Problem 20 - Thomas Calculus

Question

Which do you think will be larger, the average value of \(f ( x , y ) = x y\) over the square \(0 \leq x \leq 1,0 \leq y \leq 1 ,\) or the average value of \(f\) over the quarter circle \(x ^ { 2 } + y ^ { 2 } \leq 1\) in the first quadrant? Calculate them to find out.

Step-by-Step Solution

Answer

The average value over the square is larger.

Step 1: Define the Average Value Formula for a Function over a Region

The average value \( \bar{f} \) of a function \( f(x, y) \) over a region \( D \) is given by \[ \bar{f} = \frac{1}{\text{Area of } D} \int \int_{D} f(x, y) \ dA. \] For different regions, the integration limits and area will differ.

Step 2: Calculate the Average Value over the Square

For the square region defined by \( 0 \leq x \leq 1 \) and \( 0 \leq y \leq 1 \), the area is 1. So, the average value is \[ \bar{f}_{\text{square}} = \int_{0}^{1} \int_{0}^{1} xy \, dy \, dx. \] Compute the inner integral: \( \int_{0}^{1} xy \, dy = \frac{1}{2}x \). Then, the outer integral: \[ \int_{0}^{1} \frac{1}{2}x \, dx = \frac{1}{4}. \] Thus, \( \bar{f}_{\text{square}} = \frac{1}{4}. \)

Step 3: Calculate the Average Value over the Quarter Circle

For the region defined by the quarter circle \( x^2 + y^2 \leq 1 \), the area is \( \frac{\pi}{4} \). Thus, the average value is \[ \bar{f}_{\text{circle}} = \frac{1}{\frac{\pi}{4}} \int_{0}^{1} \int_{0}^{\sqrt{1-x^2}} xy \, dy \, dx. \] Compute the inner integral: \( \int_{0}^{\sqrt{1-x^2}} xy \, dy = \frac{x}{2} (1-x^2) \). Then the outer integral: \[ \int_{0}^{1} \frac{x}{2} (1-x^2) \, dx = \frac{1}{8}. \] Thus, \( \bar{f}_{\text{circle}} = \frac{1}{8} \cdot \frac{4}{\pi} = \frac{1}{2\pi}. \)

Step 4: Compare the Average Values

We found the average value over the square to be \( \frac{1}{4} \) and the average value over the quarter circle to be \( \frac{1}{2\pi} \). Since \( \frac{1}{4} \approx 0.25 \) and \( \frac{1}{2\pi} \approx 0.159 \), the average over the square is larger.

Key Concepts

Multiple Integrals

When dealing with functions of two or more variables, multiple integrals come into play. These integrals extend the idea of integration to higher dimensions. Think of a single integral as summing along a line, like the length of a rope, while double integrals sum over an area, like finding the weight of a sheet.

In the exercise, we dealt with a function of two variables, represented as \(f(x, y) = xy\), which we integrated over two different regions — a square and a quarter circle. To find the average value, we applied double integrals to these regions.

Here, the double integral is expressed as \(\int\int_D f(x, y) \, dA\) where \(dA\) represents a small element of area in our region of integration \(D\). Such multiple integrals help us find things like mass, volume, and average value over a specified region.

Calculus

Calculus, a fundamental branch of mathematics, allows us to study change and motion through derivatives and integrals. In this context, integrations help us evaluate the total accumulation of quantities across dimensions.

For calculating areas or evaluating average values over regions in two-dimensional spaces, calculus gives us tools like double integrals. These integrals allow us to evaluate the function \(f(x, y) = xy\) across different domains.

Calculus involves setting up integrals carefully with respect to the defined limits. These limits describe the region of interest, like our square \(0 \leq x, y \leq 1\) or the quarter circle \(x^2 + y^2 \leq 1\). These integrals were instrumental in solving the provided exercise correctly and understanding how the average values were derived.

Regions of Integration

To perform integrations over a function of several variables, one needs to specify the region of integration. This region creates boundaries where the function is evaluated.

In the provided exercise, we had two distinct regions of integration. The first was a square, defined by bounds from \(0\) to \(1\) on both axes. The second was a quarter circle with radius \(1\), confined within the first quadrant defined by inequality \(x^2 + y^2 \leq 1\).

Choosing the region influences the integration's difficulty and strategies. For a square, the bounds are straightforward and suited for Cartesian coordinates. The quarter circle requires switching to polar coordinates for simplification, though it was handled in Cartesian in the exercise. Defining clear regions helps apply integration techniques effectively, fostering a better understanding of results like area or average values.