Find the divergence of the fields shown on the attached image.

Sunday, February 9, 2014

Find the divergence of the fields shown on the attached image.

a) Divergence of a vector field is a scalar quantity that represents how the field spreads out, or "diverges", in different directions. It is usually denoted as

$\displaystyle{\vec{\nabla}}\cdot{\vec{{F}}}$ and is calculated as

$\displaystyle{\vec{\nabla}}\cdot{\vec{{F}}}=\frac{{{d}{F}_{{x}}}}{{{\left.{d}{x}\right.}}}+\frac{{{d}{F}_{{y}}}}{{{\left.{d}{y}\right.}}}+\frac{{{d}{F}_{{z}}}}{{{\left.{d}{z}\right.}}}$ .

In the given vector field, the components are

$\displaystyle{F}_{{x}}={x}$ , so $\displaystyle\frac{{{d}{F}_{{x}}}}{{{\left.{d}{x}\right.}}}={1}$

$\displaystyle{F}_{{y}}={{y}}^{{3}}{{z}}^{{2}}$ , so $\displaystyle\frac{{{d}{F}_{{y}}}}{{{\left.{d}{y}\right.}}}={3}{{y}}^{{2}}{{z}}^{{2}}$

$\displaystyle{F}_{{z}}={x}{{z}}^{{3}}$ , so $\displaystyle\frac{{{d}{F}_{{z}}}}{{{\left.{d}{z}\right.}}}={3}{x}{{z}}^{{2}}$ .

Thus, the divergence of the given vector field is

$\displaystyle{\vec{\nabla}}\cdot{\vec{{F}}}={1}+{3}{{y}}^{{2}}{{z}}^{{2}}+{3}{x}{{z}}^{{2}}$ .

b) The divergence of this vector field can be calculated the same way. Here,

$\displaystyle\frac{{{d}{F}_{{x}}}}{{{\left.{d}{x}\right.}}}={\cos{{y}}}$

$\displaystyle\frac{{{d}{F}_{{y}}}}{{{\left.{d}{y}\right.}}}={2}{x}{y}$

and $\displaystyle\frac{{{d}{F}_{{z}}}}{{{\left.{d}{z}\right.}}}={0}$

So the divergence is

$\displaystyle{\vec{\nabla}}\cdot{\vec{{F}}}={\cos{{y}}}+{2}{x}{y}$ .

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