Question

Suppose\({\rm{S}}\) and\({\rm{C}}\) satisfy the hypotheses of Stokes’ Theorem and \({\rm{f,g}}\) have continuous second-order partial derivatives. Use Exercises \(22\) and \(24\) in Section \(13.5\) to show the following.

(a)\(\int_{\rm{C}} {\left( {{\rm{f}}\nabla {\rm{g}}} \right) \cdot {\rm{dr = }}} \int {\int_{\rm{s}} {\left( {\nabla {\rm{f \times }}\nabla {\rm{g}}} \right) \cdot } } {\rm{dS}}\)

(b) \(\int_{\rm{C}} {\left( {{\rm{f}}\nabla {\rm{f}}} \right) \cdot {\rm{dr = }}} 0\)

(c) \(\int_{\rm{C}} {\left( {{\rm{f}}\nabla {\rm{g + g}}\nabla {\rm{f}}} \right) \cdot {\rm{dr = }}} 0\)

Short answer

(a) When \({\rm{S}}\) and \({\rm{C}}\) satisfy the hypotheses of Stokes’ Theorem and \({\rm{f,g}}\) have continuous second-order partial derivatives, then it is proved that \(\int_{\text{C}} {{\text{(f}}\nabla {\text{g)}}} \cdot {\text{dr = }}\iint_{\text{S}} {{\text{(}}\nabla {\text{f }} \times {\text{ }}\nabla {\text{g)}}} \cdot {\text{dS}}\) .

(b)When \({\rm{S}}\) and \({\rm{C}}\) satisfy the hypotheses of Stokes’ Theorem and \({\rm{f,g}}\) have continuous second-order partial derivatives, then it is proved that \(\int_{\rm{C}} {{\rm{(f}}\nabla {\rm{g)}}} \cdot {\rm{dr = 0}}\).

(c)When \({\rm{S}}\) and \({\rm{C}}\) satisfy the hypotheses of Stokes’ Theorem and \({\rm{f,g}}\) have continuous second-order partial derivatives, then it is proved that \(\int_{\rm{C}} {{\rm{(f}}\nabla {\rm{g + g}}\nabla {\rm{f)}}} \cdot {\rm{dr = 0}}\).

Step-by-step solution

Concept Introduction

According to Stoke's theorem "The surface integral of the curl of a function over a surface limited by a closed surface is equivalent to the line integral of the particular vector function around that surface,".

Calculation for first integral

(a)

Using Stokes' Theorem, it can be written –

\(\int_{\text{C}} {{\text{(f}}\nabla {\text{g)}}} \cdot {\text{dr = }}\iint_{\text{S}} {{\text{curl}}}{\text{(f}}\nabla {\text{g)}} \cdot {\text{dS}}.....{\text{(1)}}\)

Nowuse the identity –

\(\begin{aligned}{}{\rm{curl(fF) = f curl F + (}}\nabla {\rm{f) \times F}}\\{\rm{curl(f}}\nabla {\rm{g) = f curl(}}\nabla {\rm{g) + }}\nabla {\rm{f \times }}\nabla {\rm{g}}\end{aligned}\)

Since the curl of a gradient vector field is\({\rm{0}}\), it is obtained –

\({\rm{curl(f}}\nabla {\rm{g) = }}\nabla {\rm{f \times }}\nabla {\rm{g}}\)

Then, equation\({\rm{(1)}}\)can be written as –

\(\int_{\text{C}} {{\text{(f}}\nabla {\text{g)}}} \cdot {\text{dr = }}\iint_{\text{S}} {{\text{(}}\nabla {\text{f x }}\nabla {\text{g)}}} \cdot {\text{dS}}\)

Therefore, it is proved that

\(\int_{\text{C}} {{\text{(f}}\nabla {\text{g)}}} \cdot {\text{dr = }}\iint_{\text{S}} {{\text{(}}\nabla {\text{f x }}\nabla {\text{g)}}} \cdot {\text{dS}}\) .

Calculation for second integral

(b)

Using Stokes' Theorem, it can be written –

\(\int_{\text{C}} {{\text{(f}}\nabla {\text{f)}}} \cdot {\text{dr = }}\iint_{\text{S}} {{\text{curl}}}{\text{(f}}\nabla {\text{f)}} \cdot {\text{dS}}.....{\text{(1)}}\)

Now use the identity –

\({\rm{curl(fF) = f curl F + (}}\nabla {\rm{f) \times F}}\)

Where\({\rm{F}}\)will be replaced by\(\nabla {\rm{f}}\)–

\({\rm{curl(f}}\nabla {\rm{f) = f curl }}\nabla {\rm{f + }}\nabla {\rm{f \times }}\nabla {\rm{f}}\)

Since the curl of a gradient vector field is\({\rm{0}}\)and\({\rm{v \times v = 0}}\)for every vector\({\rm{v}}\), it is obtained –

\({\rm{f curl }}\nabla {\rm{f}} = 0\)and\(\nabla {\rm{f \times }}\nabla {\rm{f}} = 0\).

Then, equation\({\rm{(1)}}\)can be written as –

\(\int_{\text{C}} {{\text{(f}}\nabla {\text{f)}}} \cdot {\text{dr = }}\iint_{\text{S}} {\text{0}} \cdot {\text{dS = 0}}\)

Therefore, it is proved that \(\int_{\rm{C}} {{\rm{(f}}\nabla {\rm{g)}}} \cdot {\rm{dr = 0}}\).

Calculation for third integral

(c)

Using Stokes' Theorem, it can be written –

\(\int_{\text{C}} {{\text{(f}}\nabla {\text{g + g}}\nabla {\text{f)}}} \cdot {\text{dr = }}\iint_{\text{S}} {{\text{curl}}}{\text{(f}}\nabla {\text{g + g}}\nabla {\text{f)}} \cdot {\text{dS}}.....{\text{(1)}}\)

Now use the identity –

\({\rm{curl(F + G) = curl F + curl G}}\)

Now, it can be written –

\({\rm{curl(f}}\nabla {\rm{g + g}}\nabla {\rm{f) = curl}}\left( {{\rm{f}}\nabla {\rm{g}}} \right){\rm{ + curl}}\left( {{\rm{g}}\nabla {\rm{f}}} \right)\)

Next, use the identity –

\({\rm{curl(fF) = f curl F + (}}\nabla {\rm{f) \times F}}\)

And it is obtained that –

\({\rm{curl(f}}\nabla {\rm{g) + curl(g}}\nabla {\rm{f) = f curl}}\nabla {\rm{g + }}\nabla {\rm{f \times }}\nabla {\rm{g + g curl }}\nabla {\rm{f + }}\nabla {\rm{g \times }}\nabla {\rm{f}}\)

Since the curl of a gradient vector field is \({\rm{0}}\) and \({\rm{a \times b = - b \times a}}\) for vectors\({\rm{a,b}}\), it is obtained –

\({\rm{curl(f}}\nabla {\rm{g) + curl(g}}\nabla {\rm{f) = 0}}\)

Then, equation\({\rm{(1)}}\)can be written as –

\(\int_{\text{C}} {{\text{(f}}\nabla {\text{g + g}}\nabla {\text{f)}}} \cdot {\text{dr = }}\iint_{\text{S}} {\text{0}} \cdot {\text{dS = 0}}\)

Therefore, it is proved that \(\int_{\rm{C}} {{\rm{(f}}\nabla {\rm{g + g}}\nabla {\rm{f)}}} \cdot {\rm{dr = 0}}\).