Problem 15 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate each of the iterated integrals. $$ \int_{0}^{\pi} \int_{0}^{3} y \cos ^{2} x d y d x $$

Step-by-Step Solution

Answer

The value of the iterated integral is \(\frac{9\pi}{4}\).

Step 1: Evaluate the Inner Integral

Begin by evaluating the inner integral with respect to \(y\): \[ \int_{0}^{3} y \, d y \]Since \(y\) is the variable of integration and \(\cos^2 x\) acts as a constant with respect to \(y\), the integral simplifies to:\[ \int_{0}^{3} y \, d y = \left[ \frac{1}{2} y^2 \right]_{0}^{3} = \frac{1}{2} (3^2) - \frac{1}{2} (0^2) = \frac{9}{2} \]Thus, the inner integral evaluates to \(\frac{9}{2}\).

Step 2: Formulate the Outer Integral

Use the result from Step 1 to formulate the outer integral. The problem now simplifies to evaluating:\[ \int_{0}^{\pi} \cos^2 x \cdot \frac{9}{2} \, d x \]Factor the constant \(\frac{9}{2}\) out of the integral:\[ \frac{9}{2} \int_{0}^{\pi} \cos^2 x \, d x \]

Step 3: Use Trigonometric Identity

Recall that \(\cos^2 x\) can be rewritten using the trigonometric identity:\[ \cos^2 x = \frac{1 + \cos(2x)}{2} \]Substitute this identity into the outer integral:\[ \frac{9}{2} \int_{0}^{\pi} \frac{1 + \cos(2x)}{2} \, d x \]This simplifies to:\[ \frac{9}{4} \int_{0}^{\pi} (1 + \cos(2x)) \, d x \]

Step 4: Evaluate the Simplified Integral

Split the integral into two simpler integrals:\[ \frac{9}{4} \left( \int_{0}^{\pi} 1 \, d x + \int_{0}^{\pi} \cos(2x) \, d x \right) \]Calculate each one separately:1. \(\int_{0}^{\pi} 1 \, d x = \left[ x \right]_{0}^{\pi} = \pi \)2. \(\int_{0}^{\pi} \cos(2x) \, d x = \left[ \frac{1}{2} \sin(2x) \right]_{0}^{\pi} = 0 \)Thus, the combined result is \(\pi\).

Step 5: Compute the Final Result

Substitute the evaluated integrals back into the equation from Step 4:\[ \frac{9}{4} (\pi + 0) = \frac{9\pi}{4} \]This is the final result of the iterated integral.

Key Concepts

Trigonometric Identities

Trigonometric identities play a crucial role in simplifying expressions and solving integrals in calculus. In the context of the exercise provided, we encounter the trigonometric identity for the square of the cosine function: \[ \cos^2 x = \frac{1 + \cos(2x)}{2} \]This identity is particularly helpful because it allows us to transform a squared trigonometric function into a simpler form that is easier to integrate. By expressing \( \cos^2 x \) in terms of \( \cos(2x) \) and a constant, the integration process becomes more manageable.

• Such transformations reduce complex expressions, decreasing computational difficulty.
• The identity transforms squaring \( \cos x \) into a linear combination of 1 and \( \cos(2x) \), simplifying the integration.

Using trigonometric identities efficiently can significantly streamline solving problems involving multiple integrals by breaking them into simpler, easier-to-handle parts.

Definite Integrals

Definite integrals allow us to evaluate the area under a curve between two specified bounds. In the given iterated integral, our main task was to find the area under the curve defined by the iterated function over the interval. With definite integrals, it’s crucial to:

• Understand the limits of integration, as they precisely define the interval under consideration.
• Recognize that the outcome is sensitive to these limits, typically producing a numerical value representing the accumulated quantity.

When solving iterated integrals, the inner integral evaluates the function over one interval, and the result becomes a constant factor or function to be used in the outer integral. Each integral adjusts the bounds in its own dimension, effectively capturing the total volume or area for multi-dimensional spaces.

Calculus

Calculus is the mathematical study of continuous change and involves two principal branches: differentiation and integration. When discussing iterated integrals, as presented in the example, calculus allows us to compute the accumulation of quantities in a stepwise manner. Iterated integrals are particularly useful in multivariable calculus for:

• Calculating areas, volumes, and other results in higher dimensions by addressing integrals over multiple variables successively.
• Breaking down complex integration tasks into simple, manageable parts through systematic evaluation of nested integrals.

The dual nature of calculus is reflected in iterated integrals, where differentiation sets up the relationships between changing quantities, and integration captures the total change over a given domain. This combination is vital for dealing with complex problems involving multi-dimensional systems.