Problem 90 - Calculus Early Transcendentals

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 \(z=a^{2}-x^{2}-y^{2}\) over the region \(R=\left\\{(x, y): x^{2}+y^{2} \leq a^{2}\right\\},\) where \(a>0\).

Step-by-Step Solution

Answer

Answer: The average value of the function is (a^3)/6.

Step 1: Find the area of region R

The region R is given by \(\{(x, y): x^2 + y^2 \leq a^2\}\). This represents the disk of radius \(a\) centered at the origin (0,0). In other words, the region is a circle with radius \(a\). The area of a circle with a radius \(a\) is given by \(A = \pi a^2\). So the area of the region R is \(\pi a^2\).

Step 2: Set up the double integral

Now we have to set up the double integral of the function \(f(x, y) = a^2 - x^2 - y^2\) over the region R. We will use polar coordinates for the integration since the region is a disk centered at the origin. To convert to polar coordinates, we use the following transformations: \(x = r \cos (\theta)\) \(y = r \sin (\theta)\) \(dA = r dr d\theta\) The limits of integration for \(r\) and \(\theta\) will be \(0 \leq r \leq a\) and \(0 \leq \theta \leq 2\pi\), respectively, covering the entire disk of radius \(a\). Therefore, the double integral is given by: \(\iint_{R} f(x, y) dA = \int_{0}^{2\pi} \int_{0}^{a} (a^2 - r^2 \cos^2(\theta) - r^2 \sin^2(\theta))r dr d\theta\)

Step 3: Evaluate the double integral

Now we are ready to evaluate the double integral: \(\int_{0}^{2\pi} \int_{0}^{a} (a^2 - r^2 \cos^2(\theta) - r^2 \sin^2(\theta))r dr d\theta = \int_{0}^{2\pi} \int_{0}^{a} (a^2 - r^2)r dr d\theta\) First, we integrate with respect to r: \(\int_{0}^{a} (a^2 - r^2)r dr = \left[\frac{1}{3}a^2r^3 - \frac{1}{4}r^4\right]_0^a = \frac{1}{3}a^5 - \frac{1}{4}a^5 = \frac{a^5}{12}\) Next, we integrate with respect to \(\theta\): \(\int_{0}^{2\pi} \frac{a^5}{12} d\theta = \frac{a^5}{12}\left[\theta\right]_{0}^{2\pi} = \frac{a^5}{12}(2\pi - 0) = \frac{a^5 \pi}{6}\) So the double integral \(\iint_{R} f(x, y) dA\) has the value \(\frac{a^5 \pi}{6}\).

Step 4: Compute the average value

Now we can compute the average value using the formula \(\bar{f} = \frac{1}{\text{area of } R} \iint_{R} f(x, y) dA\). We found the area of region R to be \(\pi a^2\), and the value of the double integral as \(\frac{a^5 \pi}{6}\). Therefore, the average value is: \(\bar{f} = \frac{1}{\pi a^2} \cdot \frac{a^5 \pi}{6} = \frac{a^3}{6}\) So, the average value of the function \(z = a^2 - x^2 - y^2\) over the region R is \(\frac{a^3}{6}\).

Key Concepts

Double Integral

A double integral is a powerful tool in calculus used to compute the volume under a surface over a given region. It extends the concept of a single integral, which represents the area under a curve in one dimension, to two dimensions. In the exercise, the double integral \(\iint_{R} f(x, y) dA\) determines the volume under the surface described by the function \(z = a^{2} - x^{2} - y^{2}\) over the disk region \(R\).

Calculating a double integral involves integrating a function of two variables, first with respect to one variable while keeping the other constant, and then integrating the result with respect to the second variable. This yields the accumulated value - in this case, a volume - considering every point over the region of integration. It's like adding up an infinite number of infinitely thin slices of the volume to get the total volume.

The actual calculation requires setting up proper limits of integration that encompass the entire region of interest. Often, these limits are determined by the geometry or by the bounds set within the problem.

Polar Coordinates

Polar coordinates are an alternative to the standard Cartesian coordinate system, where points are defined by an angle and a distance from the origin, rather than by x and y positions. A point \( (x, y) \) in Cartesian coordinates can be translated into polar coordinates \( (r, \theta) \) where \( r \) is the radius - the distance from the origin to the point – and \( \theta \) is the angle between the positive x-axis and the line segment from the origin to the point. For the exercise with the disk region \(R\), polar coordinates are especially useful because the region is a circle, which is naturally described in terms of radii and angles.

By converting to polar coordinates, the integration becomes simpler because the bounds for \( r \) and \( \theta \) are constants reflecting the geometry of the circle. The coordinate transformation also changes the area element \(dA\) from \(dxdy\) to \(r dr d\theta\), which must be included in the integral to account for the 'stretching' of the space as one moves further from the origin.

Disk Integration

Disk integration, a method often used in polar coordinate systems, is ideal for finding the volume under a surface above regions that are disks or part of a disk. This technique relies on the symmetry of the region - since a disk is uniform in all radial directions from its center, it makes integrating over such regions more efficient and straightforward when polar coordinates are used.

In this exercise, the double integral was simplified using disk integration because the region \(R\) is a disk of radius \(a\). The bounds for \(r\) go from 0 to \(a\), covering the distance from the center to any point on the circle, and the bounds for \(\theta\) cover the full rotation around the circle, from 0 to \(2\pi\).

Disk integration exploits this rotational symmetry by allowing the integrals to be set up over a clear range of radii and angles, neatly capturing the disk's area. It's like mentally slicing the disk into concentric rings and summing their contributions to get the total. This method streamlines the integration process, ensuring that all points within the region are accounted for without complex adjustments for the shape's boundaries.