Question

Let \(R=\\{(x, y): 0 \leq x \leq a, 0 \leq y \leq a\\}\) Find the value of \(a>0\) such that the average value of the following functions over \(R\) is zero. $$f(x, y)=x+y-8$$

Short answer

Answer: \(a = 8\).

Step-by-step solution

Finding the expression for the average value of a function over a region

Given the region \(R=\\{(x,y): 0\leq x\leq a, 0\leq y\leq a\\}\) and the function \(f(x, y) = x + y - 8\), the average value of this function over the region \(R\) can be expressed as: $$\overline{f(x,y)} = \frac{1}{A}\int_0^a \int_0^a (x + y - 8) dy dx$$ where \(A\) is the area of the region \(R\). Since \(R\) is a square with side length \(a\), the area of the region is simply \(A = a^2\). Now we need to evaluate the integral to find the average value of \(f(x, y)\) over the region \(R\).

Evaluating the integral

We will evaluate the given iterated integral first with respect to \(y\) and then with respect to \(x\). $$\overline{f(x,y)} = \frac{1}{a^2}\int_0^a \left(\int_0^a (x + y - 8) dy \right) dx $$ First, we find the inner integral with respect to \(y\): $$\int_0^a (x + y - 8) dy = \left[xy + \frac{1}{2}y^2 - 8y\right]_0^a = ax + \frac{1}{2}a^2 - 8a$$ Now, we substitute this result back into the average value expression: $$\overline{f(x,y)} = \frac{1}{a^2}\int_0^a (ax + \frac{1}{2}a^2 - 8a) dx$$ Next, we find the outer integral with respect to \(x\): $$\overline{f(x,y)} = \frac{1}{a^2}\left[\frac{1}{2}ax^2 + \frac{1}{2}a^2 x - 8ax\right]_0^a$$ Evaluating this expression at the limits gives us: $$\overline{f(x,y)} = \frac{1}{a^2}\left(\frac{1}{2}a^3 + \frac{1}{2}a^3 - 8a^2\right) = \frac{1}{a^2}(a^3 - 8a^2)$$

Setting the average value equal to zero and solving for \(a\)

To find the value of \(a\) such that the average value of the function \(f(x, y)\) is zero, we set the expression we derived in Step 2 equal to zero and solve for \(a\): $$\frac{1}{a^2}(a^3 - 8a^2) = 0$$ Multiplying both sides by \(a^2\) gives us: $$a^3 - 8a^2 = 0$$ We can factor out an \(a^2\) term: $$a^2(a - 8) = 0$$ From this expression, we see that \(a\) can either be \(0\) or \(8\). However, we are given that \(a>0\). Thus, the value of \(a\) that makes the average value of \(f(x, y)\) equal to zero over the region \(R\) is: $$a = \boxed{8}$$

Key concepts

Double Integral

The concept of a double integral is crucial for finding the average value of a function over a two-dimensional region. A double integral \[ \int \int_R f(x, y) \, dA \] calculates the accumulated values of a function over a specified area in the xy-plane. Here, the double integral is used to evaluate a function over a region "R". The process involves integrating the function first in terms of one variable, then in terms of another, covering the entire area.
Double integrals are useful because they allow us to handle functions of two variables, accounting for the change across both axes. Particularly in this problem, they enable us to compute an average of the function across the region, useful for applications like center of mass and probability density distributions.

Region of Integration

The region of integration, often designated as "R", is the area over which the function is integrated. In this case, the region \( R = \{(x, y): 0 \leq x \leq a, 0 \leq y \leq a\} \) is a square with side length \( a \), bounded in the xy-plane.

• This problem explores how changing the size of the region (adjusting \( a \)) influences the outcome of complex calculations like averages.
• Determining this region is crucial because it sets the boundaries over which the function is evaluated. The shape and size directly influence the integral's results and, therefore, the eventual calculation of the function's average value.

Understanding the region of integration helps in visualizing the space affected by the integral, ensuring that the mathematical model corresponds to the real-world system or scenario being analyzed.

Factorization

Factorization is a mathematical process in which an expression is rewritten as a product of simpler expressions. In the solution, it emerges when solving for \( a \) by setting the expression to zero and factoring \[ a^3 - 8a^2 = 0 \].This is rewritten as \[ a^2(a - 8) = 0 \]. Factorization reveals the roots or solutions of the equation.

• In our problem, by factoring, we determine that the potential solutions for \( a \) are \( 0 \) or \( 8 \), and due to the constraints \( a > 0 \), \( a \) must be \( 8 \).
• Factorization simplifies complex algebraic expressions, making it easier to identify the key values which affect the solution.

Recognizing when and how to factor effectively transforms complicated expressions into manageable pieces, often revealing solutions that are not immediately evident.

Iterated Integral

The iterated integral refers to solving a double integral by performing the integration in a sequence, one variable at a time. In this exercise, the double integral is split into two separate single integrals, making it an iterated integral.Here, we first integrate with respect to \( y \), and then \( x \):

• Inner integral: \[ \int_0^a (x + y - 8) \, dy = ax + \frac{1}{2}a^2 - 8a \]
• Outer integral: \[ \int_0^a \left(ax + \frac{1}{2}a^2 - 8a\right) \, dx \]

This method breaks complex two-dimensional integration into manageable one-dimensional problems. This simplification is powerful, allowing us to progress step-by-step, each integration peeling away one dimension at a time.
Being proficient with iterated integrals is essential for evaluating functions over specific regions, especially when dealing with real-world problems modeled through multivariable calculus.