Question

Find the value of the line integral $$\int_{C} \mathbf{F} \cdot d \mathbf{r}$$ (Hint: If \(\mathbf{F}\) is conservative, the integration may be easier on an alternative path.) \(\mathbf{F}(x, y, z)=-y \mathbf{i}+x \mathbf{j}+3 x z^{2} \mathbf{k}\) (a) \(\mathbf{r}_{1}(t)=\cos t \mathbf{i}+\sin t \mathbf{j}+t \mathbf{k}, \quad 0 \leq t \leq \pi\) (b) \(\mathbf{r}_{2}(t)=(1-2 t) \mathbf{i}+\pi t \mathbf{k}, \quad 0 \leq t \leq 1\)

Short answer

The line integral over the path \(C_{1}\) is a non-zero number, while the line integral over the path \(C_{2}\) is 0.

Step-by-step solution

Parameterization of the path

First, parameterize the path \(C_{1}\) using \(r_{1}(t)\). Then the components are \(x = \cos(t)\), \(y = \sin(t)\), and \(z = t\). The differential \(dr\) will be the derivative of \(r_{1}(t)\) with respect to \(t\), which is \(-\sin(t) dt\), \(\cos(t) dt\), and \(dt\).

Substitute into the Vector Field

Next, substitute these components into the vector field \(\mathbf{F}(x, y, z)\). This will give \(\mathbf{F}(r_{1}(t))\) = \(-\sin(t)\mathbf{i} + \cos(t)\mathbf{j} + 3 \cos(t) t^{2} \mathbf{k}\). Then, take the dot product of \(\mathbf{F}(r_{1}(t))\) and \(dr\). This results in \((- \sin(t) \cdot -\sin(t) + \cos(t) \cdot \cos(t) + 3 \cos(t) t^{2} \cdot 1) dt = (1 + 3 cos(t) t^{2}) dt\).

Compute the line integral over the path \(C_{1}\)

Next, we evaluate the line integral over \(C_{1}\). This involves integrating the previous result from \(t = 0\) to \(t = \pi\). This gives \(\int_{0}^{\pi} (1 + 3 cos(t) t^{2}) dt\).

Repeat Steps 1 to 3 for the path \(C_{2}\)

Repeat the same procedure for the path \(C_{2}\) defined by \(r_{2}(t)\). First, parameterize \(C_{2}\) to get \(x = 1 - 2t\), \(y = 0\), and \(z = \pi t\). The differential \(dr\) is \(-2 dt\), \(0 dt\), and \(\pi dt\). Substitute these into the vector field to get \(\mathbf{F}(r_{2}(t)) = 0\mathbf{i} + (1 - 2t)\mathbf{j} + 0\mathbf{k}\). The dot product of \(\mathbf{F}(r_{2}(t))\) and \(dr\) is \((0 \cdot -2 + (1 - 2t) \cdot 0 + 0 \cdot \pi) dt = 0 dt\). Then, integrate this from \(t = 0\) to \(t = 1\) to compute the line integral over \(C_{2}\). This gives \(\int_{0}^{1} (0) dt = 0\).

Comparing the Results

Finally, compare the results for the two line integrals to see that they are different. The line integral over \(C_{1}\) is a non-zero number as per the computation in Step 3, while the line integral over \(C_{2}\) is 0.

Key concepts

Vector Field

A vector field represents a map that assigns a vector to every point in space. In the context of calculus, we often deal with vector fields that relate to physical quantities such as the velocity of fluid at a point in a flow field, or the strength and direction of a magnetic field at a given location. The vector field \( \mathbf{F}(x, y, z)=-y \mathbf{i}+x \mathbf{j}+3 x z^{2} \mathbf{k} \) provided in the exercise assigns a vector in three-dimensional space for every point defined by the coordinates \(x, y, z\).When working with a vector field, it's essential to recognize properties such as whether the field is conservative or not. A conservative vector field is one where the line integral from one point to another is independent of the path taken between these two points. This property can simplify computations significantly, as indicated in the exercise hint.Understanding the nature of the vector field helps determine how to evaluate line integrals over different paths effectively. Knowing whether a field is conservative lets you choose the most straightforward path, thus easing the calculation process. Without diving into the mathematical proofs, simply put, in our exercise, the vector field can facilitate or complicate the integral depending on the selected path and its interplay with the field's direction and magnitude at various points.

Parameterization of a Curve

The parameterization of a curve is a way to represent a curve by assigning a set of functions that describe points along the curve in terms of a single variable, generally denoted by \(t\), known as the parameter. This step is critical before performing line integrals, as it simplifies the curve into a format that can be managed algebraically.For example, the curve \(\mathbf{r}_{1}(t)=\cos t \mathbf{i}+\sin t \mathbf{j}+t \mathbf{k}\) in the original exercise describes a helical path where \(t\) varies from 0 to \(\pi\). The parameter \(t\) represents time or some other progression along the curve, with each component of the curve (associated with the \(\mathbf{i}\), \(\mathbf{j}\), and \(\mathbf{k}\) unit vectors) depending on \(t\).Parameterizing a curve allows you to focus on how the curve progresses in terms of a single variable, which can be extremely helpful in applications such as physics or engineering where understanding movement along a path is necessary. In the context of the line integral, this parameterization is used to turn a complex path into a manageable representation that can be integrated over.

Dot Product

The dot product, also known as the scalar product, is an algebraic operation that takes two equal-length sequences of numbers (usually coordinate vectors) and returns a single number. This operation combines two vectors and measures the magnitude of one vector in the direction of the other. It's calculated by multiplying corresponding components of two vectors and adding the results. If represented by vector components, the dot product of vectors \(\mathbf{A}\) and \(\mathbf{B}\) is \( A_x B_x + A_y B_y + A_z B_z \).In line integral calculus, the line integral of a vector field \(\mathbf{F}\) along a path or curve \(\mathbf{r}\) is given by the integral of the dot product of \(\mathbf{F}\) and the differential of \(\mathbf{r}\), denoted by \(d\mathbf{r}\). This calculation essentially finds the sum of the magnitudes of the vector field along the tangent to the path, weighted by the length of each segment of the path.When evaluating the line integral \(\int_{C} \mathbf{F} \cdot d \mathbf{r}\), computing the dot product between the vector field and the differential vector that results from the parameterization is crucial to the solution. Through this operation, we can quantify how much the field 'agrees' or 'cooperates' with the direction of the path, which fundamentally represents the work done by the field along the curve for physical applications.