Editing Contour integration (section)

===Example 2===
Let the [[vector field]] <math>\mathbf{F}=u^4\mathbf{e}_u+x^5\mathbf{e}_x+y^6\mathbf{e}_y+z^{-3}\mathbf{e}_z</math>, and remark that there are 4 parameters in this case. Let this [[vector field]] be bounded by the following:
<math display=block>{0\leq x\leq 1} \quad {-10\leq y\leq 2\pi} \quad {4\leq z\leq 5} \quad {-1\leq u\leq 3}</math>

To evaluate this, we must utilize the [[divergence theorem]] as stated before, and we must evaluate <math>\nabla\cdot\mathbf{F}</math>. Let <math>dV = dx \, dy \, dz \, du</math>

{{block indent|left=1.6|{{oiiint
| preintegral =
| intsubscpt = <math>{\scriptstyle S}</math>
| integrand = <math>\mathbf{F} \cdot \mathbf{n} \, dS</math>
}}}}

<math display=block>\begin{align}
&=\iiiint_V \left(\frac{\partial F_u}{\partial u} + \frac{\partial F_x}{\partial x} + \frac{\partial F_y}{\partial y} + \frac{\partial F_z}{\partial z}\right)\,dV\\[6pt]
&=\iiiint_V \left(\frac{\partial u^4}{\partial u} + \frac{\partial x^5}{\partial x} + \frac{\partial y^6}{\partial y} + \frac{\partial z^{-3}}{\partial z}\right)\,dV\\[6pt]
&=\iiiint_V {{\frac{4 u^3 z^4 + 5 x^4 z^4 + 5 y^4 z^4 - 3}{z^4}}}\,dV \\[6pt]
&= \iiiint_V {{\frac{4 u^3 z^4 + 5 x^4 z^4 + 5 y^4 z^4 - 3}{z^4}}}\,dV \\[6pt]
&=\int_{0}^{1}\int_{-10}^{2\pi}\int_{4}^{5}\int_{-1}^{3} \frac{4 u^3 z^4 + 5 x^4 z^4 + 5 y^4 z^4 - 3}{z^4}\,dV\\[6pt]
&=\int_{0}^{1}\int_{-10}^{2\pi}\int_{4}^{5}\left(\frac{4(3u^4z^3+3y^6+91z^3+3)}{3z^3}\right)\,dy\,dz\,du\\[6pt]
&=\int_{0}^{1}\int_{-10}^{2\pi}\left(4u^4+\frac{743440}{21}+\frac{4}{z^3}\right)\,dz\,du\\[6pt]
&=\int_{0}^{1} \left(-\frac{1}{2\pi^2}+\frac{1486880\pi}{21}+8\pi u^4+40 u^4+\frac{371720021}{1050}\right)\,du\\[6pt]
&=\frac{371728421}{1050}+\frac{14869136\pi^3-105}{210\pi^2}\\[6pt]
&\approx{576468.77}
\end{align}</math>

Thus, we can evaluate a contour integral with <math>n=4</math>. We can use the same method to evaluate contour integrals for any [[vector field]] with <math>n>4</math> as well.