Problem 29 - Thomas Calculus

Question

Average height of a hemisphere Find the average height of the hemisphere \(z=\sqrt{a^{2}-x^{2}-y^{2}}\) above the disk \(x^{2}+y^{2} \leq a^{2}\) in the \(x y\) -plane.

Step-by-Step Solution

Answer

Average height of the hemisphere is \(\frac{2a}{3}\).

Step 1: Understand the Problem

We need to find the average height of a hemisphere given by the function \(z = \sqrt{a^2 - x^2 - y^2}\) above the circular disk \(x^2 + y^2 \leq a^2\) in the \(xy\)-plane. The hemisphere's height at any point \((x, y)\) is \(z\), and we are tasked with finding the average of these heights over the entire disk.

Step 2: Set Up the Integral for Average Value

The average value of a function \(z\) over a region \(R\) is given by \[\frac{1}{A(R)} \int\int_R z\, dA\], where \(A(R)\) is the area of the region \(R\). Here, \(z = \sqrt{a^2 - x^2 - y^2}\) and \(R\) is the area of the circle \(x^2 + y^2 \leq a^2\). The area \(A(R)\) is \(\pi a^2\).

Step 3: Convert to Polar Coordinates

In polar coordinates, \(x = r\cos\theta\) and \(y = r\sin\theta\), with \(dA = r\, dr\, d\theta\). The region limits convert to \(0 \leq r \leq a\) and \(0 \leq \theta \leq 2\pi\). The function \(z\) becomes \(z = \sqrt{a^2 - r^2}\).

Step 4: Evaluate the Double Integral

Set up the double integral for the average value: \[\int_0^{2\pi} \int_0^a \sqrt{a^2 - r^2}\, r\, dr\, d\theta\]. First, integrate with respect to \(r\): \(\int_0^a \sqrt{a^2 - r^2}\, r\, dr\). Let \(u = a^2 - r^2\), then \(du = -2r\, dr\). Evaluating the integral, you find \(-\frac{1}{3} (a^2 - r^2)^{3/2} \bigg|_0^a = \frac{a^3}{3}\). Now, integrate with respect to \(\theta\): \(\int_0^{2\pi} \frac{a^3}{3}\, d\theta = \frac{2\pi a^3}{3}\).

Step 5: Compute the Average Height

Using the formula for average value, \[\text{Average Height} = \frac{1}{\pi a^2} \cdot \frac{2\pi a^3}{3} = \frac{2a}{3}.\] So, the average height of the hemisphere is \(\frac{2a}{3}\).

Key Concepts

Double Integral

A double integral is an extension of a regular integral to calculate areas and volumes over two-dimensional regions. Think of it as stacking up an infinite number of tiny rectangles to find the total sum. When you perform a double integral over a region, you're essentially summing up function values over a surface. For our hemisphere problem, the double integral helps calculate the sum of heights over a circular disk. This sum, when divided by the disk's area, gives the average height. In our example, the double integral setup is \( \int_{0}^{2\pi} \int_{0}^{a} \sqrt{a^2 - r^2} \, r \, dr \, d\theta\). Each element \(r \, dr \, d\theta\) represents a tiny area segment over which we average the "heights" represented by \(z = \sqrt{a^2 - r^2}\).

Polar Coordinates

Polar coordinates provide an alternative way of representing points on a plane using radius and angle, instead of \(x\) and \(y\). They are especially useful for circular regions, making integration over circular areas more straightforward. In our hemisphere integration, switching to polar coordinates simplifies the process because the region of interest is a disk, a natural circular shape. The transformation used is \(x = r \cos \theta\) and \(y = r \sin \theta\), which means any point on the disk can be defined by how far it is from the center \((r)\) and the angle \(\theta\). This change simplifies our integral limits to \(0 \leq r \leq a\) and \(0 \leq \theta \leq 2\pi\), helping us evaluate the double integral efficiently.

Hemisphere

A hemisphere is one-half of a full sphere. In our problem, the hemisphere is the part above the \(xy\)-plane. Its surface is described mathematically by \(z = \sqrt{a^2 - x^2 - y^2}\), meaning each point on it satisfies this equation. Imagine the full sphere as a perfect globe, then slicing it along the equator gives a hemisphere. The hemisphere's geometry often makes it necessary to use techniques like switching to polar coordinates or integrating over a specified region, as shown in our integration problem. The symmetry and simplicity of a hemisphere allow these mathematical strategies to be particularly efficient in deriving results, like average height, even though the shapes are curved and complex.

Calculus Problem-Solving

Calculus problem-solving is about breaking down complex problems into manageable steps using algebra and calculus principles. Solving for the average height of a hemisphere through calculus involves these steps:

• Understanding the problem: Recognizing the equation describing the hemisphere and the region over which the average is found.
• Setting up the problem: Using concepts like double integrals to express the solution mathematically.
• Transforming coordinates: Simplifying complex regions using polar coordinates.
• Solving the integrals: By performing integrations first with respect to \(r\), then \(\theta\).
• Arriving at a solution: Finding that the average height is \(\frac{2a}{3}\).

By following this structured approach, even complex geometrical problems become understandable and manageable, demonstrating the power of calculus in solving real-world problems.