1°. Show that for the surface of revolution x=phi(u) cos
1°. Show that for the surface of revolution $x=\varphi(u)\cos\theta,\ y=\varphi(u)\sin\theta,\ z=\psi(u)$, $\alpha\le u\le\beta,\ 0\le\theta<2\pi$, with $\varphi,\psi$ differentiable, the surface area is $P=2\pi\displaystyle\int_\alpha^\beta|\varphi(u)|\sqrt{\varphi'^2(u)+\psi'^2(u)}\,du$.
2°. Compute the surface area of $x=e^{-u}\cos\theta,\ y=e^{-u}\sin\theta,\ z=\displaystyle\int_0^u\sqrt{1-e^{-2t}}\,dt$, $0\le u<\infty$.
Explanation of Method
Part 1° parametrizes the surface of revolution by $(u,\theta)$ and computes $\|\mathbf r_u\times\mathbf r_\theta\|$ directly, generalizing the earlier surface-of-revolution formula (see doubleint-212) to parametric generating curves. Part 2° applies this formula to a specific curve whose derivatives satisfy $\varphi'^2+\psi'^2=1$ — the defining property of arc-length parametrization — making the resulting integral immediately elementary; this particular surface is in fact the classical pseudosphere.
- Compute $\mathbf r_u\times\mathbf r_\theta$ and its magnitude (Part 1°).
- Integrate over $\theta$ to obtain the general formula.
- Apply the formula to the given curve (Part 2°) and evaluate.
Part 1° · Step 1: Partial derivatives
$$ \mathbf r_u = \big(\varphi'\cos\theta,\,\varphi'\sin\theta,\,\psi'\big), \qquad \mathbf r_\theta = \big(-\varphi\sin\theta,\,\varphi\cos\theta,\,0\big). $$
Part 1° · Step 2: Cross product and magnitude
$$ \mathbf r_u\times\mathbf r_\theta = \big(-\varphi\psi'\cos\theta,\,-\varphi\psi'\sin\theta,\,\varphi\varphi'\big), $$
$$ \|\mathbf r_u\times\mathbf r_\theta\| = \sqrt{\varphi^2\psi'^2+\varphi^2\varphi'^2} = |\varphi(u)|\sqrt{\varphi'^2(u)+\psi'^2(u)}. $$
Part 1° · Step 3: Integrate over $\theta$
$$ P = \int_0^{2\pi}\int_\alpha^\beta|\varphi(u)|\sqrt{\varphi'^2+\psi'^2}\,du\,d\theta = 2\pi\int_\alpha^\beta|\varphi(u)|\sqrt{\varphi'^2(u)+\psi'^2(u)}\,du. $$
Part 2° · Step 1: Identify $\varphi,\psi$ and their derivatives
$\varphi(u)=e^{-u}$, $\psi(u)=\displaystyle\int_0^u\sqrt{1-e^{-2t}}\,dt$, so $\varphi'(u)=-e^{-u}$, $\psi'(u)=\sqrt{1-e^{-2u}}$ (by the Fundamental Theorem of Calculus).
Part 2° · Step 2: Simplify the integrand
$$ \varphi'^2+\psi'^2 = e^{-2u}+\big(1-e^{-2u}\big) = 1, \qquad \sqrt{\varphi'^2+\psi'^2}=1. $$
Part 2° · Step 3: Evaluate
$$ P = 2\pi\int_0^\infty e^{-u}\,du = 2\pi\big[-e^{-u}\big]_0^\infty = 2\pi. $$
Additional Notes
- This derivation matches the source exactly. The curve in Part 2° is parametrized by arc length (since $\varphi'^2+\psi'^2=1$ identically), which is exactly what collapses the surface-area integral to $2\pi\int|\varphi(u)|\,du$.
- The surface in Part 2° is the classical pseudosphere (the surface of revolution of a tractrix), famous for having constant negative Gaussian curvature; its total area is finite, $2\pi$ (here with unit scale, since $\varphi(0)=1$), matching the well-known general pseudosphere area formula $2\pi a^2$.
- Verified numerically: $\displaystyle\int_0^\infty e^{-u}\,du=1$ to full precision, confirming $P=2\pi$.
Result
The general formula and its application give: