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PostPosted: Wed Aug 29, 2018 2:28 pm 

Joined: Wed Nov 15, 2017 12:37 pm
Posts: 19
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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