Problem 23 - Thomas Calculus

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 \(36\pi\) cubic units.

Step 1: Identify the Region to Revolve

The given curves are the circle equation upper half, \(y = \sqrt{9-x^2}\), which is a semicircle with radius 3 centered at the origin, and the line \(y = 0\), which represents the x-axis. The region bounded by these curves is the upper half of the circle above the x-axis.

Step 2: Set Up the Volume Integral

We will use the disk method to find the volume generated by revolving this semicircle about the x-axis. The formula for the volume of a solid of revolution using the disk method is \[ V = \pi \int_{a}^{b} [f(x)]^2 \, dx \]For this problem, \(f(x) = \sqrt{9-x^2}\), \(a = -3\), and \(b = 3\). So, \[ V = \pi \int_{-3}^{3} (\sqrt{9-x^2})^2 \, dx \]

Step 3: Simplify the Integral Expression

Simplify the expression under the integral: \( (\sqrt{9-x^2})^2 = 9-x^2 \). So the integral becomes:\[ V = \pi \int_{-3}^{3} (9-x^2) \, dx \]

Step 4: Evaluate the Integral

Now evaluate the integral:\[ V = \pi \left[ \int_{-3}^{3} 9 \, dx - \int_{-3}^{3} x^2 \, dx \right] \]Calculate each integral separately: 1. For \(\int_{-3}^{3} 9 \, dx\):\[ 9x \Big|_{-3}^{3} = 9(3) - 9(-3) = 27 + 27 = 54 \] 2. For \(\int_{-3}^{3} x^2 \, dx\):\[ \frac{x^3}{3} \Big|_{-3}^{3} = \frac{3^3}{3} - \frac{(-3)^3}{3} = \frac{27}{3} - \frac{-27}{3} = 9 + 9 = 18 \]

Step 5: Calculate the Volume

Combine the results of the integrals to find the total volume:\[ V = \pi (54 - 18) = \pi (36) = 36\pi \]

Key Concepts

Disk Method

The disk method is a technique used in calculus to find the volume of a solid of revolution. When you revolve a shape around an axis, it creates a 3-dimensional object. Visualize slicing this object into thin disks, much like a stack of coins. Each disk has a very small thickness, represented by the infinitesimal \(dx\), and is similar to a cylinder.
To calculate the volume of such a solid using the disk method, you set up an integral that adds up the volumes of all these disks from one end to the other. The formula is:

• \( V = \pi \int_{a}^{b} [f(x)]^2 \, dx \)

Where \(f(x)\) is the radius of each disk, and \([a, b]\) is the interval over which the region extends. In our example, the disk radius is determined by the curve \(y=\sqrt{9-x^2}\), which is why it becomes the function \(f(x)\) for the volume calculation. Since the entire shape revolves around the x-axis, this function effectively describes how far each disk reaches out from the axis.

Integral Calculus

Integral calculus is a part of mathematical analysis that deals with integration and its corresponding concepts. Integration involves finding the whole by accumulating parts, which is immensely useful in calculating areas, volumes, and other quantities that involve summation.Here, integral calculus comes into play as we need to accumulate an infinite number of tiny slices (small volumes) to find the total volume of the solid of revolution. By setting up an integral with limits from \(-3\) to \(3\), we find the volume of the region described by the semicircle as it's revolved around the x-axis:

• The first integral, \(\int_{-3}^{3} 9 \, dx\), computes the accumulated way each piece of the semicircle stretches along the x-axis.
• The second part, \(\int_{-3}^{3} x^2 \, dx\), subtracts out the curvature of each segment, accounting for the changes in radius.

These integrals utilize polynomial functions which can often be simplified for easier computation, an essential skill in calculus, ensuring accurate summation of volumes over a range.

Semicircle

A semicircle is exactly as it sounds — half of a full circle. In mathematics, it is represented as a 2D geometric shape where only the upper or the lower half of the circle's boundary is used. The equation \(y = \sqrt{9 - x^2}\) defines the upper half, or semicircle, of a circle in a Cartesian plane.The center of our semicircle is the origin, \((0,0)\), and its radius is 3, derived from the equation \(x^2 + y^2 = 9\). This curvature is crucial when revolving the semicircle around an axis because it affects the resultant solid's shape.Revolving a semicircle about the x-axis, like in this exercise, forms a 3-dimensional shape known as a "cap"-like solid. Here's why:

• The boundary stretching from {\(-3 \, \leq x \, \leq \, 3\)} creates half of a spherical volume when revolved.
• This is part of a full sphere, showing how just a flat, simple curve can become a full, closed shape when in motion.

Understanding the geometry of simple shapes like semicircles is foundational for applying integral calculus concepts efficiently.