Question

Question: Use Stokes’ Theorem to evaluate \(\int\limits_{\rm{C}} {\rm{F}} {\rm{ \times dr}}\) . In each case \({\rm{C}}\)is oriented counterclockwise as viewed from above.

\({\rm{F(x,y,z) = yzi + 2xzj + }}{{\rm{e}}^{{\rm{xy}}}}{\rm{k,}}\)

\({\rm{C}}\)is the circle\({{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}{\rm{ = 16,z = 5}}\)

Short answer

By using Stokes’ theorem, we get\({\rm{80\pi }}\).

Step-by-step solution

Concept Introduction

The surface integral of a function's curl over any surface limited by a closed path is equivalent to the line integral of a given vector function around that path, according to a theorem.

Explanation of the solution

To begin, determine the curvature of the given vector field:

\({\rm{curl F = (x}}{{\rm{e}}^{{\rm{xy}}}}{\rm{ - 2x,y - y}}{{\rm{e}}^{{\rm{xy}}}}{\rm{,z)}}\)

We wish to use the surface with the task's specified border. The disc parametrized with: is the simplest to utilise.

\({\rm{\Phi (r,\theta ) = (rcos\theta ,rsin\theta ,5),}}{\mkern 1mu} {\rm{r\^I (0,4),}}{\mkern 1mu} {\rm{\theta \^I (0,2\pi )}}\)

This is a flat disc with a radius of 4 and a height of\({\rm{z = 5}}\). The integration over it is simple since its unit normal is\(\left( {0,0,1} \right)\), so the only part of the curl that passes through the integral is the \({\rm{z}}\) component of the parametrization. We have the following:

\(\begin{array}{c}\int {\int_{\rm{S}} {{\rm{curl }}} } {\rm{F \times (0,0,1)}}{\mkern 1mu} {\rm{dS = }}\int_{\rm{0}}^{{\rm{2\pi }}} {\int_{\rm{0}}^{\rm{4}} {\rm{5}} } {\mkern 1mu} {\rm{drd\theta }}\\{\rm{ = 80\pi }}\end{array}\)

Therefore, the required solution is \({\rm{80\pi }}\).