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 Post subject: Proof of the fundamental theorem of line integralsPosted: Wed Aug 29, 2018 2:28 pm

Joined: Wed Nov 15, 2017 12:37 pm
Posts: 20
Quote:
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? Though I can see it intuitively, I seek a more mathematically formal reasoning.

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