Line Integrals Examples 2

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 more examples of computing line integrals.

Example 1

Evaluate the line integral $\int_C xyz \: ds$ where $C$ is the curve given by the parametric equations $x(t) = 2 \sin t$, $y(t) = t$, and $z = -2 \cos t$ for $0 ≤t ≤ \pi$.

We note that $f(x,y,z) = xyz$ and so $f(x(t), y(t), z(t)) = -4t \sin t \cos t$. Furthermore, $\frac{dx}{dt} = 2 \cos t$, $\frac{dy}{dt} = 1$ and $\frac{dz}{dt} = 2 \sin t$. Thus we have that:

(3)

\begin{align} \quad \int_C xyz \: ds = -4 \int_0^{\pi} t\sin t \cos t \sqrt{(2 \cos t)^2 + (1)^2 + (2 \sin t)^2} \: dt = -4\sqrt{5} \int_0^{\pi} t \sin t \cos t \: dt = -4 \sqrt{5} \left [ \frac{\sin 2t}{8} - \frac{t \cos (2t)}{4} \right ]_{t=0}^{t=\pi} = \pi \sqrt{5} \end{align}

Example 2

Evaluate the line integral $\int_C z \: ds$ where $C$ is the curve given by the parametric equations $x(t) = \rho \cos t \sin t$, $y(t) = \rho \sin^2 t$, and $z(t) = \rho \cos t$ for $0 ≤ t ≤ \frac{\pi}{2}$ and $\rho > 0$.

We note that $f(x, y, z) = z$, and so $f(x(t), y(t), z(t)) = \rho \cos t$. Furthermore, $\frac{dx}{dt} = (\rho \cos^2 - \rho \sin^2 t) = \rho \cos 2t$, $\frac{dy}{dt} = 2 \rho \sin t \cos t = \rho \sin 2t$, and $\frac{dz}{dt} = -\rho \sin t$. Thus we have that:

(4)

\begin{align} \quad \int_C z \: ds = \rho \int_0^{\pi/2} \cos t \sqrt{(\rho \cos 2t)^2 + (\rho \sin 2t)^2 + (-\rho \sin t)^2} \: dt = \rho \int_0^{\pi/2} \cos t \sqrt{\rho^2 + \rho^2 \sin^2 t} = \rho^2 \int_0^{\pi/2} \cos t \sqrt{1 + \sin^2 t} \end{align}

Now to evaluate this integral, we will need to use a few techniques of integration. We'll start with substitution. Let $u = \sin t$ so that $du = \cos t$. Then the limits of integration become $0$ and $1$ and so:

(5)

\begin{align} \quad = \rho^2 \int_{0}^{1} \sqrt{1 + u^2} \: du \end{align}

Now to evaluate this integral, we will need to use trigonometric substitution. Let $u = \tan \theta$ so then $du = \sec^2 \theta \: d \theta$. The limits of integration become $0$ and $\frac{\pi}{4}$ and so:

(6)

\begin{align} \quad = \rho^2 \int_0^{\pi / 4} \sqrt{1 + (\tan \theta)^2} \sec^2 \theta \: d \theta = \rho^2 \int_0^{\pi / 4} \sqrt{\sec^2 \theta} \sec^2 \theta \: d \theta = \rho^2 \int_0^{\pi / 4} \sec^3 \theta \: d \theta \end{align}

Now this integral can be evaluated using integration by parts to get that $\int \sec^3 \theta \: d \theta = \frac{1}{2} \sec \theta \tan \theta + \frac{1}{2} \ln \mid \sec \theta + \tan \theta \mid$ and so:

(7)

\begin{align} \quad = \rho^2 \left [ \frac{1}{2} \sec \theta \tan \theta + \frac{1}{2} \ln \mid \sec \theta + \tan \theta \mid \right ]_{\theta = 0}^{\theta = \pi / 4} = \rho^2 \left (\frac{\sqrt{2}}{2} + \frac{1}{2} \ln \mid \sqrt{2} + 1 \mid \right ) = \frac{\rho^2 \sqrt{2} + \rho^2 \ln ( \sqrt{2} + 1)}{2} \end{align}