Problem 19 - Thomas Calculus

Question

In Exercises \(17-24\) , evaluate the double integral over the given region \(R .\) $$\iint_{R} x y \cos y d A, \quad R :-1 \leq x \leq 1, \quad 0 \leq y \leq \pi$$

Step-by-Step Solution

Answer

The double integral evaluates to -2.

Step 1: Understanding the Double Integral

The double integral \( \iint_R x y \cos y \, dA \) is to be evaluated over the region \( R \), defined by the bounds \(-1 \leq x \leq 1\) and \(0 \leq y \leq \pi\). The area element \(dA\) can be expressed as \(dx \, dy\). This means we are integrating with respect to \(x\) first, and then with respect to \(y\).

Step 2: Setting Up the Integral

Given the region \( R \), set up the double integral as follows: \[\int_{y=0}^{y=\pi} \int_{x=-1}^{x=1} x y \cos y \, dx \, dy\] We first integrate with respect to \(x\), treating \(y \cos y\) as a constant because it does not depend on \(x\).

Step 3: Evaluate the Inner Integral with Respect to x

Find the integral of \(x y \cos y\) with respect to \(x\): \[\int_{x=-1}^{x=1} x y \cos y \, dx \]This simplifies to \[y \cos y \int_{-1}^{1} x \, dx\]Calculate the integral \[\int_{-1}^{1} x \, dx = \left[ \frac{x^2}{2} \right]_{-1}^{1} = \frac{1}{2} - \left(-\frac{1}{2}\right) = 1\]Thus, the inner integral evaluates to \[y \cos y \cdot 1 = y \cos y\].

Step 4: Evaluate the Outer Integral with Respect to y

Now, evaluate the outer integral: \[\int_{y=0}^{y=\pi} y \cos y \, dy\]This requires integration by parts. Let \(u = y\) and \(dv = \cos y \, dy\). Then, \(du = dy\) and \(v = \sin y\).Apply integration by parts: \[uv - \int v \, du = y \sin y - \int \sin y \, dy\]Calculate the integral \(\int \sin y \, dy = - \cos y\). Thus, \[y \sin y - (- \cos y) = y \sin y + \cos y\]Evaluate from 0 to \(\pi\): \[\left[(y \sin y + \cos y)\right]_{0}^{\pi}\]Substitute the bounds: \[(\pi \sin \pi + \cos \pi) - (0 \cdot \sin 0 + \cos 0)\]Simplify to obtain: \[0 - 1 - 1 = -2\].

Step 5: Concluding the Calculation

The value of the double integral is determined by combining the results from the inner and outer integrals. Therefore, the evaluated double integral is \(-2\).

Key Concepts

Integration by Parts

Integration by parts is a useful technique to solve integrals where you're dealing with products of functions. It is derived from the product rule of differentiation and is given by the formula: \[ \int u \, dv = uv - \int v \, du \]
In our example, we apply integration by parts to evaluate the outer integral \( \int_{y=0}^{y=\pi} y \cos y \, dy \). Here, we select \( u = y \) and \( dv = \cos y \, dy \).

• Differentiate the chosen \( u \) to find \( du = dy \).
• Integrate \( dv \) to find \( v = \sin y \).

This leads to:\[ uv - \int v \, du = y \sin y - \int \sin y \, dy \]
So why do we use integration by parts here? It's because the term \( y \) (which multiplies with \( \cos y \)) becomes more manageable after differentiation, turning into a simpler integral to solve. Thus, by systematically integrating by parts, we unravel complex integrals with product functions into simpler calculations.

Iterated Integrals

Iterated integrals break down the evaluation of a double integral into two single integrals, allowing us to approach multi-variable integration step by step. This concept is crucial for solving double integrals like the one in our example.
The process involves first integrating with respect to one variable while treating the other as a constant, and then integrating the result with respect to the second variable. For instance, our integral is set up as follows: \[ \int_{y=0}^{y=\pi} \int_{x=-1}^{x=1} x y \cos y \, dx \, dy \]
We first evaluate the inner integral \( \int_{x=-1}^{x=1} x y \cos y \, dx \), where \( y \cos y \) acts as a constant since we integrate with respect to \( x \). Once calculated, the result is then used as the function of \( y \) for the outer integral \[ \int_{y=0}^{y=\pi} y \cos y \, dy \].

• The inner integral yields \( y \cos y \), considering bounds for \( x \).
• The outer integral solves the integral for \( y \) with the result from the inner integral.

Iterated integrals are powerful tools that convert complex multi-variable integrals into simpler, solvable steps, making the solution process more manageable.

Region of Integration

The region of integration is fundamental for setting the bounds of a double integral. Understanding the region well ensures you integrate over the correct area in the xy-plane.
In the given problem, the region \( R \) is defined by: \[-1 \leq x \leq 1, \quad 0 \leq y \leq \pi \]
This tells us that \( x \) varies from -1 to 1, and \( y \) ranges from 0 to \( \pi \). The choice and definition of these bounds are crucial.

• The order of integration, determined by these bounds, affects how we approach the problem — here, integrating with respect to \( x \) first and then \( y \).
• Accurate region definitions prevent calculation errors and ensure all steps proceed correctly.

When setting up your double integral, visualize the region on a graph to understand how the variables interact within their specified limits. This spatial reasoning guides how you apply integration techniques effectively, aiding in a successful solution.