Problem 5 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-6, evaluate the iterated integrals. $$ \int_{0}^{\pi} \int_{0}^{2} r \cos \frac{\theta}{4} d r d \theta $$

Step-by-Step Solution

Answer

The iterated integral evaluates to \( 4\sqrt{2} \).

Step 1: Identify the Order

We are given the iterated integral \( \int_{0}^{\pi} \int_{0}^{2} r \cos \frac{\theta}{4} \, d r \, d \theta \). The order of integration is first with respect to \( r \) and then with respect to \( \theta \). Our task is to integrate this step-by-step.

Step 2: Integrate with Respect to \( r \)

Integrate \( r \cos \frac{\theta}{4} \) with respect to \( r \). Treat \( \cos \frac{\theta}{4} \) as a constant while integrating. The integral becomes: \[ \int_{0}^{2} r \cos \frac{\theta}{4} \, dr = \left. \frac{r^2}{2} \cos \frac{\theta}{4} \right|_{0}^{2} \] Evaluating this gives: \[ \frac{4}{2} \cos \frac{\theta}{4} - 0 = 2 \cos \frac{\theta}{4} \]

Step 3: Integrate with Respect to \( \theta \)

Now integrate \( 2 \cos \frac{\theta}{4} \) with respect to \( \theta \) from \( 0 \) to \( \pi \): \[ \int_{0}^{\pi} 2 \cos \frac{\theta}{4} \, d \theta \] To solve, use the substitution \( u = \frac{\theta}{4} \), hence \( d\theta = 4 \, du \). The limits change accordingly: when \( \theta = 0, u = 0 \) and when \( \theta = \pi, u = \frac{\pi}{4} \). The integral becomes: \[ 2 \times 4 \times \int_{0}^{\frac{\pi}{4}} \cos u \, du = 8 \left[ \sin u \right]_{0}^{\frac{\pi}{4}} \] Evaluating this gives: \[ 8 (\sin \frac{\pi}{4} - \sin 0) = 8 \times \frac{\sqrt{2}}{2} = 4\sqrt{2} \]

Step 4: Final Answer

The iterated integral evaluates to \( 4\sqrt{2} \). This is the final result after integrating with respect to both \( r \) and \( \theta \).

Key Concepts

Integration Techniques

When tackling iterated integrals, such as \( \int_{0}^{\pi} \int_{0}^{2} r \cos \frac{\theta}{4} \, dr \, d\theta \), applying effective integration techniques is crucial. The integration is performed stepwise, respecting the order defined by the iteration.
Start by treating the integral with respect to \( r \). In this scenario, \( \cos \frac{\theta}{4} \) is a constant because its value depends solely on \( \theta \). By isolating one variable at a time, you simplify the process.
During integration, focus on the step-by-step:

• Set up: Identify the order and treat components not dependent on the current variable as constants.
• Calculate: Integrate the inner integral first.
• Substitute: Once done, plug results into the outer integral.

In the given problem, calculating \( \int_{0}^{2} r \cos \frac{\theta}{4} \, dr \) yields \( 2 \cos \frac{\theta}{4} \). This simplification then feeds into the next round of integration with respect to \( \theta \). Every step needs careful execution, and results from each phase of integration are pivotal for the proceeding steps.
Integration techniques for iterated integrals revolve around managing one variable at a time. By maintaining a single-variable focus, you simplify complex expressions into manageable chunks.

Double Integrals

Double integrals extend the concept of single-variable integrals to functions of two variables. They are crucial in evaluating the volume under a surface within a specific domain defined by two variables, often expressed in the form \( \int \int f(x,y) \, dx \, dy \).
Projects like finding areas, volumes, or centroids frequently deploy double integrals. The process involves integrating one variable while treating the other as constant, then switching roles.

• Begin with the innermost integral while holding outer variable constants.
• Upon solving, apply results to the outer integral. This turns the problem into a layered process.

In the exercise given:
The dynamics of \( r \) and \( \theta \) highlight this two-step integration. Due to the geometry of polar coordinates, \( r \) delineates radial distance while \( \theta \) refers to angular positioning. Double integrals compute the accumulation over areas defined in this two-variable space.
They allow flexibility in choosing the order of integration:
Sometimes it's beneficial to switch the roles, but it's practical in this setup to retain the provided sequence. The polar nature indicates iterating first through radial components is often straightforward, simplifying further calculations involving angles.

Calculus Problems

Solving calculus problems with iterated integrals requires an understanding of specific techniques. These problems can initially appear complex, yet adhere to consistent standards that, once mastered, demystify involved processes.
Analyzing \( \int_{0}^{\pi} \int_{0}^{2} r \cos \frac{\theta}{4} \, dr \, d\theta \):

• Deconstruction: Break down each component. Understand how every piece contributes and functions within the integral.
• Sequence: Follow the defined order of operations, ensuring systematic treatment of each stage.
• Result Confirmation: Cross-referencing calculated results at each step to ensure coherence and correctness.

This example utilizes techniques such as substitution for efficient calculation when integrating with respect to \( \theta \). The switched variable (from \( \theta \) to \( u \)) simplifies trigonometric expressions, easing the integration path.
The answer, ultimately, concludes at \( 4\sqrt{2} \), sharing insight into the transformation of space via integration. Problems like these underscore the precision required by calculus, while reiterating the accessibility of solutions through methodical practices. Understanding involves digesting each individual layer progressively leading to a holistic solution.