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

Example 19 :
A vector field is given by $\vec F = x^3\hat\imath + y^3\hat\jmath$. Verify Divergence theorem for a cylinder of radius 2 and height 5. The origin of the coordinate system is at the centre of the base of the cylinderand z-axis along the axis.

Solution :

The problem is obviously simple in cylindrical coordinates. The divergence can be easily seen to be $3(x^2+y^2)= 3\rho^2$. Recalling that the volume element is $\rho d\rho d\theta dz$, the integral is

\begin{displaymath}\int div\vec F dV = \int 3\rho^2 dV = 3\int_0^2 \rho^3 d\rho\int_0^{2\pi} d\theta\int_0^5dz = 120\pi\end{displaymath}

In order to calculate the surface integral, we first observe that the end faces have their normals along $\pm\hat k$. Since the field does not have any z- component, the contribution to surface integral from the end faces is zero.
We will calculate the contribution to the surface integral from the curved surface.

          Back                                                                                                    Next