Question

A vector field D =$2{\rho }^2{a}_{\rho }+z{a}_z$ exists inside a cylindrical region enclosed by the surfaces $\rho$ = 1, z = 0 and z = 5. Let S be the surface bounding this cylindrical region. The surface integral of this field on S ($∯D.ds$) is_______

• A. 78.53

Answer 1

The correct option is A 78.53
D = $2{\rho }^2{a}_{\rho }+z{a}_z$
Using Gauss-Divergence theorem
${\int }_{s}D.ds={\int }_v(\nabla .D)dV$
where $\nabla .D=\frac{1}{\rho }\frac{\partial }{\partial \rho }\left(\rho D_{\rho }\right)+\frac{1}{\rho }\frac{\partial D\varphi }{\partial \varphi }+\frac{\partial D_z}{\partial z}$
(Div in Cylindrical Coordinates)
$=\frac{1}{\rho }\frac{\partial }{\partial \rho }\left(\rho .2{\rho }^2\right)+0+1$
$=6\rho +1$
So by (1)
${\int }_v(\nabla .D)dv={\int }_{\rho =0}^1{\int }_{\varphi =0}^{2\pi }{\int }_{z=0}^5(6\rho +1)\rho d\rho d\varphi dz$
$=10\pi {[2{\rho }^3+\frac{{\rho }^2}{2}]}_0^1$
$=10\pi (2+\frac{1}{2})$
=78.53