How $r^2 \sin \varphi$ come? Taken from here
We have $\operatorname{div} \vec{f} = 3x^2+3y^2+3z^2$, which in spherical coordinates
$$
\begin{align}
x & = r \cos \theta \sin \varphi, \\
y & = r \sin \theta \sin \varphi, \\
z & = r \cos \varphi,
\end{align}
$$
for $0 \leq \theta \leq 2 \pi$ and $0 \leq \varphi \leq \pi$, becomes $\operatorname{div} \vec{f} = 3 r^2.$
Therefore, using Gauss's Theorem we obtain
$$
\begin{align}
\int\limits_{\mathbb{S}^2} \vec{f} \cdot \vec{n} \, \text{d}S & = \int\limits_{B} \operatorname{div} \vec{f} \, \text{d}V \\
 & = \int_0^{2\pi} \hspace{-5pt} \int_0^{\pi} \hspace{-5pt} \int_0^1 (3r^2) \cdot (r^2 \sin \varphi) \, \text{d} r \, \text{d} \varphi \, \text{d} \theta \\
 & .
\end{align}
$$
My confusion:  why $dV=r^2 \sin \varphi \text{d} r \, \text{d} \varphi \, \text{d} \theta?$
Im not getting  that how $r^2 \sin \varphi$  come ?
 A: Intuivitely you can think of $r^2\sin\varphi$ as the product of $r$ (which is the arc length of a radian of latitude along a meridian of a sphere with radius $r$), and $r\sin\varphi$ (which is the arc length of a radian of longitude along a parallel with the given $\varphi$ of a sphere with radius $r$).
Formally it's just the Jacobian determinant (modulo a sign that I don't care to work out) of the spherical coordinate map
$$ \begin{pmatrix} r \\ \varphi \\ \theta \end{pmatrix} \mapsto
\begin{pmatrix} r\sin\varphi\cos\theta \\
r \sin\varphi\sin\theta \\
r \cos\varphi \end{pmatrix} $$
A: The length of a circular arc is proportional to the radius of the circle, and to the angle.  Radians are a unit of angle measure defined so that the proportionality constant is $1$, i.e. $x = r\theta$.
To define a small volume element in spherical coordinates, we need a length "outward" from the origin, a length "southward", and a length "eastward".
The length outward is simply $dr$.
The length "southward" is a small arc of a great circle from the "North Pole" where the unit sphere touches the positive $z$ axis.  The angle is $\theta$ and the radius is $\rho$, the radius of the sphere.  So the infinitesimal length southward is $\rho d\theta$.
The length "eastward" is a small arc of a circle parallel to the plane $z=0$, the "equator."  That circle has a horizontal radius $r = \rho \sin \theta$.  The angle "eastward" is $\phi$, so the length of the eastward arc is $\rho \sin \theta d\phi$.
Putting together the volume element, we get $$dV = dr \cdot \rho d\theta \cdot \rho \sin \theta d\phi = \rho^2 \sin \theta d\rho d\theta d\phi.$$
