Question

Question: Use Stokes’ Theorem to evaluate

\({\rm{F(x,y,z) = ta}}{{\rm{n}}^{{\rm{ - 1}}}}{\rm{(}}{{\rm{x}}^{\rm{2}}}{\rm{y}}{{\rm{z}}^{\rm{2}}}{\rm{)i + }}{{\rm{x}}^{\rm{2}}}{\rm{yj + }}{{\rm{x}}^{\rm{2}}}{{\rm{z}}^{\rm{2}}}{\rm{k,}}\)

\({\rm{S}}\)is the cone\({\rm{x = }}\sqrt {{{\rm{y}}^{\rm{2}}}{\rm{ + }}{{\rm{z}}^{\rm{2}}}} {\rm{,0}} \le {\rm{x}} \le {\rm{2}}\)oriented in the direction of the positive \({\rm{x}}\)-axis.

Short answer

By using the Stokes’ Theorem, we get

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

Consider the Stoke’s theorem,

Where C denotes the surface's counter-clockwise border.

The supplied surface is a cone's portion.

Where C denotes the surface's counter-clockwise border.

The supplied surface is a cone's portion.

\({\rm{x = }}\sqrt {{{\rm{y}}^{\rm{2}}}{\rm{ + }}{{\rm{z}}^{\rm{2}}}} \)For which \({\rm{0}} \le {\rm{x}} \le 2\)

The border of this surface is a circle parallel to the \({\rm{yz}}\) plane, as seen in the graph below.

Take note of how big this circle is

\(\begin{array}{c}{{\rm{y}}^{\rm{2}}}{\rm{ + }}{{\rm{z}}^{\rm{2}}}{\rm{ = 4}}\\{\rm{x = 2}}\end{array}\)

The boundary's parameterization is

\({\rm{C:r(t) = 2i + 2costj + 2sintk}}\)

Therefore,

\({\rm{dr = }}\left( {{\rm{0i - 2sintj + 2costk}}} \right){\rm{dt}}\)

Let us plot the graph:

\({\rm{F(x,y) = ta}}{{\rm{n}}^{{\rm{ - 1}}}}\left( {{{\rm{x}}^{\rm{2}}}{\rm{y}}{{\rm{z}}^{\rm{2}}}} \right){\rm{i + }}{{\rm{x}}^{\rm{2}}}{\rm{yj + }}{{\rm{x}}^{\rm{2}}}{{\rm{z}}^{\rm{2}}}{\rm{k}}\)

\({\rm{F(r(t)) = ta}}{{\rm{n}}^{{\rm{ - 1}}}}\left( {{\rm{32costsi}}{{\rm{n}}^{\rm{2}}}{\rm{t}}} \right){\rm{i + 8costj + 16si}}{{\rm{n}}^{\rm{2}}}{\rm{tk}}\)

Because \({\rm{C}}\) is counter-clockwise, we'll integrate from \({\rm{0}}\) to \({\rm{2\pi }}\).

Substitute \({\rm{sint = u}}\) and \({\rm{costdt = du}}\) Limits of integration will change from \(\mathop \smallint \nolimits^0 \_2{\rm{\pi }}\)to\(\int_{\sin (2\pi )}^{\sin (0)} = \int_0^0 {} \)

The required solution is