Problem 20 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate the indicated double integral over \(R\). \(\iint_{R} x y \sqrt{1+x^{2}} d A\) \(R=\\{(x, y): 0 \leq x \leq \sqrt{3}, 1 \leq y \leq 2\\}\)

Step-by-Step Solution

Answer

The value of the double integral is \(\frac{7}{2}\).

Step 1: Set Up the Double Integral

To solve the problem, we begin by setting up the double integral over the region \(R\). This region is defined by the limits \(0 \leq x \leq \sqrt{3}\) and \(1 \leq y \leq 2\). So, the double integral is given by \[\int_{0}^{\sqrt{3}} \int_{1}^{2} xy \sqrt{1+x^{2}}\, dy\, dx.\]

Step 2: Evaluate the Inner Integral with Respect to y

First, we evaluate the inner integral with respect to \(y\). The expression inside the integral is \(xy\sqrt{1+x^2}\) and we treat \(x\) as a constant while integrating with respect to \(y\): \[\int_{1}^{2} xy \sqrt{1+x^{2}}\, dy = x \sqrt{1+x^{2}} \int_{1}^{2} y \, dy.\] The integral of \(y\) with respect to \(y\) is \(\frac{y^2}{2}\), evaluated from \(1\) to \(2\):\[= x \sqrt{1+x^{2}} \left[ \frac{y^2}{2} \right]_1^2 = x \sqrt{1+x^{2}} \left( \frac{4}{2} - \frac{1}{2} \right) = \frac{3}{2} x \sqrt{1+x^{2}}.\]

Step 3: Evaluate the Outer Integral with Respect to x

Next, we integrate with respect to \(x\) from \(0\) to \(\sqrt{3}\):\[\int_{0}^{\sqrt{3}} \frac{3}{2} x \sqrt{1+x^{2}} \, dx.\]We use a substitution method where \(u = 1+x^2\), thus \(du = 2x \, dx\) or \(x \, dx = \frac{1}{2} du\). When \(x = 0\), \(u = 1\), and when \(x = \sqrt{3}\), \(u = 4\). The integral becomes:\[\frac{3}{4} \int_{1}^{4} u^{1/2} \, du.\]Integrating \(u^{1/2}\) gives \(\frac{2}{3} u^{3/2}\), and after substitution back:\[\frac{3}{4} \left[ \frac{2}{3} u^{3/2} \right]_{1}^{4} = \frac{1}{2} \left[ u^{3/2} \right]_{1}^{4} = \frac{1}{2} (8 - 1) = \frac{7}{2}.\]

Step 4: Final Result

The value of the double integral over the region \(R\) is \(\frac{7}{2}\). Ensure all steps adhered to integration methods and use of limits of integration.

Key Concepts

Integration Techniques

Integration techniques are essential for solving double integrals, as they allow us to break down complex regions or functions into manageable computations. In this exercise, the integration process begins by identifying appropriate methods for evaluating the double integral. We apply the straight-forward method of integrating the inner integral first, followed by the outer integral, commonly known as iterated integration.

In step 2 of the solution, the inner integral is evaluated with respect to \(y\) while treating \(x\) as a constant. Since the expression has a simple form, direct integration is possible. For the outer integral, a substitution is employed, which transforms the integral into a simpler form to evaluate. Here, a trigonometric substitution wasn't necessary, as the substitution method easily handled the algebraic manipulation.

Key ideas of integration techniques also suggest prioritizing simpler methods before moving to more complex strategies. Understanding when to keep variables constant, and selecting the method for outer integration, helps efficiently solve the integral.

Change of Variables

The change of variables is a technique that involves transforming the function inside the integral into a more agreeable form, often making integration easier. Generally, especially in multivariable calculus, you might encounter functions that are cumbersome to integrate in their initial format.

In this solution, the substitution \( u = 1 + x^2 \) was used to simplify the integration process. This substitution turns a complicated algebraic expression into a basic form by transforming both the function and the limits of integration from \(x\) in terms of \(u\).

The derivative \(du = 2x \, dx\) plays a crucial role in changing variables, adjusting the differential to accommodate the new variable. After suitable transformation, integration becomes a matter of substituting back the original variables post-evaluation, simplifying the calculation process to achieve the final result.

Iterated Integrals

Iterated integrals are a fundamental concept in evaluating double integrals, allowing the computation to be split into a series of one-dimensional integrals. These involve evaluating one integral while treating the other variable as constant, switching variables through each integral step.

Here, iterated integration begins by first focusing on the inner integral with respect to \(y\), using specific limits. By isolating this part of the integral, calculations are simplified, setting the stage for tackling the outer integral. This systematic approach aid in isolating complex interactions between variables that are otherwise too challenging to address simultaneously.

Iterated integrals are not just a technical shortcut, they provide a framework for visualizing multidimensional spaces, one slice at a time, offering valuable insights that piecewise integration alone can't provide.

Limits of Integration

The limits of integration define the boundaries over which integration is performed. They are crucial for ensuring the correctness and physical relevance of the problem solution, especially in multivariable integrals like double integrals.

In this case, the region \(R\) is bound by linear limits \( 0 \leq x \leq \sqrt{3} \) and \( 1 \leq y \leq 2 \). These determine the bounds used in the iterated integrals since different boundaries may represent distinct physical or geometrical contexts.

When integrating, proper attention to the limits aids in choosing order of integration, ensuring that the variable order reflects the region specified. It's crucial to translate these limits correctly, especially after any change of variables, to maintain the integrity of the integral's solution.