Problem 13 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate the iterated integrals in Problems 1-14. \(\int_{0}^{2} \int_{0}^{\sqrt{4-x^{2}}}(x+y) d y d x\)

Step-by-Step Solution

Answer

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

Step 1: Understanding the integral

This is an iterated integral, which requires us to first integrate with respect to \(y\), and then with respect to \(x\). The integral has the limits \(y = 0\) to \(y = \sqrt{4-x^2}\) and \(x = 0\) to \(x = 2\).

Step 1: Integrate with respect to y

First, consider the integral \( \int_{0}^{\sqrt{4-x^2}} (x+y)\, dy \). Treat \(x\) as a constant and integrate with respect to \(y\).\[ \int (x + y) \,dy = xy + \frac{y^2}{2} + C \] Evaluate this expression from \(y = 0\) to \(y = \sqrt{4-x^2}\): \[ \left. xy + \frac{y^2}{2} \right|_0^{\sqrt{4-x^2}} = x(\sqrt{4-x^2}) + \frac{(\sqrt{4-x^2})^2}{2} \]

Step 3: Evaluate the inner integral

Continue evaluating the above expression:\[ = x\sqrt{4-x^2} + \frac{4-x^2}{2} \] Therefore, the result of integrating with respect to \(y\) is \[ x\sqrt{4-x^2} + 2 - \frac{x^2}{2} \].

Step 2: Integrate with respect to x

Now integrate the resulting expression with respect to \(x\): \( \int_{0}^{2} \left( x \sqrt{4-x^2} + 2 - \frac{x^2}{2} \right) \, dx \).We simplify by breaking the integral into parts:1. \( \int_{0}^{2} x \sqrt{4-x^2} \, dx \)2. \( \int_{0}^{2} 2 \, dx \)3. \( \int_{0}^{2} -\frac{x^2}{2} \, dx \).

Step 5: Integral 1: Solve \( \int x \sqrt{4-x^2} \, dx \) via substitution

Use the substitution \( u = 4-x^2 \), therefore \( du = -2x \, dx \) or \( -\frac{1}{2} du = x \, dx \).The limits change to \( u = 4 \) when \( x = 0 \) and \( u = 0 \) when \( x = 2 \).The integral becomes:\[ \int_{4}^{0} -\frac{1}{2} \sqrt{u} \, du = \frac{1}{2} \int_{0}^{4} u^{1/2} \, du \].

Step 6: Evaluate Integral 1

The integral \( \int u^{1/2} \, du = \frac{2}{3}u^{3/2} \), evaluated from 0 to 4, is:\[ \left. \frac{2}{3} u^{3/2} \right|_0^4 = \frac{2}{3}(8) - \frac{2}{3}(0) = \frac{16}{3}. \]Thus, the result is \( \frac{16}{3} \) and multiplying by \( \frac{1}{2} \) from the substitution gives \( \frac{8}{3} \).

Step 7: Integral 2: Evaluate \( \int 2 \, dx \)

This is a simple integral:\[ \int_{0}^{2} 2 \, dx = 2x \] Evaluated from 0 to 2:\[ 2(2) - 2(0) = 4. \]

Step 8: Integral 3: Evaluate \( \int -\frac{x^2}{2} \, dx \)

This integral evaluates as follows:\[ \int_{0}^{2} -\frac{x^2}{2} \, dx = -\frac{1}{2} \int_{0}^{2} x^2 \, dx = -\frac{1}{2} \left( \frac{x^3}{3} \right)_0^2 \] \[ = -\frac{1}{2} \left( \frac{8}{3} - 0 \right) = -\frac{4}{3}. \]

Step 9: Combine all results

Sum up the results of all integrals:\[ \frac{8}{3} + 4 - \frac{4}{3} \]This simplifies to:\[ \frac{8}{3} + \frac{12}{3} - \frac{4}{3} = \frac{16}{3}. \]

Step 10: Final Result

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

Key Concepts

Multivariable Calculus

In multivariable calculus, we extend the idea of calculus to functions of several variables. Instead of dealing with functions of just one variable, like in basic calculus, equations often involve multiple variables, like \(x\) and \(y\). This allows us to explore a richer set of problems that are more close to real-world situations.

• For instance, the temperature at a given point on the surface of a metal plate might depend on two coordinates on that plane.
• In this context, the concepts of partial derivatives and multiple integrals are used to handle functions of more than one variable.

When dealing with iterated integrals, like in our example, we use nested integrals to find the accumulation of values across a two-dimensional region. An iterated integral is simply taking the process of integration step-by-step for each variable separately. Understanding these integrals is crucial for solving problems in physics, engineering, and other fields where multiple variables come into play.

Definite Integral

A definite integral is an essential concept in calculus that provides the total accumulation of quantities, such as area, under a curve between two points. It is denoted by an integral sign \(\int\) with upper and lower limits. For example, in the integral \(\int_{0}^{2} \int_{0}^{\sqrt{4-x^2}} (x+y) \, d y \dx\), each integral has specific limits which define the interval over which accumulation is considered.
If you think of a definite integral as representing area, the limits determine the baseline of this area. For the inner integral, we integrate over the variable \(y\) from 0 to \(\sqrt{4-x^2}\), while for the outer integral, we consider the variable \(x\) from 0 to 2. These bounds will often correspond to physical constraints or boundaries in a real-world problem.
Definite integrals are the building blocks for calculating phenomena like displacement, volume, and mass, where the total output over a specific region is of interest.

Substitution Method

The substitution method in calculus is an effective technique used to simplify integration. When the problem involves complex expressions, substitution allows us to change variables, making them easier to integrate. Typically, it involves setting a new variable to replace a part of the original expression.

• For example, when we encounter \( x \sqrt{4-x^2} \) in our problem, a substitution like \( u = 4-x^2 \) helps simplify the integration process.
• After substituting, remember to change the limits of integration accordingly, as the bounds will refer to the new variable.

Substitution is a powerful tool. It can transform a difficult integral into something more manageable. It finds use not only in solving integrals but also in differential equations and transformations.

Limits of Integration

Limits of integration are critical for evaluating definite integrals, as they define the range over which you sum the function's values. They specify the start and end points of the integration process along the axes of the variables involved. In the context of iterated integrals, each variable will have its set of limits.

• Understanding the boundaries set by these limits helps ensure accurate computations of the integral's value.
• The limits often arise from real-world constraints, like the physical boundary of a region being modeled, such as in our example, where the region is defined by the circle \( x^2 + y^2 = 4 \).

As you solve problems involving multivariable integrals, carefully considering and setting the correct limits is essential for obtaining the right solution. These limitations are especially important when describing areas, volumes, or any quantity defined within a confined region.