Problem 3 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate the iterated integrals. $$ \int_{-1}^{3} \int_{0}^{3 y}\left(x^{2}+y^{2}\right) d x d y $$

Step-by-Step Solution

Answer

The value of the iterated integral is 240.

Step 1: Understand the Problem

The given iterated integral is \(\int_{-1}^{3} \int_{0}^{3y} \left(x^{2} + y^{2}\right) \, dx \, dy \). This means we will integrate \(x^2 + y^2\) with respect to \(x\) first, and then integrate the resulting function with respect to \(y\). The limits of integration for \(x\) are from 0 to \(3y\), and for \(y\) are from -1 to 3.

Step 2: Integrate with respect to \(x\)

The first step is to integrate \(x^2 + y^2\) with respect to \(x\) within the limits from 0 to \(3y\). The integral is: \[\int_{0}^{3y} (x^2 + y^2) \, dx\]Integrating term by term:- The antiderivative of \(x^2\) is \(\frac{x^3}{3}\).- The antiderivative of \(y^2\), in terms of \(x\), is \(y^2x\), since \(y^2\) is treated as a constant with respect to \(x\). Thus, you obtain:\[\left[ \frac{x^3}{3} + y^2x \right]_{0}^{3y}\]

Step 3: Evaluate the Integral at the Limits for \(x\)

Now substitute the upper and lower limits of \(x\) into the antiderivative:\[\left( \frac{(3y)^3}{3} + y^2(3y) \right) - \left( \frac{0^3}{3} + y^2(0) \right)\]Simplify the expression:- \(\frac{(3y)^3}{3} = 9y^3\)- \(y^2(3y) = 3y^3\)So the integral from 0 to \(3y\) becomes:\[9y^3 + 3y^3 = 12y^3\]

Step 4: Integrate with respect to \(y\)

The next step is integrating the resulting function \(12y^3\) with respect to \(y\) within the limits from -1 to 3:\[\int_{-1}^{3} 12y^3 \, dy\]The antiderivative of \(12y^3\) is \(12 \cdot \frac{y^4}{4} = 3y^4\). Thus, evaluate:\[\left[ 3y^4 \right]_{-1}^{3}\]

Step 5: Evaluate the Integral at the Limits for \(y\)

Now, substitute the upper and lower limits for \(y\):\[3(3)^4 - 3(-1)^4\]Calculate each term:- \(3(3)^4 = 3(81) = 243\)- \(3(-1)^4 = 3(1) = 3\)Thus, the final result is:\[243 - 3 = 240\]

Step 6: Conclusion: State the Final Value of the Integral

After evaluating the iterated integrals in sequential order, the value of the integral \(\int_{-1}^{3} \int_{0}^{3y} (x^2 + y^2) \, dx \, dy\) is 240.

Key Concepts

Double Integral

Double integrals are a key concept in calculus, allowing us to calculate the volume under a surface in a two-dimensional region.

• This involves integrating a function over two variables, typically denoted as \(x\) and \(y\).
• The process is performed iteratively, meaning you first integrate with respect to one variable, then perform the integration with respect to the other.

This exercise involves a double integral of the form \(\int_{-1}^{3} \int_{0}^{3y} (x^2 + y^2) \, dx \, dy\), which means you need to solve one integral and then move on to the next.Double integrals are widely used in finding areas, volumes, and in physics for calculating mass distributions over a region.

Antiderivative

Understanding the concept of an antiderivative is crucial when working with integrals. An antiderivative of a function \(f(x)\) is a function \(F(x)\) such that the derivative of \(F(x)\) equals \(f(x)\).

• In the context of our exercise, finding the antiderivative of \(x^2\) gave us \(\frac{x^3}{3}\), as \(\frac{d}{dx}(\frac{x^3}{3}) = x^2\).
• Similarly, since \(y^2\) is constant with respect to \(x\), its antiderivative in terms of \(x\) is \(y^2x\).

Understanding these individual antiderivatives helps in simplifying the integration process, as shown when integrating term by term in the given problem.

Integration Limits

Integration limits define the region over which you are integrating, and they are essential for solving any definite integral, including iterated integrals.

• In the given problem, \(x\) is integrated from 0 to \(3y\), which means for every fixed value of \(y\), \(x\) ranges from 0 to a line dependent on \(y\).
• Similarly, \(y\) is integrated from -1 to 3, defining the overall region along the \(y\)-axis.

These limits directly affect the result of the integral, as they dictate the actual area over which the function values are accumulated. Always replace the limits with their respective variables' values during evaluation to ensure accuracy.

Calculus Problem Solving

Solving calculus problems, such as evaluating iterated integrals, involves a systematic approach.

• Start by understanding what is being asked; breaking the problem down into manageable steps is crucial.
• Perform integration step by step—solve the inner integral first, plug in its limits, and use the result in the next integration.
• Checking your work at each step helps avoid mistakes. This particularly applies to applying limits and simplifying expressions correctly.

This structured problem-solving method not only applies to this exercise but serves as a useful approach for various problems encountered in higher mathematics. Even complex problems can become manageable with a disciplined and thorough process.