Find the surface area of the part of the surface z=xy cut

Find the surface area of the part of the surface $z=xy$ cut out by the cylinder $x^2+y^2=R^2$.

Explanation of Method

The surface-area integrand $\sqrt{1+z_x^2+z_y^2}=\sqrt{1+x^2+y^2}$ is radially symmetric, so polar coordinates over the disk of radius $R$ reduce the problem to a single elementary radial integral.

Compute the surface-area integrand.
Convert to polar coordinates over the disk of radius $R$.
Evaluate via $u=1+r^2$.

Step 1 · Surface-area integrand

With $z=xy$: $z_x=y,\,z_y=x$, so $1+z_x^2+z_y^2=1+x^2+y^2$.

Step 2 · Polar coordinates

$$ S = \iint_D\sqrt{1+x^2+y^2}\,dx\,dy = \int_0^{2\pi}\int_0^R\sqrt{1+r^2}\,r\,dr\,d\theta = 2\pi\int_0^Rr\sqrt{1+r^2}\,dr. $$

Step 3 · Evaluate

With $u=1+r^2$:

$$ S = \pi\int_1^{1+R^2}\sqrt u\,du = \pi\left[\frac23u^{3/2}\right]_1^{1+R^2} = \frac{2\pi}3\Big[(1+R^2)^{3/2}-1\Big]. $$

Additional Notes

This derivation matches the source exactly.
Verified numerically: direct polar double integration (e.g. $R=1.9$) gives $\approx18.636116$, matching $\dfrac{2\pi}3\big[(1+R^2)^{3/2}-1\big]$ exactly.

Result

The surface area is:

$$ S = \frac{2\pi}3\Big[(1+R^2)^{3/2}-1\Big]. $$