Line Integrals of Nonconservative Vector Fields

# Line Integrals of Nonconservative Vector Fields

Recall from The Fundamental Theorem for Line Integrals that if $\mathbb{F}$ is a conservative vector field and $\phi$ is a potential function of $\mathbb{F}$ such that $\mathbb{F} = \nabla \phi$, then $\mathbb{F}$ is independent of path, and if $C$ is a curve parameterized as $\vec{r}(t)$ for $a ≤ t ≤ b$ then:

(1)
\begin{align} \quad \int_C \mathbf{F} \cdot d \vec{r} = \int_C \nabla \phi \cdot d \vec{r} = \phi (\vec{r}(b)) - \phi (\vec{r}(a)) \end{align}

We will now look at using this property on nonconservative vector fields that are almost conservative.

For example, suppose that we want to evaluate the the following line integral where $C$ is the curve of intersection of the surfaces $z = \ln (1 + x)$ and $y = x$ from the point $(0, 0, 0)$ to $(1, 1, \ln 2)$:

(2)
\begin{align} \quad \int_C (2x \sin (\pi y) - e^z) \: dx + (\pi x^2 \cos (\pi y) - 3e^z) \: dy - xe^z \: dz \end{align}

Let $\mathbf{F} (x, y, z) = (2x \sin (\pi y) - e^z) \vec{i} + (\pi x^2 \cos (\pi y) - 3e^z) \vec{j} - xe^z \vec{k}$. Notice that:

(3)
\begin{align} \quad \frac{\partial P}{\partial y} =2\pi x \cos (\pi y) = \frac{\partial Q}{\partial x} \\ \quad \frac{\partial P}{\partial z} = -e^z = \frac{\partial R}{\partial x} \\ \quad \frac{\partial Q}{\partial z} = -3e^z \neq 0 = \frac{\partial R}{\partial y} \end{align}

Now consider the field $\mathbf{F}^*(x, y, z) = (2x \sin (\pi y) - e^z) \vec{i} + (\pi x^2 \cos (\pi y) - 3e^z) \vec{j} + (- xe^z - 3ye^z) \vec{k}$ then it's not hard to verify that $\frac{\partial P^*}{\partial y} = \frac{\partial Q^*}{\partial x}$, $\frac{\partial P^*}{\partial z} = \frac{\partial R^*}{\partial x}$ and $\frac{\partial Q^*}{\partial z} = \frac{\partial R^*}{\partial y}$. In fact, $\mathbf{F}^*$ is a conservative vector field. Let $\phi$ be a potential function for $\mathbf{F}^*$. Then we must have that $\frac{\partial \phi}{\partial x} = P^*$, $\frac{\partial \phi}{\partial y} = Q^*$ and $\frac{\partial \phi}{\partial z} = R^*$. From the first equation we have that:

(4)
\begin{align} \quad \frac{\partial \phi}{\partial x} = 2x \sin (\pi y) - e^z \\ \quad \int \frac{\partial \phi}{\partial x} \: dx = \int (2x \sin (\pi y) - e^z ) \: dx \\ \quad \phi (x, y, z) = x^2 \sin (\pi y) - xe^z + h(y, z) \end{align}

We will now partial differentiate both sides with respect to $y$ to get that:

(5)
\begin{align} \quad \frac{\partial \phi}{\partial y} = \pi x^2 \cos (\pi y) + \frac{\partial h}{\partial y} \end{align}

We already know that $\frac{\partial \phi}{\partial y} = Q^*(x, y, z) = \pi x^2 \cos (\pi y) - 3e^z$ which implies that:

(6)
\begin{align} \quad \frac{\partial h}{\partial y} = - 3e^z \\ \quad \int \frac{\partial h}{\partial y} \: dy = \int - 3e^z \: dy \\ \quad h(y, z) = -3ye^z + g(z) \end{align}

Therefore we have that:

(7)
\begin{align} \quad \phi (x, y, z) = x^2 \sin (\pi y) - xe^z -3ye^z + g(z) \end{align}

We now partial differentiate both sides with respect to $z$ to get that:

(8)
\begin{align} \quad \frac{\partial \phi}{\partial z} = -xe^z - 3ye^z + g'(z) \end{align}

We already know that $\frac{\partial \phi}{\partial z} = R^*(x, y, z) = -xe^z - 3ye^z$ which implies that $g'(z) = 0$ so $\int g'(z) \: dz = \int 0 \: dz$ so $g(z) = C$ and for $C = 0$ we have that a potential function for $\mathbf{F}^*$ is:

(9)
\begin{align} \quad \phi (x, y, z) = x^2 \sin (\pi y) - xe^z -3ye^z \\ \quad \phi (x, y, z) = x^2 \sin (\pi y) - (x + 3y)e^z \end{align}

Now notice that:

(10)
\begin{align} \quad \mathbf{F}(x, y, z) = \mathbf{F}^*(x, y, z) + 3ye^z \vec{k} \end{align}

Thus this implies that if we take the line integral along $C$ of both sides of the equation above, then:

(11)
\begin{align} \quad \int_C \mathbf{F} \cdot d \vec{r} = \int_C \mathbf{F}^* \cdot d \vec{r} + \int_C 3ye^z \vec{k} \cdot d \vec{r} \end{align}

However, since $\mathbf{F}^*$ is a conservative vector field, then it can be evaluated by using the potential function we computed earlier, and:

(12)
\begin{align} \quad \int_C \mathbf{F} \cdot d \vec{r} = \phi(1, 1, \ln 2) - \phi(0, 0, 0) + \int_C 3ye^z \vec{k} \cdot d \vec{r} \\ \quad \int_C \mathbf{F} \cdot d \vec{r} = (-8) - 0 + \int_C 3ye^z \vec{k} \cdot d \vec{r} \\ \quad \int_C \mathbf{F} \cdot d \vec{r} = -8 + \int_C 3ye^z \: dz \end{align}

Note that we have greatly reduced the problem of evaluating our line integral. Now we only need to evaluate $\int_C 3ye^z \: dz$. We must first parameterize the curve $C$. Recall that $C$ is the curve of intersection of $z = \ln (1 + x)$ and $y = x$. Let $x = t$. Then $y = t$ and $z = \ln (1 + t)$ and so a parameterization for $C$ is given for $0 ≤ t ≤ 1$ by:

(13)
\begin{align} \quad \vec{r}(t) = (t, t, \ln (1 + t)) \end{align}

Therefore we have that:

(14)
\begin{align} \quad \int_C \mathbf{F} \cdot d \vec{r} = -8 + \int_0^1 3te^{\ln (1 + t)} \frac{1}{1 + t} \: dt \\ \quad \int_C \mathbf{F} \cdot d \vec{r} = -8 + \int_0^1 3t \: dt \\ \quad \int_C \mathbf{F} \cdot d \vec{r} = -8 + \left [ \frac{3t^2}{2} \right ]_{t=0}^{t=1} \\ \quad \int_C \mathbf{F} \cdot d \vec{r} = -8 + \frac{3}{2} \\ \quad \int_C \mathbf{F} \cdot d \vec{r} = -\frac{13}{2} \end{align}

As you can see - we can sometimes greatly simplify the work involved in evaluating line integrals over difficult fields by breaking the original field in the sum of a conservative vector field and a "remainder" of sorts.