Question

Scalar line integrals in \(\mathbb{R}^{3}\) Convert the line integral to an ordinary integral with respect to the parameter and evaluate it. $$\begin{aligned}&\int_{C}(x+y+z) d s ; C \text { is the circle } \mathbf{r}(t)=\langle 2 \cos t, 0,2 \sin t\rangle, \text { for }\\\&y \leq t \leq 2 \pi\end{aligned}.$$

Short answer

The value of the scalar line integral is 4.

Step-by-step solution

Calculate the derivative of the vector-valued function

To find \(\mathbf{r'}(t)\), we need to take the derivative of each component of the vector-valued function with respect to \(t\). We have: $$\mathbf{r'}(t) = \frac{d\mathbf{r}(t)}{dt} = \langle\frac{d}{dt}(2\cos t), \frac{d}{dt}(0), \frac{d}{dt}(2\sin t)\rangle = \langle-2\sin t, 0, 2\cos t\rangle.$$

Calculate the magnitude of the derivative

Next, we compute the magnitude of \(\mathbf{r'}(t)\): $$\|\mathbf{r'}(t)\| = \sqrt{(-2 \sin t)^2 + (0)^2 + (2 \cos t)^2} = \sqrt{4 \sin^2 t + 4 \cos^2 t} = \sqrt{4(\sin^2 t+ \cos^2 t)}=2.$$

Replace x, y, and z in the integrand

Now, we rewrite \(x+y+z\) using the expressions for \(x\), \(y\), and \(z\) in terms of the parameter \(t\): $$x+y+z=2\cos t + 0 + 2\sin t = 2\cos t + 2\sin t.$$

Multiply the integrand by the magnitude of the derivative

We have the following expression for the scalar line integral: $$\int_C (x+y+z) ds = \int_0^{2\pi} (2\cos t + 2\sin t)\|\mathbf{r'}(t)\| dt = \int_0^{2\pi} (2\cos t + 2\sin t)(2) dt = 4\int_0^{2\pi} (\cos t + \sin t) dt.$$

Integrate with respect to t

Now we can integrate the ordinary integral with respect to \(t\): $$4\int_0^{2\pi} (\cos t + \sin t) dt = 4\left[\sin t - \cos t\right]_0^{2\pi} = 4[(\sin 2\pi-\cos 2\pi)-(\sin 0-\cos 0)]= 4(0+1)=4.$$ Hence, the scalar line integral is equal to \(4\).

Key concepts

Line Integrals in R^3

Calculating a scalar line integral in three-dimensional space, or \(\mathbb{R}^{3}\), involves the summing of values along a curve. These values are usually functions of points on the curve, such as \(x+y+z\) in the given problem, multiplied by an infinitesimal element of arc length, denoted \(ds\). The process requires a parameterization of the curve, like the vector-valued function \(\mathbf{r}(t)\), which maps a parameter \(t\) to points in space. By converting the scalar line integral to an ordinary integral with respect to this parameter, it becomes straightforward to evaluate using standard techniques of integration.

Vector-valued Functions

A vector-valued function assigns vectors to elements in its domain, usually a subset of the real numbers. In our exercise, the given curve \(C\) is defined by the vector-valued function \(\mathbf{r}(t)=\langle 2 \cos t, 0,2 \sin t\rangle\), where \(t\) varies between 0 and \(2\pi\). This function describes a circle in the \(xz\)-plane, with a radius of 2 units. Understanding how to manipulate these functions is crucial for evaluating line integrals, as they provide the connection to the geometry of the curve.

Integration with Respect to a Parameter

When we integrate with respect to a parameter, we're making a variable substitution. In our example, \(t\) is the parameter that describes how the point on the curve moves as \(t\) changes. The original integral \(\int_{C}(x+y+z)ds\) is expressed with respect to the curve's length. However, by parameterizing with \(t\), we change the integral's limits to match the parameter's range and then integrate the new function of \(t\) with respect to \(t\). This simplifies the problem to a familiar one-dimensional integration task.

Calculating Magnitude of a Derivative

The magnitude of the derivative of a vector-valued function \(\mathbf{r'}(t)\) at a particular point gives us the rate of change of \(\mathbf{r}(t)\) with respect to \(t\), which geometrically corresponds to the speed of movement along the curve. Finding the magnitude involves taking the derivative vector's components, squaring each, summing them, and then taking the square root. In our circle example \(\|\mathbf{r'}(t)\| = 2\), which is constant, meaning the speed around the circle is uniform. This value multiplies the function defined along \(C\) to yield the integrand for the scalar line integral, making it possible to evaluate the integral with respect to \(t\).