Question

Use the definition for the average value of a function over a region \(R\) (Section 1 ), \(\bar{f}=\frac{1}{\text { area of } R} \iint_{R} f(x, y) d A\). Find the average value of \(a-x-y\) over the region \(R=\\{(x, y): x+y \leq a, x \geq 0, y \geq 0\\},\) where \(a>0\)

Short answer

Answer: The average value of the function f(x, y) over the region R is (2/3)a.

Step-by-step solution

Identify the limits of the region R and find its area

To find the region \(R\) defined by the inequalities \(x+y\leq a, x\geq 0, y\geq 0\), we use the following reasoning: - \(x+y\leq a\): This is equivalent to \(y\leq a-x\). Graphically, this means we are considering the area below the line \(y = a - x\) - \(x\geq 0\): We are considering only the area where \(x\) is positive (right-hand side of the y-axis) - \(y\geq 0\): We are considering only the area where \(y\) is positive (above the x-axis) The region \(R\) is a triangle with vertices at \((0,0)\), \((a,0)\), and \((0,a)\). The area of this triangle can be calculated using the formula for the area of a triangle, which is: \(\text{Area} = \frac{1}{2} \cdot \text{base} \cdot \text{height}\). In our case, the base and the height are both equal to \(a\), so the area of the region is: \(\text{Area of R} = \frac{1}{2}a^2\).

Find the bounds for the double integral

The bounds for the double integral should be expressed in terms of \(x\) and \(y\). To find the limits for each variable, we look at the region \(R\) and the inequalities that define it: For \(x\), we have \(0 \leq x \leq a\). For \(y\), we have \(0 \leq y \leq a-x\). Now, we can express the double integral as: \(\iint_{R} f(x, y) dA = \int_{0}^{a} \int_{0}^{a-x} (a - x - y) dy dx\).

Evaluate the double integral

Now, we need to evaluate the double integral to find the average value. First, let's integrate with respect to \(y\): \(\int_{0}^{a-x} (a-x-y)dy = [(a-x)y -\frac{1}{2}y^2]_{0}^{a-x} = (a-x)(a-x)-\frac{1}{2}(a-x)^2\) Now, integrate with respect to \(x\): \(\int_{0}^{a} [(a-x)(a-x)-\frac{1}{2}(a-x)^2]dx = \left[\frac{1}{3}(a-x)^3\right]_{0}^{a} = \frac{1}{3}a^3\)

Calculate the average value

Now that we have found the double integral, we can use the formula for the average value to find the final answer: \(\bar{f} = \frac{1}{\text{Area of } R} \iint_{R} f(x, y)dA = \frac{1}{\frac{1}{2}a^2}\cdot \frac{1}{3}a^3 = \frac{2}{3}a\) Therefore, the average value of the function \(f(x, y) = a-x-y\) over the region \(R\) is \(\frac{2}{3}a\).

Key concepts

Double Integral

A double integral is a mathematical tool used to calculate the volume under a surface over a given region. Imagine a surface in a three-dimensional space. By setting up a double integral, we could "add up" tiny pieces of area under the surface across a region, which helps us understand the total volume beneath the curve over that region.

Here, our focus is on evaluating how the function \(f(x, y) = a-x-y\) behaves over the triangular region determined by the limits. Double integrals break the region down into infinitesimally small sections, summing up the value of the function multiplied by these tiny areas to get a total value.

The double integral is written as \(\iint_R f(x, y) dA\), where \(dA\) typically represents the tiny piece of area over which the function value is being summed. To solve it, we often first integrate with respect to one variable, holding the other constant, then integrate the result with respect to the other variable.

Integration Limits

Integration limits are essential components in evaluating a double integral, as they define the interval or region over which the function should be integrated. For double integrals, these limits bound the area on an \(x-y\) plane, setting the stage for where calculations begin and end.

In our exercise, we integrated over a triangular region defined by the constraints \(x+y \leq a\), \(x \geq 0\), and \(y \geq 0\). These inequalities translate into specific limits:

• \(x\) runs from 0 to \(a\)
• \(y\) is bounded by the line \(y = a-x\)

These limits allow the integral to capture exactly the desired region without exceeding into areas outside the triangle. Properly setting integration limits ensures that calculations are precise and consistent with the intended area of interest.

Geometric Region

The geometric region we considered in this problem is a triangle on the coordinate plane. Understanding its shape and properties is crucial to correctly setting up a double integral.

Defined by the inequalities \(x+y \leq a\), \(x \geq 0\), and \(y \geq 0\), this triangle has its vertices at the points \((0,0)\), \((a,0)\), and \((0,a)\). The choice of these inequalities restricts the area to the positive quadrant, bound below the line \(y = a-x\).

This geometric understanding helps us calculate the area, \(\frac{1}{2}a^2\), which influences the final result of the double integral. Knowing how to visualize the region on the graph is the first step in correctly identifying limits for the integration process.

Mathematical Analysis

Mathematical analysis provides the foundational tools for exploring functions, geometry, and their behaviors. In this exercise, analysis is about understanding how to apply the theory of integration to compute the average value of a function over a region.

• The analysis starts with deconstructing the problem into manageable parts, like identifying region limits and calculating area.
• The integration, initially with respect to \(y\) and subsequently with respect to \(x\), reflects a methodical approach to handle multiple variables and layers of calculation.

Mathematical analysis isn't just about performing calculations; it involves logical reasoning and translating abstract concepts into practical steps, such as when determining how changes in \(x\) and \(y\) affect the value of \(f(x, y)\). It transforms a seemingly complex concept into an understandable procedure, ultimately leading to the average value \(\frac{2}{3}a\) that answers the problem.