Question

Scalar line integrals in \(\mathbb{R}^{3}\) Convert the line integral to an ordinary integral with respect to the parameter and evaluate it. $$\int_{C} x e^{y z} d s ; C \text { is } \mathbf{r}(t)=\langle t, 2 t,-4 t\rangle, \text { for } 1 \leq t \leq 2.$$

Short answer

Question: Evaluate the scalar line integral $$\int_{C} x e^{y z} ds$$, where C is given by \(\mathbf{r}(t) = \langle t, 2t, -4t \rangle\) for \(1 \leq t \leq 2\). Answer: The value of the scalar line integral is $$-\frac{1}{16}\left[ e^{-32} - e^{-8} \right]$$

Step-by-step solution

Parameterize the curve C

The curve C is given by \(\mathbf{r}(t) = \langle t, 2t, -4t \rangle\) for \(1 \leq t \leq 2\). So, x = t, y = 2t, and z = -4t.

Calculate the differential, ds

To evaluate the scalar line integral, we first need to calculate the differential ds. We'll use the following expression for ds: \(\| \mathbf{r}'(t) \| dt\), where \(\mathbf{r}'(t)\) is the derivative of the parameterization vector function. We have: \(\mathbf{r}'(t) = \langle 1, 2, -4 \rangle\) \(\| \mathbf{r}'(t) \| = \sqrt{1^2 + 2^2 + (-4)^2} = \sqrt{1 + 4 + 16} = \sqrt{21}\) Therefore, \(ds = \sqrt{21} dt\).

Substitute the parameterization into the line integral

Rewrite the given line integral in terms of t: $$\int_{C} x e^{y z} ds = \int_{1}^{2} t e^{(2t)(-4t) } \sqrt{21} dt$$

Evaluate the integral

Now we can evaluate the integral: $$\int_{1}^{2} t e^{(2t)(-4t) } \sqrt{21} dt = \sqrt{21} \int_{1}^{2} t e^{-8t^2} dt$$ To solve this integral, we use substitution method: Let \(u = -8t^2\). Then, \(\frac{du}{dt} = -16t\), or \(\frac{dt}{\sqrt{21}} = -\frac{1}{16} du\). Now the integral becomes: $$-\frac{1}{16} \int_{-8}^{-32} e^u du$$ Integrating with respect to u gives: $$-\frac{1}{16} \left[ e^{-32} - e^{-8} \right]$$ So, the value of the scalar line integral is: $$\int_{C} x e^{y z} ds = -\frac{1}{16}\left[ e^{-32} - e^{-8} \right]$$

Key concepts

Parameterization

When dealing with scalar line integrals, the first step is to parameterize the curve over which you will integrate. Parameterization involves expressing the curve's path using a variable, typically denoted as "t," that changes as you move along the curve. This method allows you to transform a complex geometric path into a manageable algebraic form.

Given the exercise, the curve \( C \) is defined with the vector function \( \mathbf{r}(t) = \langle t, 2t, -4t \rangle \). This indicates:

• The \( x \)-coordinate is parameterized as \( x = t \).
• The \( y \)-coordinate as \( y = 2t \).
• The \( z \)-coordinate as \( z = -4t \).

This parameterization turns the curve into a straight line in \( \mathbb{R}^3 \), stretching from when \( t = 1 \) to \( t = 2 \). This simplification makes it easier to work with the line integral.

Differential Calculation

Calculating the differential \( ds \) is a crucial part of evaluating a line integral. This represents a small piece of the path and is derived from the rate of change or the derivative of the parameterization. By finding the magnitude of this derivative, you measure how long the curve stretches over small intervals.

For the curve \( C \), the derivative of the vector function is \( \mathbf{r}'(t) = \langle 1, 2, -4 \rangle \). To find \( ds \), calculate the magnitude:

• Compute \( \| \mathbf{r}'(t) \| = \sqrt{1^2 + 2^2 + (-4)^2} = \sqrt{21} \).
• Therefore, the differential amount of the path is represented as \( ds = \sqrt{21} \, dt \).

This magnitude gives a "scale" to the integration process, aligning with the spatial dimensions of the curve.

Substitution Method

The substitution method is a technique used to simplify the process of integration, especially when dealing with complex expressions. By substituting variables, you transform the integral into a more recognizable form.

To evaluate the given integral, substitute \( u = -8t^2 \), which simplifies the exponential function. This choice helps:

• Reconfigure the integral, \( t e^{-8t^2} \), into an expression of \( u \).
• Relate \( dt \) to \( du \) using \( \frac{du}{dt} = -16t \), providing \( dt = -\frac{1}{16t} du \).

This one-to-one correspondence between t and u allows you to integrate using known methods, streamlining what would otherwise be a cumbersome calculation. This step significantly reduces complexity and makes solving the integral manageable.

Integration

Finally, after setting up the integral with proper substitutions and differential calculations, it's time to perform the actual integration. This process connects the math with real-world accumulations and total changes.

The transformed integral becomes \( -\frac{1}{16} \int_{-8}^{-32} e^u \, du \). The steps are as follows:

• Recognize that \( \int e^u \, du = e^u \): the antiderivative of the exponential function.
• Apply the limits of integration to find \( -\frac{1}{16} \left[ e^{-32} - e^{-8} \right] \).

The solution provides the cumulative effect or total scalar line integral along the curve, reflecting the complete journey from \( t = 1 \) to \( t = 2 \). This process underscores how integral calculus evaluates the summation over complex paths.