Problem 20 - University Calculus: Early Transcendentals

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 (0.25) is larger than the average value over the quarter circle (approximately 0.159).

Step 1: Understand the Problem

We need to compare the average values of the function \(f(x, y) = xy\) over two regions: a square and a quarter circle in the first quadrant. The square is defined by \(0 \leq x \leq 1\) and \(0 \leq y \leq 1\), while the quarter circle is defined by \(x^2 + y^2 \leq 1\).

Step 2: Calculate the Average Value over the Square

To find the average value over the square, we integrate \(f(x, y) = xy\) over the square region and then divide by the area of the square. The integral is given by:\[\int_0^1 \int_0^1 xy \, dy \, dx\]Evaluating the integral,\[\int_0^1 \left( \left[\frac{1}{2}xy^2\right]_0^1 \right) \, dx = \int_0^1 \frac{1}{2}x \, dx = \left[\frac{1}{4}x^2\right]_0^1 = \frac{1}{4}\]The area of the square is \(1\), so the average value is simply \(\frac{1}{4}\).

Step 3: Calculate the Average Value over the Quarter Circle

For the quarter circle, we need to integrate \(f(x, y) = xy\) over the region \(x^2 + y^2 \leq 1\) in the first quadrant and then divide by the area of the quarter circle. The integral is:\[\int_0^1 \int_0^{\sqrt{1-x^2}} xy \, dy \, dx\]Evaluating this requires polar coordinates:Let \(x = r\cos\theta\), \(y = r\sin\theta\), where \(r\leq 1\) and \(0 \leq \theta \leq \frac{\pi}{2}\). The original integral becomes\[\int_0^{\frac{\pi}{2}} \int_0^1 r^2 \cos\theta \sin\theta \, r \, dr \, d\theta\]Simplifying and evaluating:\[\int_0^{\frac{\pi}{2}} \cos\theta \sin\theta \, d\theta \times \frac{1}{4} = \frac{1}{4} \times \frac{1}{2} = \frac{1}{8}\]The area of the quarter circle is \(\frac{\pi}{4}\), thus the average value is \(\frac{1}{8}\times \frac{4}{\pi} = \frac{1}{2\pi}\approx 0.159\).

Step 4: Compare Both Average Values

Now that we have both average values: \(\frac{1}{4} = 0.25\) from the square and approximately \(0.159\) from the quarter circle, we compare them. Clearly, \(0.25 > 0.159\).

Key Concepts

Double Integration

Double integration is a powerful technique in calculus used to evaluate the integral of a function in two-dimensional space. It involves integrating with respect to one variable, holding the other constant, and then integrating the result again with respect to the second variable.
In the context of our exercise, we calculate the average value of a function over a defined two-dimensional region. The formula for double integration over a region \( R \) for a function \( f(x,y) \) is given by:

• \( \iint_R f(x,y) \, dx \, dy \)

When carrying out double integration:

• First, select the limits of integration based on the region being considered.
• Next, integrate the function with respect to one variable, treating the other as a constant.
• Finally, integrate the result with respect to the second variable.

For instance, in our exercise, we integrated the function \( f(x, y) = xy \) over the square by integrating with respect to \( y \) first and then with respect to \( x \).
This process allows us to calculate the total quantity, such as area, mass, or in this case, the average value of a function over a particular region.

Polar Coordinates

Polar coordinates are particularly useful when dealing with circular or radial symmetric regions. In polar coordinates, a point in the plane is identified by:

• \( r \), the distance from the origin to the point
• \( \theta \), the angle from the positive x-axis to the point

In our exercise, we used polar coordinates to integrate over a quarter circle. This choice simplified the problem because:

• The quarter circle, defined by \( x^2 + y^2 \leq 1 \), translates to \( r \leq 1 \) and \( 0 \leq \theta \leq \frac{\pi}{2} \) in polar coordinates.
• The integrand involving \( x \) and \( y \) turns into a simpler expression involving \( r \) and trigonometric functions.

To convert \( f(x, y) = xy \) into polar coordinates, we substitute \( x = r\cos\theta \) and \( y = r\sin\theta \). This yields a new integrand \( r^2 \cos\theta \sin\theta \).
This change not only simplifies the integration but also leverages the symmetry of the circular region, making computations more straightforward.

Area Comparison

Area comparison involves examining the average values over different regions to draw meaningful conclusions. In our exercise, we compared the average value of \( f(x,y) \) over a square and a quarter circle.
Here's how you approach such a comparison:

• Calculate the average value over each region. This involves integrating the function over the entire region and dividing by the area of that region.
• For the square \(0 \leq x \leq 1, 0 \leq y \leq 1\), the area is straightforward to find as \(1\), and the double integration produced an average value of \(0.25\).
• For the quarter circle, after converting the integrand into polar coordinates, we found the average value was approximately \(0.159\).
• Compare the computed average values directly to determine which region exhibits a larger function average.

Through this exercise, one noticeable fact emerges: despite potential complexities in the shapes of these regions, calculating their average values can offer insightful comparisons and reveals which part tends to have higher or lower values of the function. This technique finds significant applications in areas like physics, economics, and environmental studies, where one often wishes to know where a particular function behaves most "strongly" over a region.