Riemannian metric of $3$-sphere I know this probably seems like a dumb question, I have parametrised part of the unit $3$-sphere with $(x,y,z)\to (x,y,z,(1-(x^2+y^2+z^2))^{\frac{1}{2}})$ and now I'm trying to calculate the riemannian metric gij, I thought I did it correctly but I saw it differently somewhere online and now I'm confused so just wondering if anyone can shine some light on exactly how it's done. 
 A: An expression for a metric depends on the choice of coordinate system, and the same metric can look quite different in different coordinate systems. (For instance, let $\phi$ be a general diffeomorphism of a plane region, and compute the pullback of the Euclidean metric under $\phi$.)
As TylerHG and Nicolas point out, it's impossible to give specifics without knowing the particular metric expression you've seen. That said, the "graph parametrization" in your post turns out not to be especially nice geometrically; its coordinate curves aren't mutually orthogonal, and the metric components blow up at the boundary of the $3$-ball (i.e., at the unit sphere in $\mathbf{R}^{3}$).
A far likelier choice for a formula found on-line comes from stereographic coordinates
$$
(x, y, z) \mapsto \frac{(2x, 2y, 2z, x^{2} + y^{2} + z^{2} - 1)}{x^{2} + y^{2} + z^{2} + 1}.
$$
These coordinates are defined on all of $\mathbf{R}^{3}$, they cover the complement of one point (rather than just one open hemisphere), and in these coordinates the round metric is conformally Euclidean, i.e., $g_{ij} = G(x, y, z)\, \delta_{ij}$ for some function $G$ (that turns out to be rational).
If you're perusing physics sources, you may instead have seen the metric coming from spherical coordinates
$$
(\cos\theta_{1}\cos\theta_{2}\cos\theta_{3}, \sin\theta_{1}\cos\theta_{2}\cos\theta_{3}, \sin\theta_{2}\cos\theta_{3}, \sin\theta_{3}),
$$
with $0 \leq \theta_{1} \leq 2\pi$, $-\pi/2 \leq \theta_{2} \leq \pi/2$, $-\pi/2 \leq \theta_{3} \leq \pi/2$.
