4.7 The Fundamental Theorem of Line Integrals
Recall that part of the Fundamental theorem of calculus can be written as \[\int ^b_a F'(x)\,dx = F(b) - F(a)\] where \(F'\) is continuous on \([a,b]\).
Let \(C\) be a smooth curve given by the vector function \(\overline {r}(t),\hspace {0.5cm} a\leq t \leq b\). Let \(f\) be a differentiable function of two or three variables whose gradient vector \(\nabla f\) is continuous on \(C\). Then, \[\int _C \nabla f \cdot d\overline {r} = f\big [\overline {r}(b)\big ] - f\big [\overline {r}(a)\big ]\]
If \(f\) is a function of two variables and \(C\) is a plane curve with initial point \(A(x_1,y_1)\) and terminal point \(B(x_2,y_2)\), then \[\int _C \nabla f \cdot d\overline {r} = f(x_2,y_2) - f(x_1,y_1)\]
one implication of the FTLI is that \[\int _{C_1}\nabla f \cdot d\overline {r} = \int _{C_2}\nabla f\cdot d \overline {r}\hspace {0.3cm},\hspace {0.5cm}\text {where}\hspace {0.3cm}\nabla f \hspace {0.3cm} \text {is continuous}\]
In general, if \(F\) is a continuous vector field with domain \(D\), we say that the line integral \(\displaystyle {\int _C F\cdot d\overline {r}}\) is independent of path if \[\int _{C_1}F\cdot d\overline {r} = \int _{C_2}F\cdot d\overline {r}\] for any two paths \(C_1\) and \(C_2\) in \(D\) that have the same initial and terminal points. Therefore, we say that the line integrals of conservative vector fields are independent of path.
\(\displaystyle {\int _C F\cdot d\overline {r}}\) is independent of the patht in \(D\) if and only if \(\displaystyle {\int _C F\cdot d\overline {r} = 0}\) for every closed path \(C\) in \(D\).
Find the potential function of \[F(x,y) = \big (3 + 2xy\big ) \textbf {i} + \big (x^2 - 3y^2\big )\textbf {j}\] and evaluate \(\displaystyle {\int _C F\cdot d\overline {r}}\), where \(C\) is given by \[\overline {r}(t) = e^t \sin t \textbf {i} + e^t \cos t \textbf {j}\hspace {0.3cm}, \hspace {0.5cm} 0 \leq t \leq \pi \]
Solution.
The potential function is \(f(x,y) = 3x + x^2y - y^3 +C\)
By FTLI, we have \[\int _C \nabla f \cdot d\overline {r} = f\big [\overline {r}(\pi )\big ] - f\big [\overline {r}(0)\big ]\]
\begin {align*} \overline {r}(\pi ) = 0\textbf {i} - e^{\pi }\textbf {j} & = \big ( 0, -e^{\pi }\big )\\\\ \overline {r}(0) = 0\textbf {i} + \textbf {j} & = (0,1) \end {align*}
\[\int _C F\cdot d\overline {r} = f\big (0, -e^{\pi }\big ) - f(0,1)\]
Questions on this section
Stuck on something here? Ask below and it stays attached to this topic.