Question

Question: Verify that Stokes’ Theorem is true for the given vector field \({\rm{F}}\)and surface\({\rm{S}}\).

\({\rm{F(x,y,z) = - yi + xj - 2k,}}\)

\({\rm{S}}\)is the cone \({{\rm{z}}^{\rm{2}}}{\rm{ = }}{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}{\rm{,0}} \le {\rm{z}} \le {\rm{4,}}\)that lies inside the cylinder\({{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}{\rm{ = 4}}\) , oriented downward

Short answer

The Stokes' Theorem has been proven correct:

\(\int {\int_{\rm{S}} {{\rm{curl}}} } {\rm{F \times dS = }}\int_{\rm{C}} {\rm{F}} {\rm{ \times dr = - 32\pi }}\)

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.

Stokes’ Theorem is true for the given vector field \({\rm{F}}\)and surface\({\rm{S}}\)

The Stokes’ Theorem is true for the given vector field \({\rm{F}}\)and surface\({\rm{S}}\)

Compute \(\int_{\rm{C}} {\rm{F}} {\rm{ \times dr}}\)

\({\rm{S}}\)has a circle with parametric representation as its boundary.

\({\rm{r = }} < {\rm{4cost,4sint,4 }} > {\rm{,0}} < {\rm{t}} \le {\rm{2\pi }}\)r=⟨4cost,4sint,4⟩,0<t≤2π

\(d{\bf{r}} = < {\rm{ - 4sint,4cost,0 > dt}}\)

Given that

\({\rm{F = - yi + xj - 2k}}\)

Substitute the values of \({\rm{x,y,}}\)To get

\({\rm{F}}\left( {{\rm{r}}\left( {\rm{t}} \right)} \right) = < {\rm{ - 4sint,4cost, - 2}} > \)

We'll integrate from \({\rm{2\pi }}\) to 0 because the right hand rule indicates that the \({\rm{C}}\) should be rotated clockwise.

\(\begin{array}{c}\int_{\rm{C}} {\rm{F}} {\rm{ \times dr}} = \int_{{\rm{2\pi }}}^{\rm{0}} {{\rm{ < - 4sint,4cost, - 2}}} {\rm{ > \times }} < {\rm{ - 4sint,4cost,0}} > {\rm{dt}}\\ = \int_{2\pi }^0 1 6{\sin ^2}t + 16{\cos ^2}tdt\\ = 16\int_{2\pi }^0 {dt = - 32\pi } \end{array}\)

Compute

If\({\rm{F = Pi + Qj + Rk,}}\)

\({\rm{curlF = }}\left( {{{\rm{R}}_{\rm{y}}}{\rm{ - }}{{\rm{Q}}_{\rm{z}}}} \right){\rm{i + }}\left( {{{\rm{P}}_{\rm{z}}}{\rm{ - }}{{\rm{R}}_{\rm{x}}}} \right){\rm{j + }}\left( {{{\rm{Q}}_{\rm{x}}}{\rm{ - }}{{\rm{P}}_{\rm{y}}}} \right){\rm{k}}\)

Where \({{\rm{Q}}_{\rm{y}}}\)is partial derivative of \({\rm{Q}}\)with respect to \(y\)

Given that

\({\rm{F = - yi + xj - 2k}}\)

Therefore

\({\rm{curlF = (0 - 0)i + (0 - 0)j + (1 - ( - 1))k}}\)

\({\mathop{\rm curl}\nolimits} {\bf{F}} = 2{\bf{k}}\)

Explanation of the Solution

We will use,

Where \({\rm{G = Pi + Qj + Rk}}\)

When the surface is oriented upwards, this formula is employed

however, when the surface is oriented downwards, this formula is used.

As a result, the result will be the inverse of what we discover using this technique.

\({\rm{z = g(x,y) = }}\sqrt {{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}} \;\;\;\)We're talking about the top of the cone

\(\begin{array}{c}\frac{{\partial g}}{{\partial x}}{\rm{ = }}\frac{{\rm{x}}}{{\sqrt {{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}} }}\frac{{\partial g}}{{\partial y}}\\{\rm{ = }}\frac{{\rm{y}}}{{\sqrt {{{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}} }}\end{array}\)

\({\rm{D}}\)is the \({\rm{xy}}\) plane projection of the surface,

\({\rm{D}}\)Denotes the area within the circle \({{\rm{x}}^{\rm{2}}}{\rm{ + }}{{\rm{y}}^{\rm{2}}}{\rm{ = }}{{\rm{4}}^{\rm{2}}}{\rm{,z = 0}}\)

Also \(\smallint \smallint \_{\rm{DdA}}\) is the area of the circle, which is \({\rm{16\pi }}\)

Therefore,

\(\begin{array}{c}{\rm{2}}\int {\int_{\rm{D}} {\rm{d}} } {\rm{A = 2 \times 16\pi }}\\{\rm{ = 32\pi }}\end{array}\)

However, because we used a formula for an upward-oriented surface, the result is \({\rm{ - 32\pi }}\)

Since then, Stoke's theorem has been confirmed\(\int {\int_{\rm{S}} {{\rm{curl}}} } {\rm{F \times dS = }}\int_{\rm{C}} {\rm{F}} {\rm{ \times dr = - 32\pi }}\)

The Stokes' Theorem has been proven correct

\(\int {\int_{\rm{S}} {{\rm{curl}}} } {\rm{F \times dS = }}\int_{\rm{C}} {\rm{F}} {\rm{ \times dr = - 32\pi }}\)