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\begin{center}
\textbf{Vector Calculus}

\textit{\textbf{Grad, Div and Curl}}
\end{center}

\textbf{Question}


Calculate $\textbf{\textrm{div}F}$ and $\textbf{\textrm{curl}F}$ for
the vector field

$\un{F}(r, \theta) = r\un{i} + \sin \theta\un{j}$

Given that $(r,\theta)$ are polar coordinates in the plane.


\textbf{Answer}

Since $x=r\cos\theta$ and $y=r\sin\theta$, we have $r^2=x^2+y^2$.

So
\begin{eqnarray*}
\frac{\partial r}{\partial x} & = & \frac{x}{r} = \cos \theta\\
\frac{\partial r}{\partial y} & = & \frac{y}{r} = \sin \theta\\
\frac{\partial}{\partial x}\sin\theta & = & \frac{\partial}{\partial
x}\frac{y}{r} = -\frac{xy}{r^3}\\
& = & -\frac{\cos\theta \sin\theta}{r}\\
\frac{\partial}{\partial y} \sin \theta & = & \frac{\partial}{\partial
y} \frac{y}{r} = \frac{1}{r} - \frac{y^2}{r^3} = \frac{x^2}{t^3}\\
& = & \frac{\cos^2\theta}{r}\\
\frac{\partial}{\partial x} \cos\theta & = & \frac{partial}{\partial
x} \frac{x}{r} = \frac{1}{r} - \frac{x^2}{r^3}= \frac{y^2}{r^3}\\
& = & \frac{\sin^2 \theta}{r}\\
\frac{\partial}{\partial y} \cos \theta & = & \frac{\partial}{\partial
y} \frac{x}{r} = -\frac{xy}{r^3}\\
& = & -\frac{\cos\theta \sin \theta}{r}
\end{eqnarray*}
$\Rightarrow$

\begin{eqnarray*}
\textrm{div}\un{F} & = & \frac{\partial r}{\partial x} 
+\frac{\partial}{\partial y}\sin\theta
= \cos\theta + \frac{\cos^2\theta}{r}\\
\textrm{curl}\un{F} & = & \left | \begin{array}{ccc} \un{i} & \un{j} &
\un{k}\\ \frac{\partial}{\partial x} & \frac{\partial}{\partial y} &
\frac{\partial}{\partial z}\\
r & \sin\theta & 0
\end{array}
\right |\\
& = & \left ( -\frac{\sin\theta \cos\theta}{r} -\sin\theta \right ) \un{k}
\end{eqnarray*}

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