Line Integrals along a Space Line
with Respect to the Arc Length
The definition of a space line is nearly identical to that of a plane line. When the line can be parametrized, such as $C:\gamma(t)=0$ for $a\leq t\leq b$, we have the following for $f:C\subseteq\mathbb{R}^3\to\mathbb{R}$:
\[\int_Cfds = \int_a^bf(\gamma(t))|\gamma'(t)|dt=\int_a^bf(x(t),y(t),z(t))\sqrt{(x'(t))^2+(y'(t))^2+(z'(t))^2}dt.\]where $\gamma(t)=(x(t),y(t),z(t))$.
with Respect to the $x$, $y$, and $z$
The line integral with respect to the $x$, $y$, and $z$ can be evaluated as follows:
\[\int_Cfdx=\int_a^b f(\gamma(t))x'(t)dt,\] \[\int_Cfdy=\int_a^b f(\gamma(t))y'(t)dt,\]and
\[\int_Cfdx=\int_a^b f(\gamma(t))z'(t)dt.\]Let $F=Pi+Qj+Rk$. The three line integrals $\int_CPdx$, $\int_CQdy$, and $\int_CRdz$ often appear together, resulting in the following notation:
\[\int_CPdx+\int_CQdy+\int_CRdz = \int_CPdx+Qdy+Rdz\]Work
The work is an example of those three integrals appearing simultaneously. Suppose we have a force field $F(x,y,z)=P(x,y,z)i+Q(x,y,z)j+R(x,y,z)k$. If an object is moving along a straight line $L$ with the unit directional vector $\vec{v}$, then the work of the force acting on this object is:
\[\text{work}=\text{force along }\vec{v}\cdot\text{distance}=(F\cdot \vec{v})\cdot|L|\]where the dot between $F$ and $\vec{v}$ is the inner product of vector. Thus, the work along a curve $C:\gamma(t)=0$ for $a\leq t\leq b$ can be calculated by
\[\int_CF(\gamma(t))\cdot\frac{\gamma'(t)}{|\gamma(t)|}ds.\]| Moreover, we know $ds = \sqrt{x’(t)^2+y’(t)^2}dt = | \gamma’(t) | dt$, so we can conclude that |
We can always remember $F$ as $Pi+Qj$ and $d\gamma$ as $dx i +dyj$, so $\int Fd\gamma=\int pdx+qdy$.
The Fundamental Theorem for Line Integrals
Theorem. Let $C$ be a smooth curve given by the vector function $\gamma(t)$, $a\leq t\leq b$. Let $f$ be a differentiable function of two or three variables whose gradient vector $\nabla f$ is continuous on $C$. Then
\[\int_C\nabla f\cdot d\gamma= f(\gamma(b))-f(\gamma(a)).\]proof. The proof of this theorem is an application of the Chain rule. We have
\[\begin{align*} \int_C\nabla f\cdot d\gamma &= \int_a^b \nabla f(\gamma(t))\gamma'(t)dt\\\\ &= \int_a^b\frac{\partial f}{\partial x}\frac{dx}{dt}+\frac{\partial f}{\partial y}\frac{dy}{dt})dt\\\\ &=\int_a^b\frac{d}{dt}(f(\gamma(t)))dt\\\\ &=f(\gamma(b))-f(\gamma(a)). \end{align*}\]This theorem states that under appropriate conditions, the line integral is independent of the path. A function is deemed path-independent if its line integral does not depend on the path, as specified by the theorem. Suppose $C_1$ and $C_2$ are two piecewise-smooth curves that have the same initial point and terminal point, and let $F$ be path-independent, i.e. $\int_{C_1}Fd\gamma =\int_{C_2}Fd\gamma$. Then, we have
\[\int_{C_1-C_2}Fd\gamma=\int_{C_1}Fd\gamma-\int_{C_2}Fd\gamma = 0.\]Note that $C_1-C_2$ is a path from the initial point to the terminal point, and then back to the initial point. Therefore, it forms a loop. We refer to any such loop as a closed curve. This observation leads to the following lemma.
Lemma. $\int_C F\cdot d\gamma$ is independent of path in $D$ if and only if $\int_C F\cdot d\gamma=0$ for every closed path $C$ in $D.$
We define the following in the following discussion:
- A path C is simple if it doesn’t cross itself.
- A region D is open if for any point $p\in D$ there exists a disc in $D$ containing $p$.
- A region D is connected if we can connect any two points in the region with a path that lies completely in D.
We aim to demonstrate that if $\int_C F\cdot d\gamma=0$, then $F$ is almost always conservative. Together with the lemma and the first theorem, we arrive at the following conclusion:
Theorem. If $F$ is a conservative vector field, then $\int_CF\cdot d\gamma$ is path independent.
The converse of this statement is not true (refer to the worksheet). However, if we limit $F$ to a “nice” domain, we can derive the inverse statement.
Now, let’s delve into the specifics of this inverse statement.
Theorem. Suppose $D$ is an open, connected domain, and $F$ is a vector field on $D$ that is of class $C^1$. If $\int_C F\cdot d\gamma$ is path independent for every path $C$ in $D$, then $F$ is a conservative vector field, that is, $F=\nabla f$ for some potential function $f$ on $D$.
We will prove this result after we introduce Green’s theorem.