Problem 6 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate each of the iterated integrals. $$ \int_{-1}^{1} \int_{1}^{2}\left(x^{2}+y^{2}\right) d x d y $$

Step-by-Step Solution

Answer

The iterated integral evaluates to \(\frac{16}{3}\).

Step 1: Understand the Problem

We need to evaluate the iterated integral \( \int_{-1}^{1} \int_{1}^{2}(x^{2}+y^{2}) \, dx \, dy \). This involves integrating the function \(x^2 + y^2\) first with respect to \(x\), and then with respect to \(y\).

Step 2: Integrate with Respect to x

The inner integral is \( \int_{1}^{2} (x^2 + y^2) \, dx \). We treat \(y^2\) as a constant as we integrate with respect to \(x\). The integral becomes: \[\int_{1}^{2} x^2 + y^2 \, dx = \left[\frac{x^3}{3} + y^2 x\right]_{1}^{2}.\]

Step 3: Evaluate the Inner Integral

Substitute the limits into the integrated function: - When \(x = 2\), the expression is \(\frac{2^3}{3} + y^2 \, (2) = \frac{8}{3} + 2y^2\).- When \(x = 1\), the expression is \(\frac{1^3}{3} + y^2 \, (1) = \frac{1}{3} + y^2\).Subtract the lower limit from the upper limit: \[\left(\frac{8}{3} + 2y^2\right) - \left(\frac{1}{3} + y^2\right) = \frac{7}{3} + y^2.\]

Step 4: Integrate with Respect to y

Now, integrate the result \(\frac{7}{3} + y^2\) with respect to \(y\) from -1 to 1: \[\int_{-1}^{1} \left(\frac{7}{3} + y^2\right) \, dy = \left[\frac{7}{3}y + \frac{y^3}{3}\right]_{-1}^{1}.\]

Step 5: Evaluate the Outer Integral

Substitute the limits into the integrated function: - When \(y = 1\), the expression is \(\frac{7}{3}(1) + \frac{1^3}{3} = \frac{7}{3} + \frac{1}{3} = \frac{8}{3}\).- When \(y = -1\), the expression is \(\frac{7}{3}(-1) + \frac{(-1)^3}{3} = -\frac{7}{3} - \frac{1}{3} = -\frac{8}{3}\).Subtract the lower limit from the upper limit: \[\frac{8}{3} - (-\frac{8}{3}) = \frac{8}{3} + \frac{8}{3} = \frac{16}{3}.\]

Step 6: Final Result

The value of the iterated integral is \(\frac{16}{3}\).

Key Concepts

Integration Techniques

In the context of evaluating integrals, especially iterated ones, it's important to utilize specific integration techniques. An iterated integral involves multiple integration processes, often nested one within another. Here, we focus on the function \(x^2 + y^2\), integrating first with respect to \(x\), treating \(y\) as a constant.
This can be challenging because you need to think about how to simplify the expression during each integration step. Techniques often involve:

• Integration by parts, helpful when a product of functions is involved.
• Substitution, useful to simplify complex expressions.
• Recognizing and using patterns for known integrals, like polynomial or trigonometric functions.

In this problem, recognizing that \(y^2\) is constant relative to \(x\) allows straightforward integration of \(x^2 + y^2\) with respect to \(x\). Using these techniques properly can simplify the process significantly.

Limits of Integration

The limits of integration define the region over which the function is integrated. They are crucial as they determine the boundaries for the variables involved. In an iterated integral, these limits are often given in a nested manner, reflecting the order of integration.
For our problem, the inner integral \(\int_{1}^{2} (x^2 + y^2) \, dx\) is evaluated with respect to \(x\) from 1 to 2. The outer integral \(\int_{-1}^{1} \, dy\) is evaluated with respect to \(y\) from -1 to 1. These limits dictate:

• The specific segment of the \(x\)-axis (1 to 2) over which \(y\) acts as a constant during integration.
• The section of the \(y\)-axis (-1 to 1) over which the results from the \(x\) integration are integrated.

Understanding these boundaries helps in setting up the integrals correctly and evaluating them systematically.

Multivariable Calculus

In multivariable calculus, the study extends beyond single-variable functions to incorporate functions of multiple variables. This branch of calculus is essential when dealing with real-world problems where several independent variables affect the dependent variable.
Iterated integrals are part of this field, used to evaluate functions over specified regions, often leading to volume calculations under surfaces. Here, our function \(x^2 + y^2\) depends on both \(x\) and \(y\), requiring integration respect to both independently.
Multivariable calculus principles include:

• Handling partial derivatives and integrals, where you treat all but one variable as constant.
• Using the gradient, divergence, and curl in vector fields, which extend calculus concepts to multi-dimensional spaces.
• Applying theorems like Green’s, Stokes’, or the Divergence Theorem to simplify complex integrals.

These concepts allow us to solve intricate problems by breaking them down systematically into simpler, one-variable integrals.

Definite Integral

A definite integral evaluates the accumulation of quantities, such as area or volume, over a specific interval. Unlike indefinite integrals, which yield antiderivatives plus a constant, definite integrals provide numerical values by considering the upper and lower limits of integration.
In our exercise, the definite integral \(\int_{-1}^{1} \int_{1}^{2}(x^{2}+y^{2}) \, dx \, dy\) requires two evaluations:

• The inner integral evaluates the expression \(x^2 + y^2\) from \(x = 1\) to \(x = 2\), resulting in a function solely in terms of \(y\).
• The outer integral then evaluates this new expression from \(y = -1\) to \(y = 1\).

Upon completion, the value \(\frac{16}{3}\) signifies the cumulative total within the defined region. This is a concrete numerical result that represents the entire area under the function within the stated bounds.