Module 1 : A Crash Course in Vectors
Lecture 4 : Gradient of a Scalar Function

Using the coordinate transformation to cylindrical

\begin{displaymath}x=\rho\cos\theta\ \ \ y=\rho\sin\theta\end{displaymath}

and

\begin{eqnarray*} \hat\imath &=& \hat\rho\cos\theta-\hat\theta\sin\theta\\ \hat... ...&=& \hat\rho\sin\theta+\hat\theta\cos\theta\\ \hat k &=& \hat k \end{eqnarray*}

Using these

\begin{displaymath}\vec F = \rho^3(\cos^4\theta+\sin^4\theta)\hat\rho +\rho^3(\sin^3\theta\cos\theta-\cos^3\theta\sin\theta)\hat\theta\end{displaymath}

The area element on the curved surface is $R d\theta dz\hat\rho$, where $R$is the radius. Thus the surface integral is

\begin{eqnarray*} \int F_\rho dS &=& R^4\int_0^{2\pi}(\cos^4\theta+\sin^4\theta)d\theta \int_0^5 dz\\ &=& 16. \frac{3\pi}{2}.5= 120\pi \end{eqnarray*}

where we have used $\int_0^{2\pi}\sin^4\theta d\theta = \int_0^{2\pi} \cos^4\theta d\theta = 3\pi/4$.

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