Calculate the surface area of the part of the paraboloid

Calculate the surface area of the part of the paraboloid $2z=x^2+y^2$ cut by the cylinder $x^2+y^2=1$.

Explanation of Method

With $z=\frac12(x^2+y^2)$, the surface-area integrand $\sqrt{1+z_x^2+z_y^2}=\sqrt{1+x^2+y^2}$ is radially symmetric, so polar coordinates reduce the problem to a single elementary radial integral.

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

Step 1 · Surface-area integrand

$z_x=x,\,z_y=y$, so $\sqrt{1+z_x^2+z_y^2}=\sqrt{1+x^2+y^2}$.

Step 2 · Polar coordinates

$$ S=\int_0^{2\pi}\int_0^1\sqrt{1+r^2}\,r\,dr\,d\theta = 2\pi\int_0^1r\sqrt{1+r^2}\,dr. $$

Step 3 · Evaluate

With $u=1+r^2$:

$$ \int_0^1r\sqrt{1+r^2}\,dr = \frac13\big[2^{3/2}-1\big] = \frac{2\sqrt2-1}3. $$

$$ S = 2\pi\cdot\frac{2\sqrt2-1}3 = \frac{2\pi\big(2\sqrt2-1\big)}3. $$

Additional Notes

This derivation matches the source exactly.
Verified numerically: direct polar double integration gives exactly $\dfrac{2\pi(2\sqrt2-1)}3$.

Result

The surface area is:

$$ S = \frac{2\pi\big(2\sqrt2-1\big)}3. $$