Problem 30 - Calculus an Applied Approach

Question

Find the average value of \(f(x, y)\) over the region \(R .\) \(f(x, y)=x y\) \(R:\) rectangle with vertices \((0,0),(4,0),(4,2),(0,2)\)

Step-by-Step Solution

Answer

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

Step 1: Define the limits of integration

The rectangle \(R\) is defined by the vertices \((0,0),(4,0),(4,2),(0,2)\). This implies that the limits for \(x\) are \(0\) to \(4\) and the limits for \(y\) are \(0\) to \(2\).

Step 2: Setup the double integral

The average value of a function \(f(x, y)\) over a region \(R\) can be found using the formula \(\frac{1}{\text{Area of R}} \int \int_R f(x,y) dA\). The area of the rectangular region \(R\) is \(4*2=8\). Substituting the values, we get \(\frac{1}{8} \int_0^4 \int_0^2 (x y) dy dx\).

Step 3: Evaluate the inner integral

Integrate the inner integral with respect to \(y\) by treating \(x\) as a constant: \(\int_0^2 (x y) dy\). The result will be \(\frac{1}{8} \int_0^4 (x[\frac{y^2}{2}])_0^2 dx\), which simplifies to \(\frac{1}{8} \int_0^4 (2x) dx\).

Step 4: Evaluate the outer integral

Next, integrate the outer integral with respect to \(x\) using the simplified expression: \(\frac{1}{8} \int_0^4 (2x) dx\). The result will be \(\frac{1}{8} ([x^2]_0^4)\), which simplifies to \(2\).

Key Concepts

Double Integral

When dealing with functions of multiple variables, such as the function \(f(x, y) = xy\), double integrals are a powerful mathematical tool used to find various characteristics over a particular area, like volume under a surface. In essence, a double integral allows us to accumulate a quantity over a two-dimensional area.
For our exercise, we are tasked with finding the average value of \(f(x, y)\) over a rectangular region. The double integral for this is set up in two parts: one integral for each variable, \(x\) and \(y\).

• The outer integral is often related to the first variable \(x\), while the inner integral takes care of the second variable \(y\).
• The process involves first evaluating the inner integral with respect to \(y\) and then treating \(x\) as a constant.
• The result of this inner integral is then used to evaluate the outer integral with respect to \(x\).

Through double integration, we accumulate the product \(xy\) over the area defined by \(R\), which in this exercise is a rectangle, to find the average.

Limits of Integration

Limits of integration refer to the boundaries within which we calculate our integral, guiding the extent to which our function is evaluated. This exercise focuses on using these limits to solve a double integral over a rectangular area in the \(xy\)-plane.
We use limits to clearly define this area:

• For the \(x\) variable, the integration bounds are from 0 to 4, aligning with the horizontal sides of the rectangle.
• For the \(y\) variable, the bounds extend from 0 to 2, matching the vertical edges of the rectangle.

By setting these limits, we capture exactly the area of interest and compute the integral precisely within these boundaries. This specific setting ensures that the integration process covers the entire region \(R\) comprehensively.

Area of a Region

The area of a region plays a pivotal role in computing the average value of a function over a specified space. In our scenario, the area is derived easily because the region \(R\) is a simple rectangle.

• The vertices provided, \((0,0), (4,0), (4,2), (0,2)\), outline a clean rectangular shape, making the calculation straightforward.
• Area is determined by multiplying the length of the rectangle by its width: \(4 \times 2 = 8\).

Upon finding that area, we compute the average of the function by taking the product of this region’s area and the integrated function, divided by the total area. This gives us the required average value, making the calculation accurate and efficient. Understanding the importance of this area is essential when applying the average value formula for functions over given regions.