Find the surface area of the part of the surface

Find the surface area of the part of the surface $z=\dfrac{x+y}{x^2+y^2}$ cut out by the cylinders $x^2+y^2=1$ and $x^2+y^2=4$, lying in the first octant.

Step 1 · Surface-area integrand

With $z=\dfrac{x+y}{x^2+y^2}$: $z_x=\dfrac{y^2-x^2-2xy}{(x^2+y^2)^2},\,z_y=\dfrac{x^2-y^2-2xy}{(x^2+y^2)^2}$. Writing $A=x^2-y^2,\,B=2xy$, the numerators are $\mp(A+B)$ and $A-B$, so $z_x^2+z_y^2=\dfrac{2(A^2+B^2)}{(x^2+y^2)^4}=\dfrac{2(x^2+y^2)^2}{(x^2+y^2)^4}=\dfrac2{(x^2+y^2)^2}$ (using $A^2+B^2=(x^2+y^2)^2$). Hence:

$$ 1+z_x^2+z_y^2 = 1+\frac2{(x^2+y^2)^2}. $$

Step 2 · Polar coordinates

$$ S = \int_0^{\pi/2}\int_1^2\sqrt{1+\frac2{r^4}}\,r\,dr\,d\theta = \frac\pi2\int_1^2\frac{\sqrt{r^4+2}}r\,dr. $$

Step 3 · Evaluate via $u=r^2$

With $u=r^2$: $S=\dfrac\pi4\displaystyle\int_1^4\dfrac{\sqrt{u^2+2}}u\,du$. Using $\displaystyle\int\frac{\sqrt{u^2+2}}u\,du = \sqrt{u^2+2}+\sqrt2\ln\frac u{\sqrt{u^2+2}+\sqrt2}+C$:

$$ \int_1^4\frac{\sqrt{u^2+2}}u\,du = \Big[3\sqrt2 - \tfrac{\sqrt2}2\ln2\Big] - \Big[\sqrt3-\sqrt2\ln(\sqrt2+\sqrt3)\Big] = 3\sqrt2-\sqrt3+\sqrt2\ln\frac{\sqrt2+\sqrt3}{\sqrt2}. $$

Therefore:

$$ S = \frac\pi4\left[3\sqrt2-\sqrt3+\sqrt2\ln\frac{2+\sqrt6}2\right]. $$

Additional Notes

• A note on the source computation: the source's antiderivative and its two endpoint evaluations were correct (matching Step 3's bracketed values above exactly). The error appears only in the final combination step: the source needed to simplify $-\frac{\sqrt2}2\ln2+\sqrt2\ln(\sqrt2+\sqrt3)=\sqrt2\left[\ln(\sqrt2+\sqrt3)-\frac12\ln2\right]=\sqrt2\ln\dfrac{\sqrt2+\sqrt3}{\sqrt2}$, but instead wrote this as $\sqrt2\ln\dfrac{(\sqrt2+\sqrt3)^2}2$ — effectively squaring the argument of the logarithm without compensating by halving the coefficient, i.e. treating $\ln A-\frac12\ln2=\ln\frac{A^2}2$ instead of the correct $\ln\frac{A^2}2=2\big(\ln A-\frac12\ln2\big)$. This inflates the final answer by roughly the wrong log term.
• This same integral (with the same bounds) also appears — and is evaluated correctly — in the later problem doubleint-233, which is the identical surface $z=\dfrac{x+y}{x^2+y^2}$ restricted to the identical domain $1\le r\le2$, first quadrant, just phrased differently ("$(x^2+y^2)z=x+y$"). That problem's closed form, $\dfrac\pi4\Big[3\sqrt2-\sqrt3+\sqrt2\ln\dfrac{\sqrt2+\sqrt3}{\sqrt2}\Big]$, is algebraically identical to the corrected result here (since $\frac{\sqrt2+\sqrt3}{\sqrt2}=\frac{2+\sqrt6}2$), confirming the correction independently.
• Verified numerically: direct polar double integration gives $S\approx2.859992$, matching the corrected closed form exactly, while the source's originally claimed value evaluates to $\approx3.748171$ — a confirmed discrepancy.