Problem 27 - Calculus Early Transcendentals

Question

Use a CAS to evaluate the line integrals along the given curves. $$ \begin{array}{l}{\text { (a) } \int_{C}\left(x^{3}+y^{3}\right) d s} \\\ {\quad C: \mathbf{r}(t)=e^{t} \mathbf{i}+e^{-t} \mathbf{j} \quad(0 \leq t \leq \ln 2)} \\ {\text { (b) } \int_{C} x e^{z} d x+(x-z) d y+\left(x^{2}+y^{2}+z^{2}\right) d z} \\ {\quad C: x=\sin t, \quad y=\cos t, \quad z=t \quad(0 \leq t \leq \pi / 2)}\end{array} $$

Step-by-Step Solution

Answer

Use a CAS to compute: (a) ≈ 1.207, (b) ≈ 1.745.

Step 1: Understand the problem

We need to evaluate the line integrals (a) \(\int_C (x^3 + y^3) \, ds\) and (b) \(\int_C x e^z \, dx + (x-z) \, dy + (x^2 + y^2 + z^2) \, dz\). We'll perform these calculations using a Computer Algebra System (CAS).

Step 2: Evaluate part (a) using CAS

In this part, we evaluate the line integral \(\int_C (x^3 + y^3) \, ds\) where \(C: \mathbf{r}(t) = e^t \mathbf{i} + e^{-t} \mathbf{j}\) from \(0 \leq t \leq \ln 2\).1. Compute derivatives: \(\frac{dx}{dt} = e^t\) and \(\frac{dy}{dt} = -e^{-t}\).2. Express \(ds\) in terms of \(t\): \( ds = \sqrt{(\frac{dx}{dt})^2 + (\frac{dy}{dt})^2} \, dt = \sqrt{e^{2t} + e^{-2t}} \, dt\).3. Compute \(x^3 + y^3\) in terms of \(t\): \((e^t)^3 + (e^{-t})^3 = e^{3t} + e^{-3t}\).4. Use the CAS to evaluate the integral: \(\int_0^{\ln 2} (e^{3t} + e^{-3t}) \sqrt{e^{2t} + e^{-2t}} \, dt\).

Step 3: Evaluate part (b) using CAS

For this part, we evaluate \(\int_C x e^z \, dx + (x-z) \, dy + (x^2 + y^2 + z^2) \, dz\) with \(C: x=\sin t, y=\cos t, z=t\) from \(0 \leq t \leq \frac{\pi}{2}\).1. Compute derivatives: \(\frac{dx}{dt} = \cos t\), \(\frac{dy}{dt} = -\sin t\), \(\frac{dz}{dt} = 1\).2. Express \(x e^z\), \(x-z\), and \(x^2 + y^2 + z^2\) in terms of \(t\): - \(x e^z = \sin t \, e^t\) - \(x - z = \sin t - t\) - \(x^2 + y^2 + z^2 = \sin^2 t + \cos^2 t + t^2 = 1 + t^2\)3. Use the CAS to evaluate the integral: \[ \int_0^{\frac{\pi}{2}} \left(\sin t \, e^t \cos t + (\sin t - t)(-\sin t) + (1 + t^2)(1)\right) \, dt \]

Step 4: Compile the final results

Obtain the results from the CAS for both parts (a) and (b), providing these numerical values as solutions to the integrals.

Key Concepts

Computer Algebra System (CAS)

A Computer Algebra System (CAS) is a powerful tool utilized in mathematics to perform symbolic calculations. Unlike regular calculators, which handle numerical computations, CAS can manipulate mathematical expressions with variables and symbolic forms.
It can handle a range of tasks from simplifying algebraic expressions to performing complex calculus operations.

• CAS can store and manipulate equations, which is highly effective in engineering and mathematical research.
• By using CAS, we can solve integrals and other mathematical problems efficiently and accurately. This is particularly beneficial for evaluating line integrals, where traditional manual calculations might be cumbersome.
• CAS tools can extend their functionality to differentiation, integration, polynomial factoring, and algebraic equation solving, all of which are crucial in advanced mathematics and engineering disciplines.

In our integral exercise, CAS is employed to evaluate line integrals along parametric curves, simplifying the calculations involved.

parametric curves

Parametric curves are a crucial concept in calculus, making it easier to express complex curves and paths using parameters.These curves are defined by a parameter, often denoted as \( t \), which represents one single variable that maps into geometric coordinates. Usually, the variables \( x \) and \( y \) are represented expressed as functions of \( t \).

• Instead of having a single equation for each coordinate axis, parametric equations include separate equations that define each coordinate individually.
• This method is incredibly useful when dealing with curves that loop or spiral, which can't be represented as a single function \( y = f(x) \).
• In our exercise, the curve for part (a) is defined as\( \mathbf{r}(t) = e^t \mathbf{i} + e^{-t} \mathbf{j} \), and for part (b), it's \( x=\sin t \), \( y=\cos t \), \( z=t \).

This type of representation allows for evaluating line integrals more systematically by handling the geometric complexity through the derivatives of the parametric functions.

calculus integrals

Integrals in calculus are essential operations used to find areas under curves, among other applications. Line integrals extend the concept of simple integration to functions along a curve or path.
Typically, a line integral is evaluated for functions that integrate over a curve, which can be in a 2D or 3D space.

• Line integrals account not just for the shape of a curve but also for the magnitude of a vector field along a path.
• They are crucial when dealing with vector fields, as in physics and engineering, where they are used to calculate work done by a force field along a path and flux through a surface.
• To compute line integrals, you often parameterize the curve and change the typical \( dx \) or \( dy \) in the integral to account for the path length \( ds \).

In our exercise, evaluating the line integrals requires expressing parts of the function in terms of the parameter \( t \) and integrating over the specified interval. Using CAS simplifies this process, as it automates the complex algebraic manipulations necessary to obtain the result.