The Fundamental Theorem for Line Integrals

# The Fundamental Theorem for Line Integrals

Recall from The Fundamental Theorem of Calculus Part 2 page that if $f$ is a continuous function on the interval $[a, b]$ then $\int_a^b f(x) \: dx = F(b) - F(a)$ where $F$ is any antiderivative of $f$ (that is $F' = f$).

The following theorem known as The Fundamental Theorem for Line Integrals or the Gradient Theorem is an analogue of the Fundamental Theorem of Calculus Part 2 for line integrals.

Theorem 1 (The Fundamental Theorem for Line Integrals / The Gradient Theorem): Let $C$ be a smooth curve that is parameterized by the vector equation $\vec{r}(t)$ for $a ≤ t ≤ b$, and suppose that $f$ is a differentiable function and the gradient of $f$ is continuous on the curve $C$. Then $\int_C \nabla f \cdot \: d \vec{r} = f(\vec{r}(b)) - f(\vec{r}(a))$. |

We will prove Theorem 1 in the cases where $f$ is a function of two variables and $f$ is a function of three variables. Of course, Theorem 1 holds for functions with more than $3$ variables. Nevertheless, the proofs are rather simpler and follow the same format.

**Proof (Two Variable Case):**Let $z = f(x, y)$ be a two variable real-valued function that is differentiable and suppose that $\nabla f (x, y)$ is continuous on the smooth curve $C$ parameterized by $\vec{r}(t) = (x(t), y(t))$ for $a ≤ t ≤ b$. Then:

\begin{align} \quad \int_C \nabla f \cdot d \vec{r} = \int_a^b \nabla f(\vec{r}(t)) \cdot \vec{r'}(t) \: dt \\ \quad \int_C \nabla f \cdot d \vec{r} = \int_a^b \left ( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y} \right) \cdot \left ( \frac{dx}{dt}, \frac{dy}{dt} \right ) \: dt \\ \quad \int_C \nabla f \cdot d \vec{r} = \int_a^b \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt} \: dt \\ \end{align}

- Now notice that we can compress the integrand in the right integral by noticing that from the chain rule we have that $\frac{d}{dt} ( f(\vec{r}(t)) = \frac{d}{dt} (f(x(t), y(t)) = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt}$ and so:

\begin{align} \quad \int_C \nabla f \cdot d \vec{r} = \int_a^b \frac{d}{dt} \left ( f (\vec{r}(t)) \right ) \: dt \end{align}

- Now we apply the Fundamental Theorem of Calculus for single integrals to get that:

\begin{align} \quad \int_C \nabla f \cdot d \vec{r} = \left [ f(\vec{r}(t)) \right ]_{t=a}^{t=b} \\ \quad \int_C \nabla f \cdot d \vec{r} = f(\vec{r}(b)) - f(\vec{r}(a)) \quad \blacksquare \end{align}

**Proof (Three Variable Case):**Let $w = f(x, y, z)$ be a three variable real-valued function that is differentiable and uppose that $\nabla f(x, y)$ is continuous on the smooth curve $C$ parameterized by $\vec{r}(t) = (x(t), y(t), z(t))$ for $a ≤ t ≤ b$. Then:

\begin{align} \quad \int_C \nabla f \cdot d \vec{r} = \int_a^b \nabla f(\vec{r}(t)) \cdot \vec{r'}(t) \: dt \\ \quad \int_C \nabla f \cdot d \vec{r} = \int_a^b \left ( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right ) \cdot \left (\frac{dx}{dt}, \frac{dy}{dt}, \frac{dz}{dt} \right ) \: dt \\ \quad \int_C \nabla f \cdot d \vec{r} = \int_a^b \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt} + \frac{\partial f}{\partial z} \frac{dz}{dt} \: dt \end{align}

- Now notice that we can compress the integrand in the right integral by noticing that from the chain rule we have that $\frac{d}{dt} ( f(\vec{r}(t)) = \frac{d}{dt} (f(x(t), y(t), z(t)) = \frac{\partial f}{\partial x} \frac{dx}{dt} + \frac{\partial f}{\partial y} \frac{dy}{dt} + \frac{\partial f}{\partial z} \frac{dz}{dt}$ and so:

\begin{align} \quad \int_C \nabla f \cdot d \vec{r} = \int_a^b \frac{d}{dt} \left ( f(\vec{r}(t) \right ) \: dt \\ \end{align}

- Now we apply the Fundamental Theorem of Calculus for single integrals to get that:

\begin{align} \quad \int_C \nabla f \cdot d \vec{r} = \left [ f(\vec{r}(t)) \right ]_{t=a}^{t=b} \\ \quad \int_C \nabla f \cdot d \vec{r} = f(\vec{r}(b)) - f(\vec{r}(a)) \quad \blacksquare \end{align}

Corollary 1: Let $f$ and $g$ be scalar functions whose first partial derivatives are continuous in $D$. If $C$ is a piecewise smooth curve from the point $P$ to the point $Q$ in $D$, then $\int_C f \nabla g \cdot d \vec{r} + \int_C g \nabla f \cdot d \vec{r} = f(Q)g(Q) - f(P)g(P)$. |

**Proof:**Notice that:

\begin{align} \quad \nabla (fg) = \frac{\partial}{\partial x} fg \vec{i} + \frac{\partial}{\partial y} fg \vec{j} + \frac{\partial}{\partial z} fg \vec{k} \\ \quad \nabla (fg) = \left ( f \frac{\partial g}{\partial x} + g \frac{\partial f}{\partial x} \right ) \vec{i} + \left ( f \frac{\partial g}{\partial y} + g \frac{\partial f}{\partial y} \right ) \vec{j} + \left ( f \frac{\partial g}{\partial z} + g \frac{\partial f}{\partial z} \right ) \vec{k} \\ \quad \nabla (fg) = f \nabla g + g \nabla f \end{align}

- Therefore by applying the Fundamental Theorem for Line Integrals we have that:

\begin{align} \quad \int_C f \nabla g \cdot d \vec{r} + \int_C g \nabla f \cdot d \vec{r} = \int_C \nabla (fg) \cdot d \vec{r} \\ \quad \int_C f \nabla g \cdot d \vec{r} + \int_C g \nabla f \cdot d \vec{r} = (fg)(Q) - (fg)(P) \\ \quad \int_C f \nabla g \cdot d \vec{r} + \int_C g \nabla f \cdot d \vec{r} = f(Q)g(Q) - f(P)g(P) \quad \blacksquare \end{align}