Find the surface area of the part of the surface
Find the surface area of the part of the surface $(x^2+y^2)z=x+y$ for $1<x^2+y^2<4$, $x>0,\,y>0$.
Explanation of Method
This is the surface $z=\dfrac{x+y}{x^2+y^2}$, and the domain $1<r<2$, first quadrant, matches doubleint-227 exactly — so the surface-area integrand reduces the same way, to $\sqrt{1+2/r^4}$, and the same integral $\dfrac\pi4\int_1^4\dfrac{\sqrt{u^2+2}}u\,du$ arises after $u=r^2$. Here it is evaluated with a different (equally valid) antiderivative form.
- Compute the surface-area integrand (identical to doubleint-227's Step 1).
- Convert to polar coordinates.
- Evaluate via $u=r^2$ and the antiderivative $\sqrt{u^2+2}-\sqrt2\ln\left|\frac{\sqrt2+\sqrt{u^2+2}}u\right|$.
Step 1 · Surface-area integrand
As in doubleint-227: $1+z_x^2+z_y^2=1+\dfrac2{(x^2+y^2)^2}=\dfrac{r^4+2}{r^4}$.
Step 2 · Polar coordinates
$$ S = \int_0^{\pi/2}\int_1^2\frac{\sqrt{r^4+2}}{r^2}\cdot r\,dr\,d\theta = \frac\pi2\int_1^2\frac{\sqrt{r^4+2}}r\,dr. $$
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 $\displaystyle\int\frac{\sqrt{u^2+2}}u\,du=\sqrt{u^2+2}-\sqrt2\ln\left|\frac{\sqrt2+\sqrt{u^2+2}}u\right|+C$:
$$ \int_1^4\frac{\sqrt{u^2+2}}u\,du = \Big[3\sqrt2-\sqrt2\ln\sqrt2\Big]-\Big[\sqrt3-\sqrt2\ln(\sqrt2+\sqrt3)\Big] = 3\sqrt2-\sqrt3+\sqrt2\ln\frac{\sqrt2+\sqrt3}{\sqrt2}. $$
Therefore:
$$ S = \frac\pi4\left(3\sqrt2-\sqrt3+\sqrt2\ln\frac{\sqrt2+\sqrt3}{\sqrt2}\right). $$
Additional Notes
- This derivation matches the source exactly. Note $\dfrac{\sqrt2+\sqrt3}{\sqrt2}=\dfrac{2+\sqrt6}2$, so this closed form is algebraically identical to the corrected answer for doubleint-227 (the same surface and domain, phrased differently in that problem's statement) — a useful independent cross-check that confirms both results.
- Verified numerically: direct polar double integration gives $S\approx2.859992$, matching the boxed closed form exactly.
Result
The surface area is: