Question

Use Stokes’ Theorem to evaluate \(\int_{\rm{C}} {\rm{F}} \cdot {\rm{dr}}\),where \({\rm{F(x,y,z) = xyi + yzj + zxk}}\), and \({\rm{C}}\) is thetriangle with vertices \({\rm{(1,0,0),(0,1,0),}}\) and \({\rm{(0,0,1),}}\) oriented counter-clockwise as viewed from above.

Short answer

Stokes’ Theorem evaluation \(\int_{\rm{C}} {\rm{F}} \cdot {\rm{dr = - }}\frac{{\rm{1}}}{{\rm{2}}}\).

Step-by-step solution

Explanation of solution.

Stokes’ Theorem

\(\int_{C}{F}\cdot dr=\iint_{S}{curl}F\cdot dS\)

The boundary of the surface S is given by C

Instead of computing the line integral, the surface integral can be compute.

Compute \(\iint_{S}{F}\cdot dS\)

But what is \({\rm{S,S}}\) is any curve having boundary as \({\rm{C}}\)

Equation of the plane with intercepts \({\rm{(1,0,0}}),(0,1,0)\) and \((0,0,1)\) is

\(\begin{aligned}\frac{{\rm{x}}}{{\rm{1}}}{\rm{ + }}\frac{{\rm{y}}}{{\rm{1}}}{\rm{ + }}\frac{{\rm{z}}}{{\rm{1}}}{\rm{ = 1}}\\{\rm{z = 1 - x - y}}\end{aligned}\)

Remember that if the surface \({\rm{S}}\) is oriented upwards and the form \({\rm{z = g}}\left( {{\rm{x, y}}} \right){\rm{,}}\)

Formula used

\(\iint_{S}{F}\cdot dS=\iint_{D}{-}P\left( \frac{\partial g}{\partial x} \right)-Q\left( \frac{\partial g}{\partial y} \right)+RdA\)

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

But how can know if \({\rm{S}}\)is orientedupwards

\({\rm{C}}\) 's orientation is not given;however, it is counter-clockwise by convention.

Using Stokes’s Theorem.

Need to find the \({\rm{curlF}}\)

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

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

Here

\({\rm{F(x,y,z) = xy i + yz j + zx k}}\)

compute the Required partial derivatives,

\({\rm{P = xy}} \Rightarrow {{\rm{P}}_{\rm{y}}}{\rm{ = x}}\)and \({{\rm{P}}_{\rm{z}}}{\rm{ = 0}}\)

\({\rm{Q = yz}} \Rightarrow {{\rm{Q}}_{\rm{x}}}{\rm{ = 0}}\) and \({{\rm{Q}}_{\rm{z}}}{\rm{ = y}}\)

\({\rm{R = zx}} \Rightarrow {{\rm{R}}_{\rm{x}}}{\rm{ = z}}\) and \({{\rm{R}}_{\rm{y}}}{\rm{ = 0}}\)

Therefore,

\(\begin{array}{l}{\rm{curlF = (0 - y)i + (0 - z)j + (0 - x)k}}\\{\rm{curlF = - yi - zj - xk}}\end{array}\)

\(\begin{aligned} \iint_{S}{curl}F\cdot dS&=\iint_{D}{-}P\left( \frac{\partial g}{\partial x} \right)-Q\left( \frac{\partial g}{\partial y} \right)+RdA \\ & =\iint_{D}{-}(-y)(-1)-(-z)(-1)-xdA \\ & =\iint_{D}{-}y-z-xdA \end{aligned}\)

Note \({\rm{x + y + z = 1}}\)

\(=-\iint_{D}{d}A\)

Note that \({\rm{D}}\) is the \({\rm{xy}}\) plane projection of the surface \(S\).

\({\rm{D}}\) is the region inside the triangle with vertices \({\rm{(0,0)}}\),\({\rm{(1,0)}}\) and \({\rm{(0,1)}}\)

And \(\iint_{-}{D}dA\) is the area of that triangle, which is \(\frac{{\rm{1}}}{{\rm{2}}}{\rm{ * Height * Base = }}\frac{{\rm{1}}}{{\rm{2}}}{\rm{ *1 * 1 = }}\frac{{\rm{1}}}{{\rm{2}}}\).

Therefore,

\(\int_{C}{F}\cdot dr=\iint_{S}{curl}F\cdot dS=-\iint_{D}{d}A=-\frac{1}{2}\). .

Stokes’ Theorem evaluation \(\int_{\rm{C}} {\rm{F}} \cdot {\rm{dr = - }}\frac{{\rm{1}}}{{\rm{2}}}\).