Fundamental Theorem of Line Integrals
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Theorem
Let $\CC$ be a smooth curve given by the vector function $\map {\mathbf r} t$ for $a \le t \le b$.
Let $f$ be a differentiable function of two or three variables whose gradient vector $\nabla f$ is continuous on $\CC$.
Then:
- $\ds \int_\CC \nabla f \cdot d \mathbf r = \map f {\map {\mathbf r} b} - \map f {\map {\mathbf r} a}$
Proof
\(\ds \int_\CC \nabla f \cdot \rd \mathbf r\) | \(=\) | \(\ds \int_a^b \nabla f \cdot \map {\mathbf r'} t \rd t\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int_a^b \frac {\partial f} {\partial x} \frac {\rd x} {\rd t} + \frac {\partial f} {\partial y} \frac {\rd y} {\rd t} + \frac {\partial f} {\partial z} \frac {\rd z} {\rd t} \rd t\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \int_a^b \frac \rd {\rd t} \map f {\map {\mathbf r} t} \rd t\) | Chain Rule for Real-Valued Functions | |||||||||||
\(\ds \) | \(=\) | \(\ds \map f {\map {\mathbf r} b} - \map f {\map {\mathbf r} a}\) | Fundamental Theorem of Calculus |
$\blacksquare$