Calculate the volume delimited by the ellipsoid with equation: $$$ \dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}+\dfrac{z^2}{c^2}=1$$$
Development:
We will use Gauss' theorem. First, we will parametrize the surface. We will use the spherical coordinates, i.e.:
$$$ \begin{array}{l} x=a\sin\theta\cos\varphi \\ y=b\sin\theta\sin\varphi \\ z=c\cos\theta \end{array} \qquad \begin{array}{l} \theta\in [0,\pi] \\ \varphi\in[0,2\pi] \end{array}$$$
Let's calculate the vectors and its vector product:
$$$ \left. \begin{array}{l} r_\theta = (a\cos\theta\cos\varphi,b\sin\varphi\cos\theta,-c\sin\varphi) \\ r_\varphi=(-a\sin\theta\sin\varphi,b\sin\theta\cos\varphi,0) \end{array} \right\} \Rightarrow$$$
$$$\begin{array}{rl} \Rightarrow \ r_\theta \times r_\varphi =& \begin{vmatrix} i & j & k \\ a\cos\theta\cos\varphi & b\sin\varphi\cos\theta & -c\sin\varphi\\ -a\sin\theta\sin\varphi & b\sin\theta\cos\varphi & 0 \end{vmatrix} \\ =& \big( b\cdot c\cdot\sin^2\theta\cos\varphi, a\cdot c\cdot\sin^2\theta\sin\varphi, a\cdot b\cdot \sin\theta\cos\theta\big) \end{array}$$$
Taking the vector field $$F(x,y,z)=(0,0,z)$$, we have:
$$$ \begin{array}{rl} \text{Vol}(E) =& \iiint_E 1 \ dx \ dy \ dz= \iiint_E \text{div}(0,0,z) \ dx \ dy \ dz = \int_S (0,0,z) \ d\vec{S} \\ =& \int_0^{\pi} \int_0^{2\pi} (0,0,c\cos\theta)\cdot \big( b c \sin^2\theta\cos\varphi, a c \sin^2\theta\sin\varphi, a b \sin\theta\cos\theta\big) \ d\varphi \ d\theta \\ =& \int_0^{\pi} \int_0^{2\pi} c\ \cos\theta\ a\ b\ \sin\theta\cos\theta \ d\varphi \ d\theta = abc \int_0^{\pi} \cos^2\theta \sin\theta \Big( \int_0^{2\pi} \ d\varphi \Big)\ d\theta \\ =& abc 2\pi \int_0^{\pi} \cos^2\theta \sin\theta \ d\theta = 2\pi a b c \Big[\dfrac{-\cos^3\theta}{3}\Big]_0^\pi =\dfrac{4}{3}\pi abc \end{array}$$$
Solution:
The volume of the ellipsoid is $$\text{Vol}(E)=\dfrac{4}{3}\pi abc$$