Question

Find the average value of \(f\) over the region \(R .\) Notice how these functions and regions are related to the iterated integrals given in Exercises \(5-8\). \(f(x, y)=x^{2} y-x y^{2} ; \quad R\) is bounded by \(y=x, y=1\) and \(x=3\).

Short answer

The average value of the function over the region is \( \frac{26}{15} \).

Step-by-step solution

Define the Region R

We are given that the region \( R \) is bounded by \( y = x \), \( y = 1 \), and \( x = 3 \). This means \( y \) ranges from \( x \) to \( 1 \) for each fixed \( x \), and \( x \) ranges from \( 0 \) to \( 3 \).

Set Up the Double Integral

To find the average value of \( f(x, y) = x^{2}y - xy^{2} \) over region \( R \), we use the formula for the average value of a function over a region: \[ \frac{1}{\text{Area of } R} \iint_R f(x, y) \, dA. \] We first set up the double integral: \[ \iint_R (x^2 y - x y^2) \, dy \, dx, \] where the inner integration with respect to \( y \) is from \( y = x \) to \( y = 1 \), and the outer integration with respect to \( x \) is from \( x = 0 \) to \( x = 3 \).

Calculate the Area of Region R

The area of region \( R \) is found by integrating the constant function \( 1 \) over \( R \): \[ \int_0^3 \int_x^1 1 \, dy \, dx. \] Integrating with respect to \( y \), we get \[ \int_0^3 (1-x) \, dx. \] This gives \[ \left[x - \frac{x^2}{2}\right]_0^3 = 3 - \frac{9}{2} = \frac{3}{2}. \] Thus, the area of \( R \) is \( \frac{3}{2} \).

Evaluate the Double Integral

First, integrate \( f(x, y) = x^2 y - x y^2 \) with respect to \( y \): \[ \int_x^1 (x^2 y - x y^2) \, dy = \left[ \frac{x^2 y^2}{2} - \frac{x y^3}{3} \right]_x^1. \] Evaluating at the bounds, \[ \left( \frac{x^2}{2} - \frac{x}{3} \right) - \left( \frac{x^4}{2} - \frac{x^4}{3} \right). \] Simplifying, this results in \( \frac{x^2}{2} - \frac{x}{3} - \frac{x^4}{6}\). Now, integrate with respect to \( x \): \[ \int_0^3 \left( \frac{x^2}{2} - \frac{x}{3} - \frac{x^4}{6} \right) \, dx. \] This becomes \[ \left[ \frac{x^3}{6} - \frac{x^2}{6} - \frac{x^5}{30} \right]_0^3, \] evaluating this gives \( \frac{27}{6} - \frac{9}{6} - \frac{243}{30} = \frac{13}{5} \).

Calculate the Average Value

The average value of the function over \( R \) is \( \frac{1}{\text{Area of } R} \times \text{Integral of } f \): \[ \frac{1}{\frac{3}{2}} \times \frac{13}{5} = \frac{26}{15}. \] Thus, the average value of \( f \) over \( R \) is \( \frac{26}{15} \).

Key concepts

Iterated Integrals

An iterated integral is a powerful tool that helps us compute double integrals. It involves evaluating an integral inside another integral. When we have a region over which we want to integrate a function, iterated integrals allow us to break this process into simpler parts. Let's consider the function \( f(x, y) = x^2 y - x y^2 \). To evaluate the double integral over the region \( R \), we execute two separate integrations.

First, we integrate with respect to \( y \), keeping \( x \) constant, and then we integrate the result with respect to \( x \). The integration limits for \( y \) are typically defined by the bounding curves or lines, such as \( y = x \) to \( y = 1 \), and for \( x \), they are typically specified by vertical or horizontal lines, like \( x = 0 \) to \( x = 3 \).

It's crucial to follow the integration limits closely, as iterated integrals are sensitive to these boundaries. Practicing iterated integrals enhances understanding of how functions behave over complex regions.

Double Integral

Double integrals extend the concept of a single integral to functions of two variables. Instead of a line, we consider a bounded region over a plane. The main utility of double integrals is to find volumes under surfaces. For instance, when evaluating the double integral for \( f(x, y) = x^2 y - x y^2 \) over \( R \), we are essentially summing up all the small volumes between the surface defined by \( f(x, y) \) and the \( xy \)-plane.

The double integral is represented as \( \iint_R f(x, y) \, dA \), suggesting integration over the area \( R \). In iterated form, this becomes two single integrals executed in sequence. The choice of integration order, either \( dydx \) or \( dxdy \), depends on the given region and functions.

By tackling the inner integral first, while considering how the function behaves in relation to \( y \) within the region, and then the outer integral with \( x \), we can solve complex integrals more manageably.

Bounded Region

In calculus, particularly when dealing with integrals, a bounded region is an enclosed area defined by specific boundaries or limits. For the given problem, region \( R \) is bound by lines and curves: \( y = x \), \( y = 1 \), and \( x = 3 \).

Within this region, every pair of coordinates \((x, y)\) is constrained. The boundaries form the limits for integration in both \( x \) and \( y \) directions. Understanding the behavior of a bounded region is crucial when setting up integrals. The lines \( y = x \), \( y = 1 \) offer the limits for \( y \) while \( 0 \leq x \leq 3 \) defines how far \( x \) can extend.

Visualizing or sketching this region helps in grasping how the integration limits are derived. Recognizing how these boundaries interact allows better comprehension of how the integral accounts for area or volume enclosed by these limits.