Problem 58 - Calculus of a Single Variable

Question

Using a Sphere Find the area of the zone of a sphere formed by revolving the graph of \(y=\sqrt{r^{2}-x^{2}}, 0 \leq x \leq a\) about the \(y\) -axis. Assume that \(a

Step-by-Step Solution

Answer

The area of the zone of the sphere formed by revolving the given graph about the y-axis is \(A =2 \pi r^{2} \arcsin\left( \frac{a}{r} \right)- 2\pi a \sqrt{r^{2}-a^{2}}\).

Step 1: Understand the shape and compute the differential element

The function represents a sphere of radius r. Upon revolving it over y-axis, the volume of the revolution should be computed. However, the differential volume element here is an infinitely thin ring at a distance x from origin with thickness dx. This ring has area \(dA =2 \pi y dx\) where \(y=\sqrt{r^{2}-x^{2}}\) is the radius of the ring.

Step 2: Setup the Integral

Next, setup the integral for the area of the zone of the sphere which is \(A = \int_{0}^{a} 2\pi y dx = \int_{0}^{a} 2 \pi \sqrt{r^{2}-x^{2}} dx\)

Step 3: Solve the Integral

Evaluate the integral to find the area, this is a standard integral the solution of which is \(A =2 \pi r^{2} \arcsin\left( \frac{a}{r} \right)- 2\pi a \sqrt{r^{2}-a^{2}}\)

Key Concepts

Solid of Revolution

Understanding the concept of a solid of revolution is crucial in visualizing three-dimensional shapes created by rotating a two-dimensional curve around an axis. In our exercise, the given curve is part of a sphere's equation, specifically the upper half represented by the function \(y = \sqrt{r^{2} - x^{2}}\).

When this curve is rotated around the \(y\)-axis, it generates a spherical zone—a portion of the sphere's surface bounded by two parallel planes. To visualize this better, imagine slicing through a sphere with two parallel cuts. The band that's contained between these cuts represents the zone of the sphere involved in the area calculation. This concept is a foundation for many real-world applications, such as designing bowls or domes, and is also a key part of understanding advanced geometry and calculus.

Integration in Calculus

Integration is a fundamental tool in calculus used for finding areas, volumes, and other related quantities. It is the reverse process of differentiation, often thought of as 'anti-differentiation.'

The area problem we're tackling involves an integral that calculates the surface area of a spherical zone. Specifically, we use the definite integral to accumulate all the infinitely small rings of surface area from the final shape. The integral \(\int_{0}^{a} 2\pi y dx\) sums the area of these rings along the \(x\)-axis from 0 to \(a\), where \(a < r\) and \(r\) is the radius of the sphere.

This process of integrating to find areas is crucial in various scientific and engineering fields, such as calculating the materials needed for constructing curved surfaces or in understanding fluid dynamics.

Spherical Coordinates

Spherical coordinates are an alternative to Cartesian coordinates for representing points in three-dimensional space, especially useful for dealing with objects like spheres. Instead of using \(x, y, z\) coordinates, spherical coordinates use \(\rho, \theta, \phi\), where \(\rho\) represents the radial distance from the origin, \(\theta\) is the angle in the \(xy\)-plane from the positive \(x\)-axis, and \(\phi\) is the angle from the positive \(z\)-axis to the point.

In the context of our problem, the spherical coordinates simplify the visualization and computation of the areas and volumes for spherical objects. This understanding is instrumental in fields like physics, where spherical coordinates are often used to solve problems involving gravitational fields, and in geography for pinpointing locations on Earth.

Volume and Surface Area of Spheres

For a full sphere with radius \(r\), the formulas for volume \(V = \frac{4}{3}\pi r^{3}\) and surface area \(A = 4\pi r^{2}\) are common knowledge. However, finding the area of a specific zone or 'belt' on a sphere can be a bit more challenging and is the focus of our exercise.

The solution to the textbook problem elegantly shows how to determine this partial area. It's important to understand that while the full surface area of a sphere includes all points that are a distance \(r\) from the center, thepartial area of the sphere, or any curved surface, relies on an understanding of calculus and integration techniques to combine all the elemental ring areas formed during the rotation of the curve.

The ability to calculate such specific areas is incredibly valuable in designing and manufacturing various spherical objects, from lenses in telescopes to industrial tanks and pressure vessels.