Correct Answer - Option 4 : 55
Concept:
For a vector F = F1i + F2j + F3k
\(Div = \nabla .F = \frac{{\partial {F_1}}}{{\partial x}} + \frac{{\partial {F_2}}}{{\partial y}} + \frac{{\partial {F_3}}}{{\partial z}}\)
\(Curl = \nabla \times F = \left| {\begin{array}{*{20}{c}} i&j&k\\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\ {{F_1}}&{{F_2}}&{{F_3}} \end{array}} \right|\)
Calculation:
\(\vec A = {x^2}{\hat a_x} + 6{y^2}{\hat a_y} + {z^3}{\hat a_z}\)
The divergence is given by,
\(\nabla .\vec A = \frac{\partial }{{\partial x}}\left( {{x^2}} \right) + \frac{\partial }{{\partial y}}\left( {6{y^2}} \right) + \frac{\partial }{{\partial z}}\left( {{z^3}} \right)\)
= 2x + 12y + 3z2
At the given point P (2, 4, 1), divergence is:
= 2(2) + 12(4) + 3(1) = 55
