Find the surface area of the part of the surface z=arctan

Find the surface area of the part of the surface $z=\arctan\dfrac yx$ that projects onto the region in the plane $z=0$ bounded by the Archimedean spiral $r=\varphi$ and the axis $Ox$ (one full turn, $\varphi\in[0,2\pi]$).

Explanation of Method

The integrand $\sqrt{1+z_x^2+z_y^2}=\sqrt{1+1/r^2}$ is radially symmetric, and in polar coordinates the extra factor of $r$ from $dx\,dy$ cancels the $1/r$ inside the root, leaving simply $\sqrt{r^2+1}$. The spiral's region, $0\le r\le\theta$ for $\theta\in[0,2\pi]$, then gives an iterated integral whose inner piece is an elementary $\int\sqrt{r^2+1}\,dr$, and whose outer piece requires integrating that result (itself a function of $\theta$) once more over $\theta$.

Compute the surface-area integrand and simplify in polar coordinates.
Evaluate the inner radial integral.
Evaluate the resulting outer integral in $\theta$ (splitting into an algebraic and a logarithmic part).

Step 1 · Surface-area integrand

With $z=\arctan\dfrac yx$: $z_x=\dfrac{-y}{x^2+y^2},\,z_y=\dfrac x{x^2+y^2}$, so $1+z_x^2+z_y^2=1+\dfrac1{x^2+y^2}$, giving (in polar coordinates, with $dx\,dy=r\,dr\,d\theta$):

$$ dS = \sqrt{1+\frac1{r^2}}\cdot r\,dr\,d\theta = \sqrt{r^2+1}\,dr\,d\theta. $$

Step 2 · Set up and evaluate the inner integral

The spiral region is $0\le r\le\theta$, $\theta\in[0,2\pi]$:

$$ S = \int_0^{2\pi}\int_0^\theta\sqrt{r^2+1}\,dr\,d\theta, \qquad \int_0^\theta\sqrt{r^2+1}\,dr = \frac{\theta\sqrt{\theta^2+1}}2+\frac12\ln\!\left(\theta+\sqrt{\theta^2+1}\right). $$

Step 3 · Outer integral

Splitting $S=\frac12I_1+\frac12I_2$: with $u=\theta^2+1$,

$$ I_1=\int_0^{2\pi}\theta\sqrt{\theta^2+1}\,d\theta = \frac13\Big[(4\pi^2+1)^{3/2}-1\Big]. $$

For $I_2=\int_0^{2\pi}\ln\!\left(\theta+\sqrt{\theta^2+1}\right)d\theta=\int_0^{2\pi}\sinh^{-1}\theta\,d\theta$, integrating by parts ($u=\sinh^{-1}\theta,\,dv=d\theta$):

$$ I_2 = \Big[\theta\sinh^{-1}\theta\Big]_0^{2\pi}-\int_0^{2\pi}\frac\theta{\sqrt{\theta^2+1}}\,d\theta = 2\pi\ln\!\left(2\pi+\sqrt{4\pi^2+1}\right)-\sqrt{4\pi^2+1}+1. $$

Combining:

$$ S = \frac{(4\pi^2+1)^{3/2}-1}6 - \frac{\sqrt{4\pi^2+1}-1}2 + \pi\ln\!\left(2\pi+\sqrt{4\pi^2+1}\right). $$

Additional Notes

This derivation matches the source exactly.
Verified numerically: direct double integration over the spiral region gives $S\approx48.0458$, matching the boxed closed form exactly.

Result

The surface area is:

$$ S = \frac{(4\pi^2+1)^{3/2}-1}6 - \frac{\sqrt{4\pi^2+1}-1}2 + \pi\ln\!\left(2\pi+\sqrt{4\pi^2+1}\right). $$