Line Integrals with Respect to Specific Variables Examples 1

# Line Integrals with Respect to Specific Variables Examples 1

Recall from the Line Integrals with Respect to Specific Variables page that if $z = f(x, y)$ is a two variable real-valued function and $C$ is a smooth plane curve parameterized as $x = x(t)$ and $y = y(t)$ for $a ≤ t ≤ b$ then the line integral of $f$ along $C$ with respect to $x$ and the line integral of $f$ along $C$ with respect to $y$ are:

(1)
\begin{align} \quad \int_C f(x, y) \: dx = \int_a^b f(x(t), y(t)) x'(t) \: dt \quad \quad \int_C f(x, y) \: dy = \int_a^b f(x(t), y(t)) y'(t) \: dt \end{align}

When these integrals occur together, we also use the shortened notation:

(2)
\begin{align} \quad \int_C P(x, y) \: dx + Q(x, y) \: dy = \int_C P(x, y) \: dx + \int_C Q(x, y) \: dy \end{align}

The same notation and definitions are used for line integrals with respect to $x$, $y$, or $z$ of the three variable real-valued function $w = f(x, y, z)$.

Let's now look at some examples of computing these sort of integrals.

## Example 1

Evaluate $\int_C z \: dx + x \: dy + y \: dz$ where $C$ is the curve given parametrically by $x = t^2$, $y = t^3$, and $z = t^2$ for $0 ≤ t ≤ 1$.

We note that $x'(t) = 2t$, $y'(t) = 3t^2$ and $z'(t) = 2t$, and so, using the formula directly and we have that:

(3)
\begin{align} \quad \int_C z \: dx + x \: dy + y \: dz = \int_C z \: dx + \int_C x \: dy + \int_C y \: dz \\ \quad \int_C z \: dx + x \: dy + y \: dz = \int_0^1 t^2 (2t) \: dt + \int_0^1 t^2 (3t^2) \: dt + \int_0^1 t^3 (2t) \: dt \\ \quad \int_C z \: dx + x \: dy + y \: dz = 2\int_0^1 t^3 \: dt + 5 \int_0^1 t^4 \: dt\\ \quad \int_C z \: dx + x \: dy + y \: dz = 2 \left [ \frac{t^4}{4} \right ]_{t=0}^{t=1} + 5 \left [ \frac{t^5}{5} \right ]_{t=0}^{t=1} \\ \quad \int_C z \: dx + x \: dy + y \: dz = \frac{3}{2} \end{align}

## Example 2

Evaluate $\int_C z^2 \: dx + x^2 \: dy + y^2 \: dz$ where $C$ is the line segment that joins the points $(1, 0, 0)$ and $(4, 1, 2)$.

We need to first parameterize this line segment. Fortunately, this is easy to do. For $0 ≤ t ≤ 1$ we will have that:

(4)
\begin{align} \quad \vec{r}(t) = (1 - t)(1, 0, 0) + t(4, 1, 2) \\ \quad \vec{r}(t) = (1 - t, 0, 0) + (4t, t, 2t) \\ \quad \vec{r}(t) = (1 + 3t, t, 2t) \end{align}

Thus we have that $x(t) = 1 + 3t$, $y(t) = t$ and $z(t) = 2t$. Therefore:

(5)
\begin{align} \quad \int_C z^2 \: dx + x^2 \: dy + y^2 \: dz = \int_C z^2 \: dx + \int_C x^2 \: dy + \int_C y^2 \: dz \\ \quad \int_C z^2 \: dx + x^2 \: dy + y^2 \: dz = \int_0^1 (2t)^2 (1 + 3t)' \: dt + \int_0^1 (1 + 3t)^2 (t)' \: dt + \int_0^1 (t)^2 (2t)' \: dt \\ \quad \int_C z^2 \: dx + x^2 \: dy + y^2 \: dz = \int_0^1 (4t^2) (3) \: dt + \int_0^1 (1 + 3t)^2 (1) \: dt + \int_0^1 t^2 (2) \: dt \\ \quad \int_C z^2 \: dx + x^2 \: dy + y^2 \: dz = 14\int_0^1 t^2 \: dt + \int_0^1 (1 + 3t)^2 \: dt \\ \quad \int_C z^2 \: dx + x^2 \: dy + y^2 \: dz = 14 \left [ \frac{t^3}{3} \right ]_{t=0}^{t=1} + \left [ \frac{(1 + 3t)^3}{9} \right ]_{t=0}^{t=1} \\ \quad \int_C z^2 \: dx + x^2 \: dy + y^2 \: dz = \frac{14}{3} + \frac{64}{9} - \frac{1}{9} \\ \quad \int_C z^2 \: dx + x^2 \: dy + y^2 \: dz = \frac{42}{9} + \frac{63}{9} \\ \quad \int_C z^2 \: dx + x^2 \: dy + y^2 \: dz = \frac{105}{9} \\ \quad \int_C z^2 \: dx + x^2 \: dy + y^2 \: dz = \frac{35}{3} \end{align}