2^{n+1}, x_{1} x_{2}=2^{2 n-2}$.
$$
\begin{aligned}
\left(x_{1}-x_{2}\right)^{2} & =\left(x_{1}+x_{2}\right)^{2}-4 x_{1} x_{2}=8 \times 2^{2 n}, \\
\left(y_{1}-y_{2}\right)^{2} & =\left[\left(x_{1}-2^{n-1}\right)-\left(x_{2}-2^{n-1}\right)\right]^{2} \\
& =\left(x_{1}-x_{2}\right)^{2}=8 \times 2^{2 n} .
\end{aligned}
$$

Therefore, the chord length $d_{n}=\sqrt{\left(x_{1}-x_{2}\right)^{2}+\left(y_{1}-y_{2}\right)^{2}}$
$$
=\sqrt{16 \times 2^{2 n}}=2^{n+2} \text {. }
$$

The surface of revolution is $\frac{1}{4} S_{\text {cylinder}}$,
$$
S_{\text {cylinder}}=\pi\left(r_{\text {top }}+r_{\text {bottom }}\right) l .
$$

As shown in Figure 4, by the definition of the parabola,
$$
\begin{array}{l}
r_{\text {top }}+r_{\text {bottom }} \\
=A_{n} F_{n}+B_{n} F_{n} \\
=A_{n} B_{n}=d_{n} \text {. } \\
\therefore S_{\text {cylinder}}=\pi d_{n}^{2} \\
=\pi\left(2^{n+2}\right)^{2} \\
=16 \pi \times 2^{2 n} \text {. } \\
\therefore \frac{1}{4} S_{\text {cylinder}} \\
=4 \pi \times 2^{2 n} . \\
\end{array}
$$

The sum of the areas of all the surfaces of revolution is
$$
\begin{array}{l}
4 \pi\left(2^{2}+2^{4}+\cdots+2^{4000}\right) \\
=4 \pi \cdot \frac{2^{2}\left[\left(2^{2}\right)^{2000}-1\right]}{2^{2}-1} \\
=\frac{16}{3} \pi\left(2^{4000}-1\right) .
\end{array}
$$