4.13 Stokes’ Theorem
Let \(S\) be an oriented piece-wise smooth surface that is bounded by a simple closed piecewise-smooth boundary curve \(C\) with positive orientation. Let \(F\) be a vector field whose components have continuous partial derivatives on an open region in \(\mathbb {R}^3\) that contains \(S\). Then \[\int _C F\cdot d\overline {r} = \iint \limits _S\big (\curl \hspace {0.1cm} F\big ) dS\]
Evaluate \(\displaystyle {\int _C F\cdot d\overline {r}}\), where \[F = -y^2\textbf {i} + x\textbf {j} + z^2\textbf {k}\] and \(C\) is the curve of intersection of the plane \(y + z = 2\) and the cylinder \(x^2 + y^2 =1\).
Solution.
\(y = 2-z\)
\(\implies \hspace {0.5cm} x^2 + (2-z)^2 = 1\) \(\implies \hspace {0.5cm} x^2 + (z-2)^2 = 1\)
\begin {align*} \curl \hspace {0.1cm} F & = \begin {vmatrix} \textbf {i} & \textbf {j} & \textbf {k}\\\\ \dfrac {\partial }{\partial x} & \dfrac {\partial }{\partial y} & \dfrac {\partial }{\partial z}\\\\ -y^2 & x & z^2\\ \end {vmatrix} = (1 + 2y)\textbf {k} = \langle 0, 0, 1 + 2y\rangle \\ \end {align*}
The projection \(D\) of \(S\) onto the \(xy-\) plane is the disc \(x^2 + y^2 \leq 1\) and so using \(z = g(x,y) = 2 -y\)
We get: \begin {align*} \int _C F\cdot d \overline {r} & = \iint \limits _S\big ( \curl F\big )\cdot dS\\\\ & = \iint \limits _D \Bigg [-0 \Bigg (\dfrac {\partial g}{\partial x}\Bigg ) - 0\Bigg (\dfrac {\partial g}{\partial y}\Bigg ) + (1 + 2y)\Bigg ]dA\\\\ & = \iint \limits _D (1 + 2y)dA\\\\ & = \int ^{2\pi }_0\int ^{\pi }_0\big ( 1 + 2r\sin \theta \big ) r dr d\theta \\\\ & = \pi \\\\ \end {align*}
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