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

(7)