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Magazine Chapter 15: Problem 21
Evaluate the double integral over the given region \(R\). $$\iint_{R} \frac{x y^{3}}{x^{2}+1} d A, \quad R: \quad 0 \leq x \leq 1, \quad 0 \leq y \leq 2$$
Short Answer
Expert verified
The double integral evaluates to \(2 \ln(2)\).
Step by step solution
01
Understand the Region R
The region described by \(R\) is defined by two intervals for \(x\) and \(y\). Specifically, \(x\) ranges from 0 to 1, and \(y\) ranges from 0 to 2. This means \(R\) is a rectangle in the \(xy\)-plane with these boundaries.
02
Setup the Double Integral
To evaluate the double integral \(\iint_{R} \frac{x y^{3}}{x^{2}+1} dA\), we express it as an iterated integral \(\int_{0}^{1} \int_{0}^{2} \frac{x y^{3}}{x^{2}+1} \; dy \; dx\). We will first integrate with respect to \(y\), then with respect to \(x\).
03
Inner Integral with respect to y
Evaluate the inner integral \(\int_{0}^{2} \frac{x y^{3}}{x^{2}+1} \; dy\). Treat \(x\) as a constant while integrating with respect to \(y\). The integral becomes \(\frac{x}{x^2 + 1} \int_{0}^{2} y^3 \; dy\).
04
Integrate y^3
Compute \(\int_{0}^{2} y^3 \; dy\). The antiderivative of \(y^3\) is \(\frac{y^4}{4}\). Evaluate from 0 to 2: \(\left[\frac{2^4}{4} - \frac{0^4}{4}\right] = \frac{16}{4} = 4\).
05
Multiply by x term
Substitute back into the expression from Step 3 to find \(\frac{x}{x^2 + 1} \cdot 4\). This simplifies to \(\frac{4x}{x^2 + 1}\).
06
Outer Integral with respect to x
Now evaluate the outer integral: \(\int_{0}^{1} \frac{4x}{x^2 + 1} \; dx\). Use substitution: let \(u = x^2 + 1\), then \(du = 2x \; dx\), which implies \(x \; dx = \frac{1}{2} du\).
07
Substitution and Integration
Substitute to get the integral \(\int \frac{4}{2} \frac{1}{u} \; du = 2 \int \frac{1}{u} \; du\). This integrates to \(2 \ln|u|\).
08
Substitute Back and Evaluate
Substitute \(u = x^2 + 1\) back into the integral: \(2 \ln|x^2 + 1|\). Evaluate this from 0 to 1: \(2(\ln(2) - \ln(1)) = 2 \ln(2)\).
09
Final Result
The value of the double integral \(\iint_{R} \frac{x y^{3}}{x^{2}+1} d A\) is \(2 \ln(2)\), which represents the integral of the given function over the specified region \(R\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Iterated Integral
An iterated integral is a way to compute a double integral by splitting it into two single integrals. The key here is to first integrate with respect to one variable, while treating the other as a constant, and then integrate with respect to the second variable. For the given problem, the double integral \( \iint_{R} \frac{x y^{3}}{x^{2}+1} d A \) is expressed as \( \int_{0}^{1} \int_{0}^{2} \frac{x y^{3}}{x^{2}+1} \; dy \; dx \). This approach makes it easier to handle, as it transforms a potentially complex problem into simpler, more manageable steps.
**Key Points About Iterated Integrals:**
**Key Points About Iterated Integrals:**
- The order of integration can affect the complexity, so choose carefully based on the limits and functions.
- Initially, you integrate the innermost integral using constant limits, given by the region.
- Each integration step typically requires techniques from single-variable calculus.
Region of Integration
The region of integration, \( R \), describes the area over which the integration takes place. For this exercise, \( R \) is given as a rectangle with boundaries defined by \( 0 \leq x \leq 1 \) and \( 0 \leq y \leq 2 \). This means our double integral computes the volume under the surface defined by \( \frac{x y^{3}}{x^{2}+1} \) over this rectangular area.
**Understanding the Region of Integration:**
**Understanding the Region of Integration:**
- The limits \( 0 \leq x \leq 1 \) and \( 0 \leq y \leq 2 \) describe a rectangle in the \( xy \)-plane.
- Visual representation helps to identify the integration order and simplify the setup.
- This setup is crucial for correctly performing the operations in both iterated integrals.
Integration by Substitution
Integration by substitution is a technique used to simplify integrals, making them easier to solve. In the outer integral of this exercise, the substitution method was applied to the integral \( \int_{0}^{1} \frac{4x}{x^2 + 1} \; dx \). The substitution was \( u = x^2 + 1 \), leading to \( du = 2x \; dx \). This efficient substitution reshapes the problem, transforming the integral into \( 2 \int \frac{1}{u} \; du \), a much simpler form.
**Steps Involved in Integration by Substitution:**
**Steps Involved in Integration by Substitution:**
- Identify a substitution that simplifies the integrand, relating one variable to another.
- Change the limits of integration if the substitution affects them.
- Transform the differential accordingly, solving for \( dx \) or \( dy \) as needed.
- Perform the integration, then substitute back the original variable to find the solution.
