to prove $dx^2 + dy^2 + dz^2 = dr^2 + r^2 d(\theta)^2 + r^2(\sin\theta)^2 d\phi^2$ Given $$ x = r\sin\theta \cos\phi, \; y= r\sin\theta \sin\phi, \; z=r\cos \theta, $$ show that $$ \text{d}x^2 + \text{d}y^2 + \text{d}z^2 = \text{d}r^2 + r^2 \text{d}\theta^2 + r^2(\sin\theta)^2 \text{d}\phi^2. $$
How do I do this?
 A: Here $$x = r\sin\theta \cos\phi,\quad  y = r\sin\theta \sin\phi, \quad z=r\cos\theta $$
are functions of $(r, \theta, \phi)$ so, by the chain rule,
\begin{align}
& \begin{aligned}
\text{d}x & = \frac{\partial x}{\partial r}\text{d}r+\frac{\partial x}{\partial \theta}\text{d}\theta+\frac{\partial x}{\partial \phi}\text{d}\phi = \text{d}r \sin(\theta) \cos(\phi) + r \cos(\theta) \cos(\phi)\text{d}\theta − r \sin(\theta) \sin(\phi)\text{d}\phi, \\
\text{d}y & = \frac{\partial y}{\partial r}\text{d}r+\frac{\partial y}{\partial \theta}\text{d}\theta+\frac{\partial y}{\partial \phi}\text{d}\phi = \text{d}r \sin(\theta) \sin(\phi) + r \cos(\theta) \sin(\phi)\text{d}\theta + r \sin(\theta) \cos(\phi)\text{d}\phi, \\
\text{d}z & = \frac{\partial z}{\partial r}\text{d}r+\frac{\partial z}{\partial \theta}\text{d}\theta+\frac{\partial z}{\partial \phi}\text{d}\phi = \text{d}r \cos(\theta) − r \sin(\theta)\text{d}\theta,
\end{aligned} \\[1em]
\therefore \; & \begin{aligned}[t]
& \phantom{=} \text{d}x^2+\text{d}y^2+\text{d}z^2 \\
= & \begin{aligned}[t]
& \{\text{d}r \sin(\theta) \cos(\phi) + r \cos(\theta) \cos(\phi)\text{d}\theta − r \sin(\theta) \sin(\phi)\text{d}\phi\}^2 \\
+ \; & \{\text{d}r \sin(\theta) \sin(\phi) + r \cos(\theta) \sin(\phi)\text{d}\theta + r \sin(\theta) \cos(\phi)\text{d}\phi\}^2 \\
+ \; & \{\text{d}r \cos(\theta) − r \sin(\theta)\text{d}\theta\}^2
\end{aligned} \\[0.5em]
= & \begin{aligned}[t]
& \text{d}r^2\{\sin^2 \theta\cos^2\phi+\sin^2\theta\sin^2\phi+\cos^2\phi\} \\
+ \; & r^2 \text{d}\theta^2\{\cos^2\theta\cos^2\phi+\cos^2\theta\sin^2\phi+\sin^2\theta\} \\
+\; & r^2\text{d}\phi^2\{\sin^2\theta\sin^2\phi+\sin^2\theta\cos^2\phi\} \\
+\; & 2r^2\text{d}r\text{d}\theta\{\sin(\theta) \cos(\phi)\cos(\theta) \cos(\phi)+\sin(\theta) \sin(\phi)\cos(\theta) \sin(\phi)-\sin(\theta)\cos(\theta)\} \\
+\; & 2r^2\text{d}r\text{d}\phi\{-\sin(\theta) \cos(\phi) \sin(\theta) \sin(\phi)+\sin(\theta) \sin(\phi)\sin(\theta) \cos(\phi)\} \\
+\; & 2r^2\text{d}\theta \text{d}\phi \{-\cos(\theta) \cos(\phi)\sin(\theta) \sin(\phi)+\cos(\theta) \sin(\phi)\sin(\theta) \cos(\phi)\}
\end{aligned} \\[0.5em]
= &\;\, \text{d}r^2+r^2\text{d}\theta^2+r^2\sin^2\theta \text{d}\phi^2\text{, QED.}
\end{aligned}
\end{align}
A: Expand $$ dx = \frac{\partial x}{\partial r}dr+\frac{\partial x}{\partial \theta}d\theta+\frac{\partial x}{\partial \phi}d\phi,$$
similairily with $y$ and $z$, calculate their squares and add them together, remembering that the product of 1-forms is symmetric in this case. You need to know the derivatives of triginometric functions and use the pythagorean identity $\sin^2\alpha + \cos^2\alpha =1$.
A: I think going from cylindrical coordinates will make the derivation easier. In clyndrical coordinates,
$$dx^2+dy^2+dz^2=d\rho^2+\rho^2d\phi^2+dz^2.$$
Now, we use the identities, $\rho=r\sin\theta$ and $z=r\cos\theta:$
$$(\sin\theta dr+r\cos\theta d\theta)^2+(r\sin\theta)^2d\phi^2+(\cos\theta dr-r\sin\theta d\theta)^2$$
to get
$$dr^2+r^2d\theta^2+r^2\sin^2\theta d\phi^2.$$
I think we can geometrically interpret this as follows: The sides of the differential cube in spherical coordinates are 1) $dr$ 2) $rd\theta$ which is obtanied by a rotation  through a small bizmut angle 3) $\rho  d\phi$ which is obtained by a rotation of $xy$-plane. So, the square length of the diagonal of this cube is the sum of the squares of these sides.
