Problem 37 - Calculus Early Transcendentals: Pearson New International Edition

Question

Evaluate each iterated integral. \(\int_{-2}^{2} \int_{-1}^{1}\left|x^{2} y^{3}\right| d y d x\)

Step-by-Step Solution

Answer

The value of the integral is \( \frac{32}{9} \).

Step 1: Understanding the Integral Setup

The given integral \( \int_{-2}^{2} \int_{-1}^{1} \left|x^{2} y^{3}\right| \ dy \ dx \) represents the area under the surface \( |x^{2} y^{3}| \) over the rectangular region \( x \in [-2, 2] \) and \( y \in [-1, 1] \). This is a double integral and will be evaluated by integrating with respect to \( y \) first and then with respect to \( x \).

Step 2: Integrate with Respect to y

First, we handle the inner integral \( \int_{-1}^{1} \left|x^{2} y^{3}\right| \ dy \). Since \( x^2 \) is always non-negative and \( y^3 \) retains the sign of \( y \), the expression simplifies to \( |x^2 y^3| = x^2 |y^3| = x^2 y^2 \) since \( |y^3| = y^2 \) over the symmetric interval \([-1, 1]\). Now we integrate: \[\int_{-1}^{1} x^2 y^2 \ dy = x^2 \int_{-1}^{1} y^2 \ dy\]Evaluating the integral,\[x^2 \cdot \left[ \frac{y^3}{3} \right]_{-1}^{1} = x^2 \cdot \left( \frac{1^3}{3} - \frac{(-1)^3}{3} \right) = x^2 \cdot \frac{2}{3} = \frac{2}{3}x^2\]

Step 3: Integrate with Respect to x

Now integrate the result from the inner integral with respect to \( x \) over the range \([-2, 2]\): \[\int_{-2}^{2} \frac{2}{3} x^2 \ dx\]Factor out the constant \( \frac{2}{3} \):\[\frac{2}{3} \int_{-2}^{2} x^2 \ dx\]Evaluate the integral:\[\frac{2}{3} \left[ \frac{x^3}{3} \right]_{-2}^{2} = \frac{2}{3} \cdot \left( \frac{2^3}{3} - \frac{(-2)^3}{3} \right) = \frac{2}{3} \cdot \left( \frac{8}{3} + \frac{8}{3} \right) = \frac{2}{3} \cdot \frac{16}{3} = \frac{32}{9}\]

Step 4: Simplify the Final Result

The solution to the iterated integral is \( \frac{32}{9} \). This is the total volume under the surface \( |x^2 y^3| \) over the specified region.

Key Concepts

Iterated Integrals

In calculus, iterated integrals are an important tool for evaluating multivariable functions over a given region. They are used to calculate the area or volume under a surface. Essentially, an iterated integral represents multiple integrations, where you perform integration successively over one variable at a time.

• The double integral given involves integrating a function of two variables, \(x\) and \(y\), over a specific rectangular region.
• This is achieved by first integrating over \(y\) (keeping \(x\) fixed), which is called the inner integral, and then integrating the result over \(x\), called the outer integral.

The bounds of integration determine the limits within which each variable is integrated. By evaluating each integral one at a time, we can keep track of how the function changes with respect to each variable independently.
This two-step process helps to simplify complex calculations and extract meaningful quantitative properties from multivariable functions.

Absolute Value Integrals

Integrating expressions involving absolute values can be tricky due to their piecewise nature. The absolute value function makes the integrand non-negative, which considerably influences the integration process. In our exercise, the expression \(|x^2 y^3|\) required some simplification before integrating.

• Recognize that \(x^2\) is always non-negative since it is a square term.
• The trick is to consider \(y^3\), which affects the absolute value due to its sign changes across different intervals of \(y\).
• When \(y\) ranges from -1 to 1, \(y^3\) changes sign, but \(|y^3|\) simply yields \((y^3)^2 = y^6 = y^2\) over symmetric limits, simplifying integration.

The main task is to carefully handle these expressions to simplify the integral while respecting the bounds, ensuring all terms are accounted for correctly. This leads to simplified, often quadratic forms which are easier to integrate.

Multivariable Calculus

Multivariable calculus extends the concept of calculus to functions with more than one variable. In this exercise, we work with a function of two variables, \(x\) and \(y\), as indicative by the notation \(f(x, y) = |x^2 y^3|\).

• It encompasses dealing with limits, continuity, partial derivatives, and multiple integrals.
• Functions in multivariable calculus, such as the one given, can describe surfaces or curves in space, with applications in physics, engineering, and more.
• Double integrals, like the one here, are a method to find the volume under a surface over a region in the \(xy\)-plane.

Understanding multivariable calculus requires a good grasp of how each variable affects the function independently and together with others. This involves evaluating partial derivatives or performing iterated integrations step-by-step. Moreover, it connects to real-world problems where multiple factors influence outcomes, requiring us to calculate areas, volumes, or averages over multidimensional spaces.