Vtyjkyuk

Suppose $C$ is a smooth curve given by $vecr(t)$, $a leq t leq b$. Also suppose that $Phi$ is a function whose gradient vector, $nabla Phi=f$, is continuous on $C$. Then
$$
int_C f cdot ,mathrmdvecr
= Phi(vecr(b))-Phi(vecr(a)).
$$

To prove this, we start by rewriting the integral using the parameterization of $C$. So
$$
int_C f cdot ,mathrmdvecr
= int_a^bf(vecr(t)) cdot vecr^,prime(t) , mathrmdt
$$
Since $Phi$ is the potential function of $f$,
$$
int_a^b f(vecr(t)) cdot vecr^,prime(t) , mathrmdt
= int_a^b nabla Phi(vecr(t)) cdot vecr^,prime(t) , mathrmdt,
$$
and with the substitution $omega=r(t)$, $mathrmdomega=r^prime(t),mathrmdt$,
$$
int_a^b nabla Phi(vecr(t)) cdot vecr^,prime(t) , mathrmdt
= int_omega_1^omega_2bigl[Phi(omega)bigr]',mathrmdomega,
$$
and since $omega_1=vecr(a)$ and $omega_2=vecr(b)$, the fundamental theorem of calculus gives
$$
int_C f cdot mathrmdvecr
= Phi(omega_2) - Phi(omega_1)
= Phi(vecr(b)) - Phi(vecr(a)). quadBox
$$
Is this proof complete? Can you explain why the equality
$$
int_a^b nabla Phi(vecr(t)) cdot vecr^,prime(t) , mathrmdt
= int_omega_1^omega_2bigl[Phi(omega)bigr]',mathrmdomega,
$$
holds?

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
$$fracddtPhi(x_1(t),dots,x_n(t))=
sum_j=1^n fracpartial Phipartial x_j fracd x_j(t)dt.
$$
Now $nabla Phi = left(fracpartial Phipartial x_1,dots fracpartial Phipartial x_nright)$ and $r'(t)=(x_1'(t),dots, x_n'(t))$ so
$$nablaPhi cdot r'(t) = sum_j=1^n fracpartial Phipartial x_j fracd x_j(t)dt = [Phi(r(t))]'
$$