Problem 23 - Thomas Calculus

Question

In Exercises \(17-24\) , evaluate the double integral over the given region \(R .\) $$\iint_{R} \frac{x y^{3}}{x^{2}+1} d A, \quad R : \quad 0 \leq x \leq 1, \quad 0 \leq y \leq 2$$

Step-by-Step Solution

Answer

The double integral evaluates to \(4 \ln 2\).

Step 1: Set up the Double Integral

First, carefully examine the region \( R \): \( 0 \leq x \leq 1 \) and \( 0 \leq y \leq 2 \). The double integral can be expressed as \( \int_{0}^{1} \int_{0}^{2} \frac{x y^{3}}{x^{2}+1} \,dy \,dx \).

Step 2: Integrate Inner Integral with respect to y

Calculate the inner integral \( \int_{0}^{2} \frac{x y^{3}}{x^{2}+1} \, dy \). Consider \( \frac{x}{x^{2}+1} \) as a constant while integrating with respect to \( y \). The integral becomes \[ \int_{0}^{2} \frac{x}{x^{2}+1} y^3 \,dy = \frac{x}{x^{2}+1} \int_{0}^{2} y^3 \,dy. \]

Step 3: Evaluate Inner Integral

Solve \( \int_{0}^{2} y^3 \, dy \), which is \[ \left[ \frac{y^4}{4} \right]_{0}^{2} = \frac{2^4}{4} - \frac{0^4}{4} = 4. \] Thus, the inner integral becomes \( \frac{4x}{x^{2}+1} \).

Step 4: Integrate Outer Integral with respect to x

Substitute the result from Step 3 into the outer integral: \( \int_{0}^{1} \frac{4x}{x^{2}+1} \, dx \). Use the substitution method where \( u = x^2+1 \), then \( du = 2x \, dx \). Adjust the bounds: when \( x = 0, u = 1 \) and when \( x = 1, u = 2 \). The integral becomes \[ 2 \int_{1}^{2} \frac{1}{u} \frac{1}{2} \, du = \int_{1}^{2} \frac{1}{u} \, du. \]

Step 5: Evaluate Outer Integral

Solve \( \int_{1}^{2} \frac{1}{u} \, du \), which is \[ \left[ \ln|u| \right]_{1}^{2} = \ln 2 - \ln 1 = \ln 2. \] Therefore, the value of the outer integral is \( 4 \ln 2 \).

Step 6: Conclusion: Evaluate the Double Integral

The value of the double integral \( \iint_{R} \frac{x y^{3}}{x^{2}+1} \, dA \) over the region \( R \) is \( 4 \ln 2 \).

Key Concepts

Integral Calculus

Integral calculus is a branch of mathematics that focuses on integrals, both definite and indefinite. It is an essential area of study as it allows us to calculate areas, volumes, central points, and many other valuable things. In simple terms, integration is the reverse process of differentiation.

• A definite integral has limits that define the interval on which the integration is performed, producing a numerical value.
• An indefinite integral, meanwhile, does not have limits and represents a family of functions with an arbitrary constant, often represented as "C".

In the context of double integration, integral calculus helps evaluate the volume under a surface described by a two-variable function. The exercise you are examining involves such concepts to determine an area in the given region \( R \). When handling calculus, always consider the functions and limits given to solve integrals accurately.

Integration by Substitution

Integration by substitution is a technique to simplify the integration process. It involves changing the variable of integration to make the integral easier to solve. This method is similar to the chain rule in differentiation but in reverse.
In the example provided, the outer integral uses substitution. The original problem contains the expression \( \frac{4x}{x^2+1} \), and by letting \( u = x^2+1 \), it simplifies the integration process. The differential \( du \) is calculated as \( 2x \, dx \), aiding in rewriting the integral in terms of \( u \) rather than \( x \). This leads to a much simpler integral \( \int \frac{1}{u} \, du \), easy to evaluate. Over time, recognizing when to apply substitution will become intuitive, particularly for functions involving compound or nested expressions.

Evaluation of Integrals

Evaluating integrals is about calculating the exact value of an integral or understanding the geometrical significance of the regions under a curve. When dealing with definite integrals, the focus is on finding the area between the function and the x-axis, within the specified limits.
Double integrals, like the one in your problem, require evaluating two nested integrals. The method involves integrating an inner integral, treating other variables as constants, then substituting the result into the outer integral.

• First, solve the inner integral, which generally simplifies the expression, as shown with \( \int_{0}^{2} y^3 \, dy \), resulting in \( 4 \).
• Second, substitute this into the outer integral and evaluate. In this instance, the integral \( \int_{1}^{2} \frac{1}{u} \, du \) results in \( \ln 2 \).

Applying these steps systematically helps simplify complex integrals and mitigates errors.

Multivariable Calculus

Multivariable calculus extends the principles of calculus to functions with more than one variable. This field is crucial for working with systems that are too complex to describe with single-variable calculus.
Within this context, double integrals are used to compute volumes and more complicated scenarios involving two dimensions. The region \( R \) delineates the boundaries over which integration occurs, playing a key role in setting up the integral.
Understanding the geometry of the region and the function behavior is vital. For instance, the structure \( R : 0 \leq x \leq 1, 0 \leq y \leq 2 \) means integration takes place over a rectangle in the xy-plane.
Always visualize the region and break the problem into manageable parts, as this enhances understanding and accuracy when evaluating double or multiple integrals.