Evaluate the triple integral $V = \iiint dxdydz$ where $V$ is the solid ball centered at $1,2,3$ with radius $4$.
My teacher hinted at a previous example that used divergence of a vector field to find the volume of a ball with radius $1$, centered at the origin.
His previous example used vector field: $\mathbf{F} = \frac{x}{3}, \frac{y}{3}, \frac{z}{3}$.
The divergence of this vector field equaled the value $1$. So his integral was $V = \iiint 1 \, dxdydz$, which became $\frac{4}{3}\pi$.
Parameterise the ball using
$$x = 1+r\cos\phi\sin\theta \qquad y = 2 + r\sin\phi\sin\theta \qquad z = 3 + r\cos\theta$$
where $r \in [0,4]$, $\phi \in [0,2\pi)$, and $\theta \in [-\pi/2 , \pi/2]$. The Jacobian of the transformation is
$$dxdydz = |r^2\sin\theta| \, d\theta d\phi dr$$
Thus, we compute
\begin{align} V & = \iiint\, dxdydz \\ & = \int_{r=0}^{r=4}\int_{\phi=0}^{\phi=2\pi}\int_{\theta=-\pi/2}^{\theta=\pi/2}|r^2\sin\theta| \, d\theta d\phi dr \\ & = \int_{r=0}^{r=4}\int_{\phi=0}^{\phi=2\pi} 2r^2 \, d\phi dr \\ & = \int_{r=0}^{r=4} 4\pi r^2 \, dr \\ & = \frac 43 \pi \cdot 4^3 \end{align}
as expected.