Editing Divergence theorem (section)

===Example 2===

Let's say we wanted to evaluate the flux of the following [[vector field]] defined by <math> \mathbf{F}=2x^2 \textbf{i} +2y^2 \textbf{j} +2z^2\textbf{k} </math> bounded by the following inequalities:

:<math>\left\{0\le x \le 3\right\}, \left\{-2\le y \le 2\right\}, \left\{0\le z \le 2\pi\right\}</math>

By the divergence theorem,

:{{oiint
| preintegral = <math>\iiint_V\left(\mathbf{\nabla}\cdot\mathbf{F}\right) \mathrm{d}V=</math>
| intsubscpt = <math>\scriptstyle S</math>
| integrand = <math>(\mathbf{F}\cdot\mathbf{n})\, \mathrm{d}S .</math>
}}

We now need to determine the divergence of <math>\textbf{F}</math>. If <math>\mathbf{F}</math> is a three-dimensional vector field, then the divergence of <math>\textbf{F}</math> is given by <math display="inline">\nabla \cdot \textbf{F} = \left( \frac{\partial}{\partial x}\textbf{i} + \frac{\partial}{\partial y}\textbf{j} + \frac{\partial}{\partial z}\textbf{k} \right) \cdot \textbf{F}</math>.

Thus, we can set up the following flux integral {{oiint
| preintegral = <math>I = </math>
| intsubscpt = <math>{\scriptstyle S}</math>
| integrand = <math>\mathbf{F} \cdot \mathbf{n} \, \mathrm{d}S,</math>
}}
as follows:
:<math>
\begin{align}
I
&=\iiint_V \nabla \cdot \mathbf{F} \, \mathrm{d}V\\[6pt]
&=\iiint_V \left( \frac{\partial\mathbf{F_x}}{\partial x}+\frac{\partial\mathbf{F_y}}{\partial y}+\frac{\partial\mathbf{F_z}}{\partial z} \right) \mathrm{d}V\\[6pt]
&=\iiint_V (4x+4y+4z) \, \mathrm{d}V\\[6pt]
&=\int_0^3 \int_{-2}^2 \int_0^{2\pi} (4x+4y+4z) \, \mathrm{d}V
\end{align}
</math>

Now that we have set up the integral, we can evaluate it.

:<math>\begin{align}
\int_0^3 \int_{-2}^2 \int_0^{2\pi} (4x+4y+4z) \, \mathrm{d}V &=\int_{-2}^2 \int_0^{2\pi} (12y+12z+18) \, \mathrm{d}y \, \mathrm{d}z\\[6pt]
&=\int_0^{2\pi} 24 (2z+3)\, \mathrm{d}z\\[6pt]
&=48\pi(2\pi+3)
\end{align}

</math>