Question

Use the definition for the average value of \(a\) function over a region \(R \text { (Section } 13.1), \bar{f}=\frac{1}{\text { area of } R} \iint_{R} f(x, y) d A\). 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 over the given region is \(\frac{-1}{3}a\).

Step-by-step solution

Find the area of the region R

By visualizing the inequality constraints \(x+y \leq a\), \(x \geq 0\), and \(y \geq 0\) on the Cartesian plane, we can see that these constraints define the region of a right-angled triangle with vertices at (0,0), (a,0), and (0,a). The area of this triangle can be found using the formula for the area of a triangle: \(\frac{1}{2} \cdot \text{base} \cdot \text{height}\). Since the base and height are both equal to \(a\), the area of region \(R\) is: $$ \text {Area of } R= \frac{1}{2} \cdot a \cdot a = \frac{1}{2} a^2. $$

Set up the double integral

Next, we need to set up the double integral of the function \(f(x,y)=a-x-y\) over the region \(R\). Since the region is a triangle, we can use the given constraints to determine the limits of integration. The integral will be of the form: $$ \iint_{R} (a-x-y) dA = \int_{c_1}^{c_2}\int_{d_1}^{d_2}(a-x-y) dy\, dx, $$ where we need to find \(c_1\), \(c_2\), \(d_1\), and \(d_2\).

Determine the limits of integration

As we know that \(x \geq 0\), the lower limit of integration for x is 0, and the upper limit is a. For y, the limits of integration should be for \(y \geq 0\) and \(x+y \leq a\). The lower limit for y is \(0\), and the upper limit is the line \(y= -x + a\). Thus, the limits of integration are: $$ \int_{0}^{a}\int_{0}^{-x+a}(a-x-y) dy\, dx. $$

Evaluate the double integral

Now, we can evaluate the integral in the following steps: 1. Integrate with respect to y: $$ \int_{0}^{a} \left[-\frac{1}{2}(a-x-y)^2\Big|_{y = 0}^{y = -x + a}\right] dx. $$ 2. Substitute the limits of integration for y: $$ \int_{0}^{a} \left[-\frac{1}{2}(a-x-(-x+a))^2\right]dx. $$ 3. Simplify and integrate with respect to x: $$ \int_{0}^{a} \left[-\frac{1}{2}(x^2 - 2ax + 2a^2)\right] dx = -\frac{1}{6} a^3. $$

Calculate the average value

Finally, we can calculate the average value of the function over the region using the formula \(\bar{f} = \frac{1}{\text { area of } R} \iint_{R}f(x, y) dA\) : $$ \bar{f}=\frac{1}{\frac{1}{2}a^2}[-\frac{1}{6} a^3]=\frac{-1}{3}a. $$ So the average value of the function \(f(x, y) = a-x-y\) over the given region \(R\) is \(\frac{-1}{3}a.\)

Key concepts

Double Integral

Double integrals extend the concept of integration to functions of two variables, typically denoted as \(f(x, y)\). These integrals help us compute a variety of quantities over a two-dimensional region, such as area, volume, and, in this case, the average value of a function.

The double integral of a function over a region \(R\) is represented as \(\iint_R f(x, y) \, dA\), where \(dA\) indicates an infinitesimally small area element.

• The process involves first integrating with respect to one variable while considering the other variable a constant.
• Then, integrating the resulting expression with respect to the second variable.
• This allows us to cover the entire area of the region \(R\).

The ability to break up the integration process into simpler, one-dimensional integrals, makes double integrals manageable and powerful in evaluating areas under surfaces in a three-dimensional context.

Limits of Integration

The limits of integration are crucial in setting up the appropriate double integral. They define the bounds for each variable over the specific region of interest.

In our problem:

• The limits for \(x\) are from \(0\) to \(a\), because the region \(R\) starts at the y-axis and goes up to the point \((a,0)\).
• For \(y\), the limits are from \(0\) to \(-x + a\), corresponding to the line that bounds the region from the top right corner and extends down to the x-axis.

With these limits, we set the boundaries for our integration and ensure that we are correctly calculating over the desired area of the region \(R\). Understanding how to determine these limits is essential for correctly evaluating double integrals.

Area of a Region

To find the average value of a function over a region, we first need to determine the area of that region. In our scenario, the defined region \(R\) is a right-angled triangle on the Cartesian plane. Its vertices are located at \((0, 0), (a, 0),\) and \((0, a)\).

The area of this triangular region can be calculated using the formula for the area of a triangle, \(\frac{1}{2} \times \text{base} \times \text{height}\). With both base and height being equal to \(a\), the area of \(R\) comes out to be \(\frac{1}{2}a^2\).

• This area is vital as it normalizes the double integral, converting it from a total quantity into an average value.
• Without determining this area, we wouldn't be able to compute the average effectively.

When dealing with any problems involving averages over geometric regions, assessing the geometry of the region to determine its area is a crucial first step.

Cartesian Plane Visualization

Visualizing the region of integration on the Cartesian plane is an essential step in solving problems involving double integrals. The region is defined by the inequalities imposed by the problem, such as \(x + y \leq a\), \(x \geq 0\), and \(y \geq 0\).

These inequalities form a set of boundaries on the Cartesian plane, outlining the exact shape of the region \(R\). In our context:

• The inequality \(x + y \leq a\) describes a boundary line that forms the hypotenuse of the triangle.
• The inequalities \(x \geq 0\) and \(y \geq 0\) keep the region confined to the first quadrant.

By plotting these constraints, the graphical representation clarifies the problem and helps determine the limits of integration. Visual tools are invaluable for ensuring that you're accurately capturing the intended region for your calculations.