Problem 32 - Calculus Early Transcendentals

Question

Average value Compute the average value of the following functions over the region \(R\). $$f(x, y)=4-x-y ; R=\\{(x, y): 0 \leq x \leq 2,0 \leq y \leq 2\\}$$

Step-by-Step Solution

Answer

Answer: The average value of the function \(f(x, y) = 4 - x - y\) over the region R is 1.

Step 1: Set up the double integral

Based on the given region R, we can set up the double integral of the function \(f(x, y)\) as follows: $$ \frac{1}{4} \int_{0}^{2} \int_{0}^{2} (4 - x - y) dy\,dx $$

Step 2: Evaluate the inner integral

We will first evaluate the inner integral with respect to y: $$ \frac{1}{4} \int_{0}^{2} \left[\int_{0}^{2} (4 - x - y)\, dy\right]dx $$ Integration of the inner integral yields: $$ \frac{1}{4} \int_{0}^{2} \left[4y - xy - \frac{1}{2}y^2\right]_{0}^{2}dx $$ Plugging in the upper and lower limits for y, we get: $$ \frac{1}{4} \int_{0}^{2} (4 - x)\,dx $$

Step 3: Evaluate the outer integral

Now, we will evaluate the outer integral with respect to x: $$ \frac{1}{4} \left[\int_{0}^{2} (4 - x)\, dx\right] $$ Integration of the outer integral yields: $$ \frac{1}{4} \left[4x - \frac{1}{2}x^2\right]_{0}^{2} $$ Plugging in the upper and lower limits for x, we get: $$ \frac{1}{4} \left[(4(2) - \frac{1}{2}(2)^2) - (4(0) - \frac{1}{2}(0)^2)\right] $$

Step 4: Compute the average value of the function

Now we simplify the expression to get the average value of the function \(f(x, y)\) over the region R: $$ \frac{1}{4}(8 - 4) = \frac{1}{4}(4) = \boxed{1} $$ So, the average value of the function \(f(x, y) = 4 - x - y\) over the region R is 1.

Key Concepts

Average Value

In the realm of multivariable calculus, finding the average value of a function over a particular region involves integrating that function over the region and then dividing by the measure of the region.
This concept can be visualized as finding the "height" of the function that would provide an equal volume if spread evenly across the region.

• For this exercise, our function is given as \( f(x, y) = 4 - x - y \).
• The region \( R \) is defined by \( 0 \leq x \leq 2 \) and \( 0 \leq y \leq 2 \), which represents a square in the xy-plane.
• The entire area of this square is \( (2 - 0) \times (2 - 0) = 4 \).

To determine the average value, we first integrate the function over this square (i.e., use a double integral), and then divide the result by the area of the square. This gives us: \[ \frac{1}{4} \int_{0}^{2} \int_{0}^{2} (4 - x - y) dy\,dx \].
The resolution of this integral, as shown in the step-by-step solution, leads to a final average value of 1.

Region of Integration

The region of integration in multivariable calculus is crucial for setting up a double integral. It defines where the integration will be applied on the coordinate plane, specified by a set of boundaries.
For rectangular coordinates, these boundaries are usually parallel to the axes, making them easier to visualize and compute.

• In our exercise, the region \( R \) is \( \{ (x, y) : 0 \leq x \leq 2, 0 \leq y \leq 2 \} \).
• This forms a square with vertices at \((0, 0)\), \((2, 0)\), \((2, 2)\), and \((0, 2)\).
• When handling integrals, knowing the region ensures that we are computing the integral over the correct area.

The integration limits used in the double integral come directly from these bounds. The integral is first computed over one variable while treating the other as constant, and then over the remaining variable. This split in integration is facilitated by the presumed independence within rectangles or rectangular regions.

Multivariable Calculus

Multivariable calculus expands the techniques of single-variable calculus to functions of multiple variables.
This branches into concepts like partial derivatives and multiple integrals, which are pivotal for analyzing and understanding complex physical systems.

• Partial derivatives involve taking the derivative of a multivariable function with respect to one variable while keeping others constant.
• Double and triple integrals allow for calculating the cumulative effect of a function over an area (or volume) in the multivariable context.

In this exercise, we are dealing with a double integral which is a staple tool in multivariable calculus. The benefit lies in being able to calculate the average value, or the total volume under a surface, across a defined region. This helps in various fields, from physics to engineering, where analyzing changing variables across space is essential. The ability to handle and compute these integrals sets the foundation for more advanced topics in calculus and applied mathematics.