By applying Stokes theorem, the value of \(\mathop \oint \limits_C \left[ {\left( {x + y} \right)dx + \left( {2x - z} \right)dy + \left( {y + z} \right)dz} \right]\) where C is the boundary of the triangle with vertices (2, 0, 0), (0, 3, 0) and (0, 0, 6), is

Concept:

By stokes theorem:

\(\oint_{c}^{}\vec{F}.\vec{dr}=\iint_{s}^{}(∇\times\vec{F}).\hat {n}.ds\)

Calculation:

Given:

\(\vec{F}.\vec{dr}= \left[ {\left( {x + y} \right)dx + \left( {2x - z} \right)dy + \left( {y + z} \right)dz} \right]\)

\(∴\vec{F}=(x+y)\hat{i}\;+(2x-z)\hat{j}\;+(y+z)\hat{k}\)

\(∇ \times \vec F = \left| {\begin{array}{*{20}{c}} \hat{i}&\hat{j}&\hat{k}\\ {\frac{\partial }{{\partial x}}}&{\frac{\partial }{{\partial y}}}&{\frac{\partial }{{\partial z}}}\\ {x + y}&{2x - z}&{y + z} \end{array}} \right| = 2\hat{i} + \hat{k}\)

Equation of the plane through A, B, C is

\(\frac{x}{2} + \frac{y}{3} + \frac{z}{6} = 1\)

\(∴\; 3x + 2y + z = 6 \)

∴ ϕ = 3x + 2y + z - 6 represents equation of the surface.

Vector normal to the surface is given by gradient:

\(\nabla \phi=\left ( \hat{i}\frac{\partial}{\partial x}\;+\;\hat{j}\frac{\partial}{\partial y}\;+\;\hat{k}\frac{\partial}{\partial z} \right )(3x\;+\;2y\;+\;z\;-6)\)

\(\vec{n}=3\hat{i}\;+\;2\hat{j}\;+\hat{k}\)

Unit vector n̂: 

\(̂{n}=\frac{\vec{n}}{|\vec n|}\Rightarrow\frac{3\hat{i}\;+\;2\hat{j}\;+\;\hat{k}}{\sqrt{3^2\;+\;2^2\;+\;1^2}}=\frac{3\hat{i}\;+\;2\hat{j}\;+\;\hat{k}}{\sqrt{14}}\)

\(\oint_{c}^{}\vec{F}.\vec{dr}=\iint_{s}^{}(∇\times\vec{F}).\hat{n}.ds\)

\(\iint_{s}^{}(2\hat{i}\;+\;\hat{k}).\left ( \frac{3\hat{i}\;+\;2\hat{j}\;+\;\hat k}{\sqrt{14}} \right )ds\)

\(\frac{1}{\sqrt{14}}(6\;+\;1)\iint_{s}^{}ds\)

\(\rm [where \iint_{s}^{}represents\;area\;of\;triangle\;ABC]\)

A (2, 0, 0), B (0, 3, 0) and C (0, 0, 6)

\(\overrightarrow {AB} = \overrightarrow {OB} - \overrightarrow {OA} \) ⇒ [(0 - 2), (3 - 0), (0 - 0)] = (-2, 3, 0)

\(\overrightarrow {BC} = \overrightarrow {OC} - \overrightarrow {OB} \) ⇒ [(0 - 0), (0 - 3), (6 - 0)] = (0, -3, 6)

\(Area\;of\;{\rm{\Delta }}ABC \)

\(= magnitude ~of ~\frac{1}{2} \times \overrightarrow {AB} \times \overrightarrow {BC} \)

\(\therefore\;\frac{1}{2}\left| {\begin{array}{*{20}{c}} {\hat i}&{\hat j}&{\hat k}\\ {-2}&3&0\\ 0&{-3}&6 \end{array}} \right|\)

\(\therefore\;\frac{1}{2}\left( {18\hat i + 12\hat j + 6\hat k} \right)\)

\(\therefore\;9\hat i + 6\hat j + 3\hat k\)

\(\therefore\;Area = \sqrt {{9^2} + {6^2} + {3^2}} \Rightarrow \sqrt {126} = \sqrt {14 \times 9} \Rightarrow 3\sqrt {14} \)

\(\frac{1}{\sqrt{14}}(6\;+\;1)\iint_{s}^{}ds\)

\(\therefore\;\frac{7}{\sqrt{14}}\times\;3\sqrt{14}\Rightarrow21\;units\)