Question

The line integral of the vector field \(F = 5xz\hat i + \left( {3{x^2} + 2y} \right)\hat j + {x^2}z\hat k\)  along a path from \(\left( {0,0,0} \right)\;to\;\left( {1,1,1} \right)\) parametrized by \(\left( {t,\;{t^2},\;t} \right)\) is _____.

Answer 1

Explanation:

Line integral:

\(\smallint {\vec{F}} \cdot {\rm{d\vec r}} = \smallint 5{\rm{xzdx}} + \smallint\left( {3{{\rm{x}}^2} + 2{\rm{y}}} \right){\rm{dy}} + \smallint{{\rm{x}}^2}{\rm{z}} {\rm{dz}}\)

\( x = t \Rightarrow dx = dt\)

\(y = {t^2} \Rightarrow dy = 2t\;dt\)

\( z = t \Rightarrow dz = dt\)

\(\smallint {\rm{\vec F}} \cdot {\rm{d\vec r}} = \mathop \smallint \limits_0^1 5{{\rm{t}}^2}{\rm{dt}} + \left( {3{{\rm{t}}^2} + 2{{\rm{t}}^2}} \right)2{\rm{tdt}} + {{\rm{t}}^2} \cdot {\rm{t}} \cdot {\rm{dt}} = \frac{5}{3} + \frac{{10}}{4} + \frac{1}{4} = \frac{5}{3} + \frac{{11}}{4}\)

\(\therefore \smallint {\rm{\vec F}} \cdot {\rm{d\vec r}} = \frac{{53}}{{12}}{\rm{\;}} = {\rm{\;}}4.42\)