Problem 20 - Calculus Early Transcendentals: Pearson New International Edition

Question

Find the volume of the solid generated by revolving about the \(x\)-axis the region in the first quadrant bounded by the circle \(x^{2}+y^{2}=r^{2}\), the \(x\)-axis, and the line \(x=r-h, 0

Step-by-Step Solution

Answer

The volume of the spherical segment is \( \pi h (3r^2 - h^2) / 3 \).

Step 1: Identify the Bounded Region

The bounded region is determined by the circle \(x^2 + y^2 = r^2\), the line \(x = r - h\), and the \(x\)-axis. We will work in the first quadrant, meaning \(x\) ranges from \(r-h\) to \(r\), and \(y\) is found from the circle equation.

Step 2: Express y in terms of x using the Circle Equation

From \(x^2 + y^2 = r^2\), solve for \(y\): \(y = \sqrt{r^2 - x^2}\). This represents the upper semicircle in the bounded region.

Step 3: Set Up the Volume Integral Using the Disk Method

To find the volume of the solid of revolution, use the disk method: integrate \( \pi y^2 \) (the area of a disk) with respect to \(x\). The limits of integration are from \(x = r-h\) to \(x = r\). Therefore, the volume \(V\) is given by:\[ V = \pi \int_{r-h}^{r} (r^2 - x^2) \, dx \]

Step 4: Evaluate the Integral

Calculate the integral:\[ \int (r^2 - x^2) \, dx = r^2x - \frac{x^3}{3} \]Evaluate this from \(r-h\) to \(r\):\[ V = \pi \left[ \left(r^2 \cdot r - \frac{r^3}{3}\right) - \left(r^2(r-h) - \frac{(r-h)^3}{3}\right) \right] \]

Step 5: Simplify the Volume Expression

Simplify the expression obtained from evaluating the definite integral:\[ V = \pi \left( \frac{2r^3}{3} - \left( r^2(r-h) - \frac{(r-h)^3}{3} \right) \right) \]This involves expanding and combining terms to represent the volume of a spherical segment.

Key Concepts

Disk Method

The Disk Method is a straightforward technique used in calculus to find the volume of a solid of revolution. It is especially useful when the solid is generated by rotating a curve around an axis, in this case, the x-axis.

Visualize slicing the solid into thin disks perpendicular to the axis of revolution. Each of these disks has a small thickness, typically denoted as \( dx \), and a radius that varies along the axis. The formula for the volume of each disk is the area of the circle, \( \pi y^2 \), times its thickness, \( dx \).

To find the total volume, add up the volumes of all these infinitesimally thin disks from the starting boundary to the end using definite integration:

• The radius of the disk is determined by the function value, \( y \), at each point, which in this problem is \( \sqrt{r^2 - x^2} \).
• The limits of integration define the start and end of the region, which are \( x = r-h \) and \( x = r \).

This process is represented mathematically by the integral \( V = \pi \int_{r-h}^{r} y^2 \, dx \).

Spherical Segment

In geometry, a spherical segment is the part of a sphere that resides between two parallel planes cutting the sphere. The problem described involves finding the volume of such a segment by using integral calculus.

A helpful way to think of a spherical segment is to visualize it as a portion of a sphere "capped-off" by a flat surface. In this problem, you find the volume by revolving a bounded region around an axis using the identified limits and function.

The region bounded by the circle, the x-axis, and the line creates the cross-section of this spherical segment when rotated about the x-axis. Evaluating this using the disk method, you find the formula for its volume, which takes into account the spherical geometry of the problem.

Integral Calculus

Integral Calculus is a branch of mathematics focused on finding quantities like areas and volumes. It is essential to solving problems involving functions and their accumulations over intervals.

When dealing with volumes of revolutions, like in this exercise, integral calculus makes it possible to add up the infinitesimally small quantities, such as the volumes of slices or disks.

• Integration is used here to sum the areas of disks—and by extension, their volumes—across the domain set by the problem.
• This is done using a definite integral, which provides a numerical value, giving the total volume of the solid of revolution.

This method relies heavily on your ability to set up and evaluate an integral correctly.

Definite Integral

A Definite Integral is the integral evaluation between two specific points in its domain. Here, it is used to compute the total volume of a solid generated by rotating a shape around an axis.

The formula \( \int_{a}^{b} f(x) \, dx \) is utilized, where \( a \) and \( b \) are the lower and upper limits of integration, respectively. For our spherical segment problem, these limits are \( x = r-h \) to \( x = r \).

The definite integral gives a precise value for the volume by calculating the net accumulation of the function over this interval.

• The function here is \( r^2 - x^2 \), reflecting the disk's area as a function of x.
• Integrating this function within the specified bounds captures the entire volume of the solid of revolution.
• By evaluating the antiderivative at the bounds, the solution yields the exact volume of the spherical segment.

The use of definite integrals provides not just accuracy but also an exact answer to spatial problems like those in geometry and physics.