Problem 29 - College Algebra and Calculus: An Applied Approach

Question

Find the average value of \(f(x, y)\) over the region \(R\). $$ \begin{aligned} &f(x, y)=x\\\ &R \text { : rectangle with vertices }(0,0),(4,0),(4,2),(0,2) \end{aligned} $$

Step-by-Step Solution

Answer

The average value of the function \(f(x, y) = x\) over the given region is \(2\).

Step 1: Compute Area of Region R

First, calculate the area of the rectangle. The rectangle's vertices are (0,0), (4,0), (4,2), and (0,2). This is a rectangle 4 units wide and 2 units high, so the area \(A\) is \(4*2 = 8\).

Step 2: Setup and Compute the Double Integral

Now set up the double integral over the region, to integral the function \(f(x, y) = x\). The double integral is \(\int_{0}^{2}\int_{0}^{4}x \, dx \, dy\). We can compute this by first integrating inside integral with respect to \(x\) to get \(1/2 x^2\), which we evaluate at 4 and 0 to get \(8\). Then we integrate \(8\) with respect to \(y\) from 0 to 2 to get \(16\).

Step 3: Compute the average value

Finally, the average value of the function \(f(x, y)\) over the region \(R\) is given by the double integral divided by the area of the region. This is \(16/8 = 2\).

Key Concepts

Double Integral

The concept of a double integral is fundamental in multivariable calculus and is used to compute the volume under a surface defined by a function of two variables, such as the function \( f(x, y) \).

Imagine slicing up the surface into infinitesimally small rectangles, each with an area \( dA \) composed of the differential elements \( dx \) and \( dy \), which are akin to a mesh over the region of interest. The double integral, symbolized as \( \int\int f(x, y) \:dA \), essentially adds up the product of the function's value and the area of each tiny rectangle to determine the total volume.

In practice, we compute the double integral by performing the inner integral first, holding one variable constant while integrating with respect to the other, and then doing the outer integral. For example, the double integral of \( f(x, y) = x \) over area \( R \) in our exercise becomes \( \int_{0}^{2}\int_{0}^{4} x \: dx \: dy \), where we proceed 'inside out' by integrating with respect to \( x \) first and then \( y \).

Area of a Rectangle

The area of a rectangle is one of the simplest geometric concepts, and yet it is profoundly significant in the context of double integrals and multivariable calculus. The area \( A \) is determined by multiplying the length \( l \) of the rectangle by its width \( w \) resulting in the formula \( A = l \times w \).

For our rectangle in the region \( R \) with vertices at \( (0,0), (4,0), (4,2), (0,2) \), we clearly see that the length is 4 units and the width is 2 units. Applying our formula, the area is \( 4 \times 2 = 8 \) square units.

Understanding how to compute the area is pivotal when utilizing double integrals because it allows us to normalize the integral's value—essentially spreading it evenly over the region—when we seek to find an average value, such as in the original exercise.

Multivariable Calculus

Multivariable calculus is a branch of mathematics that deals with functions of more than one variable. Unlike single-variable calculus, which deals with functions of a single variable (like \( f(x) \) for some variable \( x \)), multivariable calculus explores terrain in higher dimensions—like the hills and valleys of a landscape.

In multivariable calculus, concepts such as limits, continuity, derivatives, and integrals are generalized to functions of multiple variables. For instance, the average value of a function over a region is calculated by dividing the double integral of the function over that region by the area of the region. This average value represents a sort of 'balancing point' or the height at which you could flatten the entire volume under the surface into a single, even layer covering the region.

The exercise at hand taps into the essence of multivariable calculus by extending the notion of the average value from a single line (in single-variable calculus) to an entire region in the plane. By solving the problem with double integrals, we not only compute volumes but also gain insights into the spatial distribution of the function's values over a specific area.