Volumes calculation using Gauss' theorem

As with what it is done with Green's theorem, we will use a powerful tool of integral calculus to calculate volumes, called the theorem of divergence or the theorem of Gauss.

This theorem says that if we have a solid $D$ limited by a closed surface $S$ and $\vec{F} (x, y, z)$ is a vector field, then:

$\int_{S} \vec{F} d \vec{S} = ∭_{V} div (\vec{F}) d x d y d z$

where, if $\vec{F} (x, y, z) = (P (x, y, z), Q (x, y, z), R (x, y, z))$ , then $div (\vec{F}) = P_{x} + Q_{y} + R_{z}$ is the divergence of the field.

Let's remember that to calculate the first term, if $r (u, v)$ is a parametrization of the surface,

$\int_{S} \vec{F} d \vec{S} = \iint \vec{F} (r (u, v)) \cdot (r_{u} \times r_{v}) d u d v$

where $r_{u} \times r_{v}$ denotes the vector product of the derivatives of the parametrization with respect to $u$ and $v$ .

Therefore, if we find a field whose divergence is equal to $1$ , on the right term we will have the volume of the solid and, therefore, we have a method to calculate volumes.

The most typical fields that satisfy what we need are:

$\vec{F} (x, y, z) = (x, 0, 0)$
$\vec{F} (x, y, z) = (0, y, 0)$
$\vec{F} (x, y, z) = (0, 0, z)$
$\vec{F} (x, y, z) = \frac{1}{3} (x, y, z)$

Example

For example, we are going to calculate the volume delimited by half a sphere and the equatorial plane, which is:

imagen

From Gauss’ theorem, we can integrate one of the fields given along the surface that closes the volume up.

We will use the following parametric form, using spherical coordinates:

$\begin{array}{l} x = \sin θ \cos φ \\ y = \sin θ \sin φ \\ z = \cos θ \end{array} \begin{array}{l} θ \in [0, \frac{π}{2}] \\ φ \in [0, 2 π] \end{array}$

Let's calculate the vectors and the vector product:

$\begin{array}{l} r_{θ} = (\cos θ \cos φ, \sin φ \cos θ, - \sin φ) \\ r_{φ} = (- \sin θ \sin φ, \sin θ \cos φ, 0) \end{array}} \Rightarrow$

$\begin{aligned} \Rightarrow r_{θ} \times r_{φ} = & | \begin{array}{c} i & j & k \\ \cos θ \cos φ & \sin φ \cos θ & - \sin φ \\ - \sin θ \sin φ & \sin θ \cos φ & 0 \end{array} | \\ = & (\sin^{2} θ \cos φ, \sin^{2} θ \sin φ, \sin θ \cos θ) \end{aligned}$

Taking the vector field $F (x, y, z) = (0, 0, z)$ , we have:

$\begin{aligned} \int_{S} (0, 0, z) d \vec{S} = & \int_{0}^{\frac{π}{2}} \int_{0}^{2 π} (0, 0, \cos θ) \cdot (\sin^{2} θ \cos φ, \sin^{2} θ \sin φ, \sin θ \cos θ) d φ d θ \\ = & \int_{0}^{\frac{π}{2}} \int_{0}^{2 π} \cos^{2} θ \cdot \sin θ d φ d θ = \int_{0}^{\frac{π}{2}} \cos^{2} θ \cdot \sin θ (\int_{0}^{2 π} d φ) d θ \\ = & 2 π \int_{0}^{\frac{π}{2}} \cos^{2} θ \cdot \sin θ d θ = 2 π [\frac{- \cos^{3} θ}{3}]_{0}^{\frac{π}{2}} = \frac{2}{3} π \end{aligned}$

Now we must integrate the same field on the low lid since the theorem is valid for closed surfaces. A parametrization of the circle is:

$\begin{array}{l} x = r \cos (t) \\ y = r \sin (t) \\ z = 0 \end{array} \begin{array}{l} t \in [0, 2 π] \\ r \in [0, 1] \end{array}$

Let's calculate also the vectors and its vector product:

$\begin{array}{l} r_{r} = (- r \sin (t), r \cos (t), 0) \\ r_{t} = (\cos (t), \sin (t), 0) \end{array}} \Rightarrow$

$\Rightarrow r_{t} \times r_{r} = | \begin{matrix} i & j & k \\ - r \sin (t) & r \cos (t) & 0 \\ \cos (t) & \sin (t) & 0 \end{matrix} | = (0, 0, - r^{2})$

Now, we have:

$\int_{C} (0, 0, z) d \vec{S} = \int_{0}^{2 π} \int_{0}^{1} (0, 0, 0) \cdot (0, 0, - r^{2}) d r d t = 0$

Therefore,

$\begin{aligned} Vol (E) = & ∭_{S} 1 d x d y d z = ∭_{E} div (0, 0, z) d x d y d z \\ = & \int_{S} (0, 0, z) d \vec{S} + \int_{C} (0, 0, z) d \vec{S} = \frac{2}{3} π + 0 = \frac{2}{3} π \end{aligned}$