Let's say I take the sphere $x^2 + y^2 + z^2 = 10a^2$ for some $a\in\mathbb{R}$ and consider the quarter-sphere that lies in the first octant. I'd like to find its surface area. I could say that its surface area is
$$\frac{4\pi\cdot 10a^2}{8}$$
knowing the general equation of the surface area of a sphere. I could also parameterize the surface and use a surface integral to calculate the surface area that way. But, let's say I would really like to do it "implicitly" -- i.e. as a level set. The surface is a piece of the level set for $f(x,y,z) = x^2 + y^2 + z^2 = 10a^2$. This sits above the region $R$ (a quarter disk) in the $xy$-plane nicely, so according to the calculus I'm learning in my class, I should have that
$$\displaystyle \text{surface area} = \iint_S d\sigma = \iint_R \dfrac{|\nabla f|}{|\nabla f \cdot \langle 0,0,1\rangle|} dx~dy.$$
However, since $\nabla f = \langle 2x, 2y, 2z\rangle$, its magnitude is $2\sqrt{x^2 + y^2 + z^2} = 2\cdot a\sqrt{10}$. So, this integral becomes
$$\iint_R \dfrac{2\cdot a\sqrt{10}}{2z}~dx~dy = a\sqrt{10} \iint_R \dfrac{1}{z}~ dx~dy.$$
... but $R$ sits in the $z=0$ plane. So my integrand is $1/0$? Clearly something went wrong here, and I'm not sure what it is. I should note that this same problem arises if we consider the sphere as a graph over the $xz$- or the $yz$-plane as well.
The integrand should be
$$ \frac{1}{\sqrt{10a^2-x^2-y^2}}$$
since that is the value of $z$ over the region. Doing this integral in polar coordinates gives us
$$a\sqrt{10}\int_0^{\frac{\pi}{2}} \int_0^{a\sqrt{10}} \frac{r}{\sqrt{10a^2-r^2}}\:dr\:d\theta = 5\pi a^2 = \frac{4\pi \cdot 10a^2}{8}$$