Problem 4 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate the iterated integrals. \(\int_{0}^{\pi} \int_{0}^{1-\cos \theta} r \sin \theta d r d \theta\)

Step-by-Step Solution

Answer

\(\frac{4}{3}\)

Step 1: Identify the Limits of Integration

We begin by identifying the limits of the integrals: the integral with respect to \(r\) runs from \(0\) to \(1 - \cos \theta\) and the integral with respect to \(\theta\) runs from \(0\) to \(\pi\).

Step 2: Evaluate the Inner Integral

Next, we integrate with respect to \(r\). The integrand is \(r \sin \theta\). The integral of \(r\) with respect to \(r\) is \(\frac{1}{2}r^2\), so the indefinite integral is \[\int r \sin \theta \, dr = \frac{1}{2} r^2 \sin \theta\]. Evaluating this from \(0\) to \(1 - \cos \theta\), we get:\[\left. \frac{1}{2}r^2 \sin \theta \right|_0^{1-\cos \theta} = \frac{1}{2}((1-\cos \theta)^2) \sin \theta\].

Step 3: Simplify the Expression

We simplify \((1-\cos \theta)^2\) to get \(1 - 2\cos \theta + \cos^2 \theta\). So, the inner integral becomes:\[\frac{1}{2}(1 - 2\cos \theta + \cos^2 \theta) \sin \theta\].

Step 4: Evaluate the Outer Integral

Integrate the expression with respect to \(\theta\) from \(0\) to \(\pi\):\[\int_{0}^{\pi} \frac{1}{2} (1 - 2\cos \theta + \cos^2 \theta) \sin \theta \, d\theta\].Separate into three integrals:\[\frac{1}{2} \left( \int_{0}^{\pi} \sin \theta \, d\theta - 2 \int_{0}^{\pi} \cos \theta \sin \theta \, d\theta + \int_{0}^{\pi} \cos^2 \theta \sin \theta \, d\theta \right)\].

Step 5: Solve Each Integral

Evaluate each of the three integrals separately:1. \(\int_0^{\pi} \sin \theta \, d\theta = -\cos \theta |_{0}^{\pi} = 2\).2. \(-2\int_0^{\pi} \cos \theta \sin \theta \, d\theta\): Using \(2 \sin \theta \cos \theta = \sin(2\theta)\), this is \(\frac{-1}{2}\int_0^{2\pi} \sin u \, du = 0\).3. \(\int_0^{\pi} \cos^2 \theta \sin \theta \, d\theta\), let \( u = \cos \theta \), then \( du = -\sin \theta \, d\theta\), this becomes \( \int_{1}^{-1} -u^2 \, du = \int_{-1}^{1} u^2 \, du\).Which evaluates to \(\left. \frac{u^3}{3} \right|_{-1}^{1} = \frac{1}{3}(1 - (-1)) = \frac{2}{3}\).

Step 6: Combine Results for Final Answer

Now, substitute the results back into the broken-down integral:\[\frac{1}{2} \left( 2 + 0 + \frac{2}{3} \right) = \frac{1}{2} \cdot \frac{8}{3} = \frac{4}{3}\].

Step 7: Final Answer

The result of evaluating the iterated integrals is \(\frac{4}{3}\).

Key Concepts

Limits of Integration

Understanding the limits of integration is a foundational step in evaluating iterated integrals. These limits outline the boundaries within which the integration is performed.
In our exercise, the iterated integral \[ \int_{0}^{\pi} \int_{0}^{1-\cos \theta} r \sin \theta \, d r \, d \theta \] features two limits:

• The inner integral's limits are from \( r = 0 \) to \( r = 1 - \cos \theta \). This means the integration with respect to \( r \) takes place over a range determined by \( \theta \).
• The outer integral's limits are from \( \theta = 0 \) to \( \theta = \pi \). It shows that \( \theta \) varies in this range while evaluating the entire integral.

Identifying these correctly helps set the stage for solving the problem. Always ensure your limits match the given problem context and boundaries to achieve accurate results.

Inner Integral

The concept of the inner integral involves focusing on integrating with respect to one variable while keeping others constant. In our exercise, the inner integral is \[ \int_{0}^{1-\cos \theta} r \sin \theta \, d r \].
Let's break this down:

• The integrand is \( r \sin \theta \), and \( \sin \theta \) is considered unchanged at this stage because it's a function of \( \theta \), not \( r \).
• The integral of \( r \) with respect to \( r \) is \( \frac{1}{2}r^2 \), resulting in the expression \( \frac{1}{2} r^2 \sin \theta \).
• After integrating, you substitute the limits from \( 0 \) to \( 1-\cos \theta \) to get \[ \frac{1}{2}((1-\cos \theta)^2) \sin \theta \].

Completing the inner integration successfully provides the function that will be used in the outer integration.

Outer Integral

With the inner integral solved, the resulting expression becomes the input for the outer integral. In our case, we move to \[ \int_{0}^{\pi} \frac{1}{2} (1 - 2\cos \theta + \cos^2 \theta) \sin \theta \, d \theta \].
Important points here are:

• Breaking down the expression makes it easier to manage. It's separated into three parts: \( \frac{1}{2} \int_{0}^{\pi} \sin \theta \, d\theta \), \( - \frac{1}{2} \cdot 2 \int_{0}^{\pi} \cos \theta \sin \theta \, d\theta \), and \( \frac{1}{2} \int_{0}^{\pi} \cos^2 \theta \sin \theta \, d\theta \).
• Each part is treated as a separate integral, allowing for easier calculation and combination at the end.

Evaluating these integrals step-by-step allows you to accurately solve the entire expression, ensuring you handle trigonometric identities and substitutions where necessary.

Trigonometric Integration

Trigonometric integration plays a crucial role when dealing with functions involving sine and cosine. Each sub-integral in the exercise makes use of different techniques:

• For \( \int_{0}^{\pi} \sin \theta \, d\theta \), the basic antiderivative is \( -\cos \theta \), which simplifies quickly.
• The integral \( -2 \int_{0}^{\pi} \cos \theta \sin \theta \, d\theta \) utilizes the identity \( 2 \sin \theta \cos \theta = \sin(2\theta) \). Using a trigonometric substitution or identity here transforms the integration process and simplifies the solution.
• Lastly, when working with \( \int_{0}^{\pi} \cos^2 \theta \sin \theta \, d\theta \), a substitution such as \( u = \cos \theta \) is used. Here, \( du = -\sin \theta \, d\theta \), changing the problem to a familiar polynomial form.

Mastering these trigonometric techniques can significantly ease solving integrals involving sinusoids.