Problem 33 - Thomas Calculus

Question

In polar coordinates, the average value of a function over a region \(R\) (Section 15.3\()\) is given by $$\frac{1}{\text { Area(R) }} \iint_{R} f(r, \theta) r d r d \theta$$ Average height of a hemisphere Find the average height of the hemispherical surface \(z=\sqrt{a^{2}-x^{2}-y^{2}}\) above the disk \(x^{2}+y^{2} \leq a^{2}\) in the \(x y\) -plane.

Step-by-Step Solution

Answer

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

Step 1: Convert to Polar Coordinates

Start by expressing the hemisphere and the disk in polar coordinates. Given a hemisphere expressed as \(z = \sqrt{a^2 - x^2 - y^2}\), convert \(x\) and \(y\) using \(x = r\cos\theta\) and \(y = r\sin\theta\). The disk \(x^2 + y^2 \leq a^2\) becomes \(r \leq a\) in polar coordinates.

Step 2: Set Up the Integral

The problem requires finding the average value of the height \(z\) over the disk. The integral becomes \(\iint_R z \cdot r \, dr \, d\theta\) where \(z = \sqrt{a^2 - r^2}\). Set up the integral over \(0 \leq r \leq a\) and \(0 \leq \theta \leq 2\pi\).

Step 3: Calculate the Area of the Region

First, calculate the total area of the disk \(R\) in the \(xy\)-plane. Since the disk is a circle with radius \(a\), its area is given by \(\pi a^2\).

Step 4: Evaluate the Double Integral

Evaluate the integral \(\int_0^{2\pi} \int_0^a \sqrt{a^2 - r^2} \cdot r \, dr \, d\theta\). Solve the inner integral with respect to \(r\) first, followed by the outer integral with respect to \(\theta\).

Step 5: Calculate the Average Value

Find the average value by dividing the result of the double integral by the area of the region. This is done using the formula \(\frac{1}{\pi a^2} \int_0^{2\pi} \int_0^a \sqrt{a^2 - r^2} \cdot r \, dr \, d\theta\).

Key Concepts

Hemispherical Surface

In mathematics, a hemispherical surface represents half of a sphere. Imagine slicing a sphere along its diameter; the resulting surface is called a hemisphere. These surfaces are often encountered in geometry and calculus when dealing with three-dimensional objects.
In this exercise, we are considering the upper half of a sphere, which forms a hemisphere. The hemisphere has an equation given by \( z = \sqrt{a^2 - x^2 - y^2} \), where \(z\) is the height at any point \((x, y)\). The equation is derived from the general equation for a sphere: \(x^2 + y^2 + z^2 = a^2\). By solving this for \(z\), we find the hemisphere's surface.
This example focuses on the hemisphere above a circular disk on the \(xy\)-plane, making it a perfect candidate for using polar coordinates, due to the symmetry and simplicity in calculation.

Average Value

The average value of a function provides a simple way to compute the typical value of a set of numbers or a continuous function over a region. When applied to functions in calculus, the formula for the average value involves integrating the function over a region and then dividing by the region's area.
For the hemispherical surface, the goal is to find the average height \(z\) of the hemisphere above the disk. The average value formula is \(\frac{1}{\text{Area(R)}} \iint_{R} f(r, \theta) \, r \, dr \, d\theta\), where \(R\) is the region and \(f(r, \theta)\) is the height function in polar coordinates. This helps in examining the spread of different heights across the region and finding one representative height value.
Computing this average height involves setting up a double integral and dividing it by the area of the circular disk.

Double Integral

A double integral extends the concept of integration to two dimensions, allowing for the evaluation of functions over a defined area. In the case of polar coordinates, these integrals are often set up with \(r\) and \(\theta\) as variables.
In this exercise, the double integral takes the form \(\iint_R z \cdot r \, dr \, d\theta\), where \(z = \sqrt{a^2 - r^2}\). The function \(z\) describes the height of the hemisphere. Essentially, this double integral computes the total sum of heights over the disk-shaped region \(R\).
The limits for \(r\) range from 0 to \(a\), and for \(\theta\), from 0 to \(2\pi\). The process involves integrating first with respect to \(r\) within these bounds, then \(\theta\), ensuring that each part contributes properly to finding the integral over the entire region.

Polar Coordinate Transformation

Polar coordinate transformation is a powerful tool that simplifies problems involving circular or spherical symmetry. It is especially useful when working with regions defined by circles or arcs, such as the hemisphere in this exercise.
In polar coordinates, any point in the plane is described by a distance from the origin, \(r\), and an angle from a reference direction, \(\theta\). This transformation turns Cartesian coordinates \((x, y)\) into polar coordinates \((r\cos\theta, r\sin\theta)\).
For the given problem, transforming the domain into polar coordinates made it easier to manage. The disk \(x^2 + y^2 \leq a^2\) becomes \(r \leq a\). This transformation also alters the computation of integrals, as the Cartesian \(dx \times dy\) differential turns into \(r \, dr \, d\theta\), acknowledging the distance element \(r\) in polar coordinates. This often simplifies the integration process, as seen in the solution.