4.7 The Fundamental Theorem of Line Integrals
Part of the fundamental theorem of calculus says that \[\int ^b_a F'(x)\,dx = F(b) - F(a)\] when \(F'\) is continuous on \([a,b]\). The analogue for line integrals replaces \(F'\) by a gradient.
Theorem 4.7.1 (Fundamental theorem of line integrals). Let \(C\) be a smooth curve given by the vector function \(\mathbf r(t)\), \(a\leq t \leq b\). Let \(f\) be a differentiable function of two or three variables whose gradient vector \(\nabla f\) is continuous on \(C\). Then \[\int _C \nabla f \cdot d\mathbf r = f(\mathbf r(b)) - f(\mathbf r(a)) .\]
Proof. By the chain rule, \(\frac {d}{dt}f(\mathbf r(t)) = f_x x'(t) + f_y y'(t) + f_z z'(t) = \nabla f(\mathbf r(t))\cdot \mathbf r'(t)\). So by the fundamental theorem of calculus, \[\int _C \nabla f \cdot d\mathbf r = \int _a^b \nabla f(\mathbf r(t))\cdot \mathbf r'(t)\,dt = \int _a^b\frac {d}{dt}f(\mathbf r(t))\,dt = f(\mathbf r(b)) - f(\mathbf r(a)) .\] The result extends to piecewise smooth curves by adding over the pieces, the intermediate values cancelling. □
If \(f\) is a function of two variables and \(C\) is a plane curve from \(A(x_1,y_1)\) to \(B(x_2,y_2)\), the theorem reads \[\int _C \nabla f \cdot d\mathbf r = f(x_2,y_2) - f(x_1,y_1) .\] In particular, the line integral of a gradient depends only on the endpoints. In general, if \(\mathbf F\) is a continuous vector field with domain \(D\), the line integral \(\int _C \mathbf F\cdot d\mathbf r\) is said to be independent of path if \[\int _{C_1}\mathbf F\cdot d\mathbf r = \int _{C_2}\mathbf F\cdot d\mathbf r\] for any two paths \(C_1\) and \(C_2\) in \(D\) with the same initial and terminal points. Theorem 4.7.1 says that line integrals of conservative vector fields are independent of path.
Theorem 4.7.2. The line integral \(\int _C \mathbf F\cdot d\mathbf r\) is independent of path in \(D\) if and only if \(\oint _C \mathbf F\cdot d\mathbf r = 0\) for every closed path \(C\) in \(D\). If \(D\) is open and connected and the integral is independent of path, then \(\mathbf F\) is conservative on \(D\).
Proof. Suppose the integral is independent of path, and let \(C\) be closed. Choose two points \(A\) and \(B\) on \(C\), splitting it into a path \(C_1\) from \(A\) to \(B\) and a path \(C_2\) from \(B\) back to \(A\). Then \(C_1\) and \(-C_2\) both run from \(A\) to \(B\), so \[\oint _C\mathbf F\cdot d\mathbf r = \int _{C_1}\mathbf F\cdot d\mathbf r + \int _{C_2}\mathbf F\cdot d\mathbf r = \int _{C_1}\mathbf F\cdot d\mathbf r - \int _{-C_2}\mathbf F\cdot d\mathbf r = 0 .\] Conversely, if \(C_1\) and \(C_2\) both run from \(A\) to \(B\), then \(C_1\) followed by \(-C_2\) is closed, so \(\int _{C_1} - \int _{C_2} = 0\).
For the last statement, fix \((a,b)\) in \(D\) and define \(f(x,y) = \int _{(a,b)}^{(x,y)}\mathbf F\cdot d\mathbf r\), along any path in \(D\), which is unambiguous by independence of path. To compute \(f_x\) at \((x,y)\), take a path to a nearby point \((x_1,y)\) followed by the horizontal segment to \((x,y)\). Along the segment \(dy = 0\), so \(f(x,y) = \text {constant} + \int _{x_1}^x P(t,y)\,dt\), and \(f_x = P\) by the fundamental theorem of calculus. Similarly \(f_y = Q\), so \(\nabla f = \mathbf F\). □
This also completes the proof of Theorem 4.2.2 in the plane: on a simply connected region, \(P_y = Q_x\) makes the circulation around every simple closed curve zero, by Green’s theorem in the next subsection, and so \(\mathbf F\) is conservative.
Example 4.7.3. Find a potential function for \(\mathbf F(x,y) = (3 + 2xy)\,\mathbf i + (x^2 - 3y^2)\,\mathbf j\), and evaluate \(\int _C \mathbf F\cdot d\mathbf r\), where \(C\) is the curve \(\mathbf r(t) = e^t \sin t \,\mathbf i + e^t \cos t \,\mathbf j\), \(0 \leq t \leq \pi \).
Solution. Since \(P_y = 2x = Q_x\) on the whole plane, \(\mathbf F\) is conservative. From \(f_x = 3 + 2xy\), \(f = 3x + x^2y + g(y)\); then \(f_y = x^2 + g'(y) = x^2 - 3y^2\), so \(g(y) = -y^3\) and \[f(x,y) = 3x + x^2y - y^3 + C .\] The curve runs from \(\mathbf r(0) = (0,1)\) to \(\mathbf r(\pi ) = (0,-e^{\pi })\). By Theorem 4.7.1, \[\int _C \mathbf F\cdot d\mathbf r = f(0, -e^{\pi }) - f(0,1) = e^{3\pi } - (-1) = e^{3\pi } + 1 .\] Evaluating the line integral directly along this spiral would be much harder.
Example 4.7.4. Find the work done by the gravitational field \(\mathbf F = -\frac {mMG}{|\mathbf x|^3}\mathbf x\) of Example 4.1.5 in moving a particle from \((3,4,12)\) to \((2,2,0)\) along any piecewise smooth path avoiding the origin.
Solution. The field is \(\nabla f\) with \(f(\mathbf x) = \frac {mMG}{|\mathbf x|}\), so by Theorem 4.7.1 the work is \[f(2,2,0) - f(3,4,12) = mMG\left (\frac {1}{\sqrt 8} - \frac {1}{13}\right ) = mMG\left (\frac {1}{2\sqrt 2} - \frac 1{13}\right )\approx 0.277\,mMG ,\] whatever the path: the particle moves closer to \(M\), and gravity does positive work.
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