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Suppose $C$ is a smooth curve given by $\vec{r}(t)$, $a \leq t \leq b$. Also suppose that $\Phi$ is a function whose gradient vector, $\nabla \Phi=f$, is continuous on $C$. Then $$ \int_C f \cdot \,\mathrm{d}\vec{r} = \Phi(\vec{r}(b))-\Phi(\vec{r}(a)). $$

To prove this, we start by rewriting the integral using the parameterization of $C$. So $$ \int_C f \cdot \,\mathrm{d}\vec{r} = \int_a^bf(\vec{r}(t)) \cdot \vec{r}^{\,\prime}(t) \, \mathrm{d}t $$ Since $\Phi$ is the potential function of $f$, $$ \int_a^b f(\vec{r}(t)) \cdot \vec{r}^{\,\prime}(t) \, \mathrm{d}t = \int_a^b \nabla \Phi(\vec{r}(t)) \cdot \vec{r}^{\,\prime}(t) \, \mathrm{d}t, $$ and with the substitution $\omega=r(t)$, $\mathrm{d}\omega=r^{\prime}(t)\,\mathrm{d}t$, $$ \int_a^b \nabla \Phi(\vec{r}(t)) \cdot \vec{r}^{\,\prime}(t) \, \mathrm{d}t = \int_{\omega_1}^{\omega_2}\bigl[\Phi(\omega)\bigr]'\,\mathrm{d}\omega, $$ and since $\omega_1=\vec{r}(a)$ and $\omega_2=\vec{r}(b)$, the fundamental theorem of calculus gives $$ \int_C f \cdot \mathrm{d}\vec{r} = \Phi(\omega_2) - \Phi(\omega_1) = \Phi(\vec{r}(b)) - \Phi(\vec{r}(a)). \quad\Box $$ Is this proof complete? Can you explain why the equality $$ \int_a^b \nabla \Phi(\vec{r}(t)) \cdot \vec{r}^{\,\prime}(t) \, \mathrm{d}t = \int_{\omega_1}^{\omega_2}\bigl[\Phi(\omega)\bigr]'\,\mathrm{d}\omega, $$ holds?

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All you need is the chain rule and the definition of scalar product. Setting $r(t)=(x_1(t),\dots, x_n(t))$ we have $$\frac{d}{dt}\Phi(x_1(t),\dots,x_n(t))= \sum_{j=1}^n \frac{\partial \Phi}{\partial x_j} \frac{d x_j(t)}{dt}. $$ Now $\nabla \Phi = \left(\frac{\partial \Phi}{\partial x_1},\dots \frac{\partial \Phi}{\partial x_n}\right)$ and $r'(t)=(x_1'(t),\dots, x_n'(t))$ so $$\nabla\Phi \cdot r'(t) = \sum_{j=1}^n \frac{\partial \Phi}{\partial x_j} \frac{d x_j(t)}{dt} = [\Phi(r(t))]' $$

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