Line Integrals Examples 1

# Line Integrals Examples 1

Recall from the Line Integrals page that if $z = f(x, y)$ is a two variable real-valued function and $C$ is a smooth plane curve defined by the parametric equations $x = x(t)$ and $y = y(t)$ then the line integral of $f$ along $C$ is given by:

(1)
\begin{align} \quad \int_C f(x, y) \: ds = \int_a^b f(x(t), y(t)) \sqrt{\left ( \frac{dx}{dt} \right )^2 + \left ( \frac{dy}{dt} \right )^2} \: dt = \int_a^b f(\vec{r}(t)) \| \vec{r''}(t) \| \: dt \end{align}

Similarly if $z = f(x, y, z)$ is a three variable real-valued function and $C$ is a smooth space curve defined by the parametric equations $x = x(t)$, $y = y(t)$ and $z = z(t)$ then the line integral of $f$ along $C$ is given by:

(2)
\begin{align} \quad \int_C f(x, y, z) \: ds = \int_a^b f(x(t), y(t), z(t)) \sqrt{\left ( \frac{dx}{dt} \right )^2 + \left ( \frac{dy}{dt} \right)^2 + \left ( \frac{dz}{dt} \right )^2} \: dt = \int_a^b f(\vec{r}(t)) \| \vec{r'}(t) \| \: dt \end{align}

We will now look at some examples of computing line integrals.

## Example 1

Evaluate the line integral $\int_C x - y \: ds$ where $C$ is the line segment from the point $(1, 3)$ to $(5, -2)$.

In order to evaluate this line integral, we will need to parameterize this line segment. We can parameterize this line segment as $\vec{r}(t) = (1 - t)(1, 3) + t(5, -2)$ which yields the parametric equations $x(t) = 1 + 4t$ and $y(t) = 3 - 5t$ for $0 ≤ t ≤ 1$. Therefore $\frac{dx}{dt} = 4$ and $\frac{dy}{dt} = -5$.

Furthermore we have that $f(x(t), y(t)) = (1 + 4t) - (3 - 5t) = -2 + 9t$. So:

(3)
\begin{align} \quad \int_C x - y \: ds = \int_0^1 (-2 + 9t) \sqrt{(4)^2 + (-5)^2} \: dt = \int_0^1 (-2 + 9t) \sqrt{41} \: dt = \sqrt{41} \left [ -2t + \frac{9t^2}{2}\right]_0^1 = \frac{5 \sqrt{41}}{2} \end{align}

## Example 2

Evaluate the line integral $\int_C y^3 \: ds$ where $C$ is the curve given by the parametric equations $x(t) = t^3$ and $y(t) = t$ for $0 ≤ t ≤ 2$.

We first note that $f(x, y) = y^3$ and so $f(x, y) = f(x(t), y(t)) = f(t^3, t) = t^3$. Also note that $\frac{dx}{dt} = 3t^2$ and $\frac{dy}{dt} = 1$. Applying the formula for evaluating line integrals for two variable functions and we get that:

(4)
\begin{align} \quad \int_C y^3 \: ds = \int_0^2 t^3 \sqrt{(3t^2)^2 + (1)^2} \: dt = \int_0^2 t^3 \sqrt{9t^4 + 1} \: dt \end{align}

To solve this integral, we will need to use substitution. Let $u = 9t^4 + 1$. Then $du = 36t^3 \: dt$ and $\frac{1}{36} du = t^3 \: dt$, and thus for some $\alpha, \beta \in \mathbb{R}$ we have that:

(5)
\begin{align} \quad = \frac{1}{36} \int u^{1/2} \: du = \frac{1}{54} [u^{3/2}]_{u=\alpha}^{u=\beta} = \frac{1}{54} [(9t^4 + 1)^{3/2}]_{t=0}^{t=2} = \frac{145^{3/2} - 1}{54} \end{align}

## Example 3

Evaluate the line integral $\int_C y \: ds$ where $C$ is the upper half semicircle with radius $2$ and centered at the origin.

We first need to parameterize $C$. We note the parametric equations $x(t) = 2 \cos t$ and $y(t) = 2 \sin t$ for $0 ≤ t ≤ \pi$ parameterize this curve. We can verify this since:

(6)
\begin{align} \quad x^2 + y^2 = (2\cos t)^2 + (2 \sin t)^2 = 4\cos^2 t + 4 \sin^2 t = 4 \end{align}

Now $f(x(t), y(t)) = 2 \sin t$, $\frac{dx}{dt} = 2 \sin t$ and $\frac{dy}{dt} = -2 \cos t$ so:

(7)
\begin{align} \quad \int_C y \: ds = \int_0^{\pi} 2 \sin t \sqrt{(2 \sin t)^2 + (-2 \cos t)^2} \: dt = 2 \int_0^{\pi} \sin t \sqrt{4 \sin^2 t + 4 \cos^2 t} \: dt = 4 \int_0^{\pi} \sin t \: dt = 4 \left [ - \cos t \right]_{t=0}^{t=\pi} = 4 (2) = 8 \end{align}