Problem 21 - Thomas Calculus in SI Units

Question

Find the volumes of the solids generated by revolving the regions bounded by the lines and curves about the \(x\) -axis. $$y=\sqrt{9-x^{2}}, \quad y=0$$

Step-by-Step Solution

Answer

The volume is \(36\pi\).

Step 1: Understand the Problem

We are tasked with finding the volume of a solid created by revolving the region bounded by the curve \(y = \sqrt{9-x^2}\) and the line \(y = 0\) about the \(x\)-axis. This curve is a semi-circle of radius 3 centered at the origin.

Step 2: Identify the Boundaries

The equation \( y = \sqrt{9 - x^2} \) describes the upper half of a circle with radius 3, centered at the origin. The line \( y = 0 \) is the \(x\)-axis which forms the boundary along with the curve. The interval for \(x\) is from -3 to 3 as derived from the full circle equation \(x^2 + y^2 = 9\).

Step 3: Set Up the Volume Integral

We use the disk method to set up the integral to find the volume. The volume \( V \) is given by \[ V = \pi \int_{-3}^{3} (\sqrt{9 - x^2})^2 \, dx \]Simplifying this, we get:\[ V = \pi \int_{-3}^{3} (9 - x^2) \, dx \]

Step 4: Evaluate the Integral

Calculate the definite integral:\[ V = \pi \left[ 9x - \frac{x^3}{3} \right]_{-3}^{3} \]First, evaluate at the upper limit (3):\[ 9(3) - \frac{3^3}{3} = 27 - 9 = 18 \]Next, evaluate at the lower limit (-3):\[ 9(-3) - \frac{(-3)^3}{3} = -27 + 9 = -18 \]

Step 5: Calculate the Final Volume

Subtract the evaluated boundaries:\[ V = \pi (18 - (-18)) = \pi \times 36 = 36\pi \]

Key Concepts

Disk Method

When dealing with the volume of solids generated by revolving a region around an axis, the Disk Method is an essential tool. This method simplifies finding volumes by graphical revolution into a manageable calculus problem.

The process involves slicing the solid into thin disks perpendicular to the axis of revolution. Each disk has a thickness, \Delta x\, and a radius that corresponds to the value of the function at a particular point. The volume of each disk resembles that of a cylinder and can be calculated using \pi r^2 \Delta x\, where \pi r^2\ is the area of the circle representing the face of the disk.

By integrating, we sum up the volumes of these infinitesimally thin disks over the given interval. For our specific problem:

• The function, \( y = \sqrt{9 - x^2} \), represents the radius of each disk.
• The interval of integration is from \[ -3 \] to \[ 3 \].

This method leads to an integral expression that, once evaluated, gives the total volume of the solid.

Integral Calculus

Integral calculus is the branch of mathematics focused on the accumulation of quantities and the areas under and between curves. It is the backbone of solving problems involving the Disk Method.

In our context, integral calculus helps us find the volume by summing continuously varying quantities. By setting up an integral, we compute the total volume of a solid formed by revolution. The function within the integral represents the geometry of the cross-sections, and the limits of the integral represent the interval over which this revolution occurs.

Key elements of integral calculus include:

• The integral sign (∫) which denotes the operation of integration.
• The differential, \dx\, indicating the variable of integration.
• The integrand, which is the function being integrated. In this case, \( (9-x^2) \) after simplifying \( (\sqrt{9-x^2})^2 \).

This framework allows us to compute aspects of continuous geometric figures, like volumes, comprehensively.

Definite Integral

A definite integral is a concept in calculus that computes the net area under a curve between two specific limits. It is key to solving problems involving rotated solids, like the one in this exercise.

To find the volume of the solid, we use the definite integral from \( -3 \) to \( 3 \). The purpose of this integral is to calculate the total accumulated value of the function \( (9 - x^2) \) over this range. This range represents the bounds of the revolution around an axis.

Important characteristics of a definite integral:

• It is bounded by upper and lower limits (here, \( -3 \) and \( 3 \)) which specify the interval of interest.
• The evaluation involves finding the antiderivative of the function and then substituting the limits to find the difference.
• In this exercise, the integration and subsequent evaluation yields a concrete number: the volume \( 36\pi \).

A definite integral not only gives the total net area (or equivalent measure, like volume) but also ensures precision in the calculated results.

Geometry of Solids

Understanding the Geometry of Solids is essential for visualizing and solving problems involving volumes of revolution. These problems often require one to think about three-dimensional shapes resulting from rotating two-dimensional figures.

In our case, the semi-circular region described by \( y = \sqrt{9-x^2} \) when revolved around the \( x \)-axis forms a solid of revolution. This solid is a full sphere with a radius of 3, derived from the equation of the semi-circle. This transformation into a three-dimensional sphere showcases the interplay between the shape, area, and volume.

Key aspects to consider:

• The original shape provides the necessary geometry for revolution.
• Revolutionary patterns determine the solid's nature (e.g., semi-circle leads to a sphere).
• Envisioning the complete 3D structure helps in setting up the correct integral and solving it for volume.

The sphere's geometric nature allows us to predict its volume as \( \frac{4}{3}\pi r^3 \), aligning with our calculated volume, thereby enhancing our confidence in the Disk Method.