Problem 34 - University Calculus: Early Transcendentals

Question

In polar coordinates, the average value of a function over a region \(R\) (Section 14.3 ) is given by $$\frac{1}{\operatorname{Area}(R)} \iint_{R} f(r, \theta) r d r d \theta$$ Find the average height of the (single) cone \(z=\sqrt{x^{2}+y^{2}}\) above the disk \(x^{2}+y^{2} \leq a^{2}\) in the \(x y\)-plane.

Step-by-Step Solution

Answer

The average height is \( \frac{2a}{3} \).

Step 1: Understand the Problem Statement

We need to find the average height of the cone defined by \( z=\sqrt{x^{2}+y^{2}} \) above a disk given by \( x^{2}+y^{2} \leq a^{2} \) in the \( xy \)-plane. The average value is calculated using the given formula for regions described in polar coordinates.

Step 2: Convert Cartesian to Polar Coordinates

In polar coordinates, \( x = r\cos\theta \) and \( y = r\sin\theta \). The region \( R \) is the disk \( x^{2}+y^{2} \leq a^{2} \), which translates to \( 0 \leq r \leq a \) and \( 0 \leq \theta \leq 2\pi \) in polar coordinates. The function \( z = \sqrt{x^2 + y^2} \) simplifies to \( z = r \) in polar coordinates.

Step 3: Determine the Area of Region R

The area of the disk \( R \) in polar coordinates is \( \operatorname{Area}(R) = \pi a^2 \), since it's a disk of radius \( a \).

Step 4: Set Up the Double Integral

To find the average value, use the formula: \[\frac{1}{\pi a^2} \iint_{R} r \cdot r \, d r \, d \theta = \frac{1}{\pi a^2} \int_{0}^{2\pi} \int_{0}^{a} r^2 \, d r \, d \theta.\]

Step 5: Evaluate the Integral with Respect to r

Compute the inner integral:\[\int_{0}^{a} r^2 \, dr = \left[ \frac{r^3}{3} \right]_{0}^{a} = \frac{a^3}{3}.\]

Step 6: Evaluate the Integral with Respect to θ

Substitute the result from Step 5 into the outer integral:\[\frac{1}{\pi a^2} \int_{0}^{2\pi} \frac{a^3}{3} \, d\theta = \frac{a^3}{3\pi a^2} \int_{0}^{2\pi} \, d\theta = \frac{a}{3\pi} \cdot 2\pi.\]

Step 7: Simplify the Expression

Simplify the result:\[\frac{2a}{3}.\]

Step 8: Conclude the Solution

The average height of the cone above the disk is \( \frac{2a}{3} \).

Key Concepts

Polar Coordinates

Polar coordinates offer a unique way of describing points in a plane using a distance and an angle. Unlike Cartesian coordinates which use \(x\) and \(y\) to denote positions, polar coordinates use \(r\) (the radial distance from the origin) and \(\theta\) (the angle from the positive x-axis). This system is especially useful when dealing with problems involving circular or rotational symmetry.

The conversion between Cartesian and polar coordinates is based on these relationships:

• \(x = r \cos \theta\)
• \(y = r \sin \theta\)

These equations allow us to translate any point from one coordinate system to the other.
In multivariable calculus, polar coordinates simplify the computation of areas and integrals over circular regions because the descriptions become more straightforward. This can make integrations "over a disk" a lot easier to handle because the circle's geometry aligns with the natural radial and angular components of polar coordinates.

Double Integral

Double integrals extend the idea of summation across two dimensions, allowing us to calculate volumes, average values, and areas across an entire surface rather than just a line or a curve.

These integrals build up the total value over a defined area in the plane by breaking it down into infinitesimally small parts. For a function \(f(x, y)\) over a region \(R\), the double integral is represented as:\[\iint_R f(x, y) \, dA\]where \(dA\) can be thought of as a tiny piece of area over which the function is defined.
Double integrals can also be translated into polar coordinates when dealing with circular regions. Here \(dA\) becomes \(r \, dr \, d\theta\), and the function under the integral changes from \(f(x, y)\) to \(f(r, \theta)\).

In solving exercises like finding the average value of a function over a disk, the double integral helps by calculating the aggregate effect of the function over every tiny part of that area.

Average Value of a Function

The average value of a function over a region gives a sense of the typical "height" or "output" of that function when spread out over the specified region.

The formula for the average value in polar coordinates over a region \(R\) involves dividing the total volume calculated by a double integral by the area of the region itself:\[\text{Average value} = \frac{1}{\text{Area}(R)} \iint_R f(r, \theta) \, dA\]This concept is similar to averaging a list of numbers but extends to functions over a plane.

In the context of the example exercise, the average height of a cone over a disk translates into efficiently evaluating how the height changes across the disk's area and equating that to a single constant value which best represents all those variations. After setting up the double integral to find this, computation of two integrals (one for \(r\) and one for \(\theta\)) is required to determine the exact average.

Integration in Polar Coordinates

Integration in polar coordinates is particularly useful when dealing with circular or spherical regions. The change of variables from cartesian \((x, y)\) to polar \((r, \theta)\) coordinates transforms the integration approach to fit the symmetry of the problem more naturally.

In polar coordinates, the differential element \(dA\) changes to \(r \, dr \, d\theta\), which matches the circular geometry better:

• \(r\) is the radial component representing how far from the center each point is.
• \(dr\) and \(d\theta\) represent the small changes in radius and angle, respectively.

For example, when integrating to find the average height of a cone defined by \(z = r\) over a disk, the integration process becomes better aligned with the shape of the region when using \(\int_0^{2\pi} \int_0^a r^2 \, dr \, d\theta\).

This kind of setup simplifies what can be a complex multidimensional problem into more manageable parts, allowing us to effectively handle such integrals with respect to circular symmetry.