Question

A vector field D = 2ρ²a_ρ + z a_z exists inside a cylindrical region enclosed by the surfaces ρ = 1, z = 0 and z = 5. Let S be the surface bounding this cylindrical region. The surface integral of this field on \(S\left( {\mathop{{\displaystyle\int \!\!\!\!\displaystyle\int}\mkern-21mu\bigcirc} {_S} D.ds} \right)\) is _______.

Answer 1

Concept:

The Divergence theorem states that: \(\mathop{{\int\!\!\!\ \!\!\int}\mkern-21mu \ \bigcirc} {A.ds = \iiint_V {\left( {\nabla .A} \right)\;dV}}\)

Also, the Divergence of a vector field in Cylindrical Coordinates is given by:

 \(\nabla .A = \frac{1}{{\rm{\rho }}}\frac{{\partial \left( {\rho {A_\rho }} \right)}}{{\partial \rho }} + \frac{1}{{\rm{\rho }}}\frac{{\partial \left( {{A_\phi }} \right)}}{{\partial \phi }} + \frac{{\partial \left( {{A_z}} \right)}}{{\partial z}}\)

Calculation:

Given vector field D = 2ρ²a_p +z.a_z

Closed Surface S is defined as:

ρ=1, z=0, and z=5

We have to calculate\(\mathop{{\int\!\!\!\ \!\!\int}\mkern-21mu \ \bigcirc}_S D.ds \).

Applying Divergence Theorem,

\(\mathop{{\int\!\!\!\ \!\!\int}\mkern-21mu \ \bigcirc} {D.ds = \iiint_V {\left( {\nabla .D} \right)\;dV}}\), where volume V is given by:

ρ → 0 to 1

z → 0 to 5

ϕ → 0 to 2π

dV = p.dρ.dϕ.dz

Divergence of D is calculated as:

\(\nabla .D = \frac{1}{{\rm{\rho }}}\frac{{\partial \left( {\rho {D_\rho }} \right)}}{{\partial \rho }} + \frac{1}{{\rm{\rho }}}\frac{{\partial \left( {{D_\phi }} \right)}}{{\partial \phi }} + \frac{{\partial \left( {{D_z}} \right)}}{{\partial z}}\)

With D = 2p²a_p +za_z

∇.D = 6p + 1

\(\mathop \smallint \nolimits_v^\; \left( {\nabla .{\rm{D}}} \right){\rm{\;dV}} \) is given by:

\(= \mathop \smallint \nolimits_{\rho = 0}^{\rho = 1} \mathop \smallint \nolimits_{\phi = 0}^{\phi = 2\pi } \mathop \smallint \nolimits_{z = 0}^{z = 5} \left( {\nabla .D} \right)\;\rho \;d\rho \;d\phi \;dz\; \)

\(= \mathop \smallint \nolimits_{\phi = 0}^{2\pi } \mathop \smallint \nolimits_{z = 0}^5 \mathop \smallint \nolimits_{\rho = 0}^1 \left( {6p + 1} \right)\;\rho \;d\rho \;d\phi \;dz\)

\( = \mathop \smallint \nolimits_{\phi = 0}^{2\pi } \mathop \smallint \nolimits_{z = 0}^5 \left[ {2{\rho ^3} + \frac{{{\rho ^2}}}{2}} \right]_0^1d\phi \;dz \)

\(= \frac{5}{2} \times 2\pi \times 5 = 25\pi \)

So, \(\mathop{{\int\!\!\!\ \!\!\int}\mkern-21mu \ \bigcirc}_S D.ds \mathop \smallint \nolimits_v^\; \left( {\nabla .{\rm{D}}} \right){\rm{dV}} = 25{\rm{\pi }}\)

= 78.539