Problem 44 - Calculus for Biology and Medicine

Question

Find the volumes of the solids obtained by rotating the region bounded by the given curves about the \(x\) -axis. In each case, sketch the region together with a typical disk element. \(y=\sqrt{1-x^{2}}, y=1, x=1\) (in the first quadrant)

Step-by-Step Solution

Answer

The volume of the solid is \(\frac{\pi}{3}\).

Step 1: Understand the Region

The given curves are \(y=\sqrt{1-x^2}\), \(y=1\), and \(x=1\). We need to find the region in the first quadrant. The curve \(y=\sqrt{1-x^2}\) is the top half of a circle with radius 1 centered at the origin. The line \(y=1\) is a horizontal line, and \(x=1\) is a vertical line. We consider the region in the first quadrant above the x-axis that is bounded by these curves.

Step 2: Set Up the Integral For Volume

To find the volume of the solid formed by rotating this region about the x-axis, we use the disk method. The outer radius of the disk will be \(y=1\) from \(x=0\) to \(x=1\) and the inner radius is the curve \(y=\sqrt{1-x^2}\). The volume \(V\) is given by the integral \[ V = \pi \int_{0}^{1} (1)^2 - (\sqrt{1-x^2})^2 \, dx \].

Step 3: Simplify the Integrand

Simplify the integrand: \[(1)^2 - (\sqrt{1-x^2})^2 = 1 - (1 - x^2) = x^2\]. Thus, our volume integral becomes \[ V = \pi \int_{0}^{1} x^2 \, dx \].

Step 4: Calculate the Integral

Evaluate the integral \(\int_{0}^{1} x^2 \, dx\). Using the power rule for integration, we get \[ \int x^2 \, dx = \frac{x^3}{3} + C \]. Evaluating from 0 to 1, we find \[ \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3} \].

Step 5: Find the Volume

Now multiply the result by \(\pi\) to find the volume: \[ V = \pi \times \frac{1}{3} = \frac{\pi}{3} \]. Thus, the volume of the solid is \(\frac{\pi}{3}\).

Key Concepts

Volume of Solids

Volume of solids refers to the space a three-dimensional object occupies. In geometry, particularly in calculus, finding the volume of solids of revolution is a common task. When a planar region is revolved around a line (often an axis), it creates a solid with a certain volume. Calculating these volumes involves understanding the boundaries of the region being revolved and using calculus techniques to integrate over these boundaries.

A solid of revolution is obtained by rotating a two-dimensional shape around an axis. In our exercise, the region bounded by curves in the first quadrant is rotated around the x-axis to form such a solid. The volume of this solid is found by determining the infinitesimally thin slice areas and integrating these areas with respect to the axis of rotation.

Disk Method

The disk method is a technique for calculating the volume of a solid of revolution. It is particularly useful when the solid is generated by revolving a region around one of the coordinate axes. In our example, the region is revolved around the x-axis.

### How Disk Method Works

• The disk method involves slicing the solid into a series of thin circular disks.
• The volume of each disk is the area of its circular face times its infinitesimal thickness (dx or dy, depending on the axis).
• The radius of each disk is determined by the function from which the region is generated.

In this scenario, the disk method is applied by subtracting the inner radius from the outer radius, squaring them, and integrating along the axis of rotation. This helps in obtaining the total volume by summing the volume of all the disks.

Integral Calculus

Integral calculus is a branch of calculus that focuses on accumulation, helping in determining areas, volumes, and overall sums. It is based on the concept of integration, which involves summing up infinitely small quantities to find a total amount.

### Application to Volume Calculation

• The technique is used to set up an integral that represents the volume of a solid by summing the infinite disks or washers.
• In our exercise, after finding the function expressions for the inner and outer boundaries of the region, the integrand can be simplified using algebraic manipulation.
• This simplification results in a function that can be easily integrated over the required bounds.

In specific, by applying the integral calculus as in the given problem, we integrate the function representing the disk's cross-sectional area to compute the entire solid's volume. This exemplifies how integral calculus serves as a tool to transform geometric problems into analytical ones that can be solved mathematically.