To perform the same computation using (one-dimensional) surface elements, we note that the surface is given by
\[
z - f(x,y) = z - \sqrt{a^2 - x^2 - y^2} = 0.
\]

We know from lectures that the surface elements can be projected onto the $xy$-plane via the formula
\[
\de{S} = \frac{\de{A}}{\lvert \cos\alpha \rvert} = \frac{\de{A}}{\lvert \mb{n} \cdot \mb{k}\rvert},
\]

where $\alpha$ is the angle between the normal of the surface element and the vertical axis, and $\mb{n}$ is the unit normal of the surface. However, we also know that the unit normal of the surface is given by
\[
\mb{n} = \frac{\nabla f}{\lvert \nabla f \rvert} = \frac{[-f_x, -f_y, 1]}{\sqrt{1+f_x^2 + f_y^2}}.
\]

Thus, if we wish to project onto the $xy$-plane, then formula for the conversion from $\de{S}$ to $\de{A} = \de{x} \, \de{y}$ is
\[
\de{S} = \sqrt{1 + f_x^2 + f_y^2} \, \de{x} \de{y},
\]

and for the case of a hemisphere of radius, $a$,
\[
\de{S} = \frac{a}{\sqrt{a^2-x^2-y^2}}
\]

and so, we have thus far,
\[
\iint_S \de{S} = \iint_A \frac{a}{\sqrt{a^2-x^2-y^2}} \, \de{x} \de{y},
\]

where $A$ is the circle of radius $a$ in the xy-plane. It is easier to perform this integration in polar coordinates. We set $x = r\cos\theta$ and $y = r\sin\theta$, along with the Jacobian conversion of
\[
\de{x} \, \de{y} = r \de{r} \de{\theta}.
\]

Then finally, the surface integral becomes
\[
\iint_S \de{S}
= \int_0^{2\pi}\int_0^a \frac{a}{\sqrt{a^2 - r^2}} \, r \, \de{r} \de{\theta} = 2\pi a^2,
\]

which is confirmed. Note that it is a common mistake to automatically set $r = a$ into the integrand. Remember that although you are evaluating along the surface of the sphere, where $|\mb{r}| = a$, once this surface is projected onto the xy-plane, then you are now concerned with allowing the polar radius, $r$ to vary from $r = 0$ to $r = a$. Finally, we note that there is one additional way to compute the normal instead of evaluating $\nabla f$.

Note that the normal out of a sphere is given by the radial vector. So
\[
\mb{n} = \frac{[x,y,z]}{\sqrt{x^2 + y^2 + z^2}}.
\]

Thus,
\[
\mb{n} \cdot \mb{k} = \frac{z}{\sqrt{x^2 + y^2 + z^2}} = \frac{\sqrt{a^2 - x^2 - y^2}}{a},
\]

which gives us the same result as the one which used the gradient of $f$ to determine the normal.