Problem 39 - Calculus: Early Transcendentals

Question

The region bounded by the given curves is rotated about the specified axis. Find the volume of the resulting solid by any method. \( y^2 - x^2 = 1 \) , \( y = 2 \) ; about the x-axis

Step-by-Step Solution

Answer

The volume is \( 4\pi\sqrt{3} \).

Step 1: Identifying the Bounded Region

The given curves are a hyperbola, \( y^2 - x^2 = 1 \), and a horizontal line, \( y = 2 \). We need to find their intersection points by substituting \( y = 2 \) into the hyperbola equation: \( 2^2 - x^2 = 1 \), which simplifies to \( x^2 = 3 \), giving \( x = \pm\sqrt{3} \). Hence, the region is bounded by \( y = 2 \), \( y = \sqrt{x^2 + 1} \), \( x = -\sqrt{3} \), and \( x = \sqrt{3} \).

Step 2: Setting Up the Integral for Volume

Since we are rotating around the x-axis, we use the disk method. The volume \( V \) is given by the integral \( V = \pi \int_{-\sqrt{3}}^{\sqrt{3}} (R^2 - r^2) \, dx \), where \( R = 2 \) is the outer radius (from \( y = 2 \) to the x-axis), and \( r = \sqrt{x^2 + 1} \) is the inner radius (from the hyperbola to the x-axis).

Step 3: Calculating the Integral

Plugging into the volume integral, we have: \( V = \pi \int_{-\sqrt{3}}^{\sqrt{3}} \left( 4 - (x^2 + 1) \right) \, dx \). Simplifying this gives \( V = \pi \int_{-\sqrt{3}}^{\sqrt{3}} (3 - x^2) \, dx \).

Step 4: Evaluating the Integral

Integrate \( 3 - x^2 \): \( \int (3 - x^2) \, dx = 3x - \frac{x^3}{3} \). Evaluate this from \( -\sqrt{3} \) to \( \sqrt{3} \): \[ V = \pi \left[ 3x - \frac{x^3}{3} \right]_{-\sqrt{3}}^{\sqrt{3}} \].

Step 5: Calculating Definite Integral Values

First, substitute \( \sqrt{3} \): \( 3(\sqrt{3}) - \frac{(\sqrt{3})^3}{3} = 3\sqrt{3} - \frac{3\sqrt{3}}{3} = 3\sqrt{3} - \sqrt{3} = 2\sqrt{3} \). Next, substitute \( -\sqrt{3} \): \( 3(-\sqrt{3}) - \frac{(-\sqrt{3})^3}{3} = -3\sqrt{3} + \sqrt{3} = -2\sqrt{3} \). The result of the definite integral is \( 2\sqrt{3} - (-2\sqrt{3}) = 4\sqrt{3} \).

Step 6: Final Calculation of Volume

Multiply the result by \( \pi \) to obtain the volume: \( V = \pi \times 4\sqrt{3} = 4\pi\sqrt{3} \). This is the volume of the solid formed by rotating the region around the x-axis.

Key Concepts

Disk Method

The disk method is a technique used in calculus to find the volume of a solid of revolution. When a region in the xy-plane is revolved around an axis, it forms a three-dimensional solid. The disk method treats the solid as a collection of infinitely small disks stacked together from one end of the region to the other.

• Each small disk has a thickness, typically denoted as dx or dy, depending on the axis of rotation.
• The volume of each disk is calculated using the formula: \( \pi (\text{radius})^2 \times \text{thickness} \).

To find the total volume of the solid, you integrate over the range of the bounded region. In this specific problem, we rotate around the x-axis and use \( \pi \int (R^2 - r^2) \, dx \), where R is the external radius and r is the internal radius. This method is straightforward and essential for problems involving solids of revolution.

Hyperbola

A hyperbola is a type of conic section that is characterized by its distinctive open shape, formed by the difference of squares of the coordinates. The general form of a hyperbola is \( y^2 - x^2 = c \), where c is a constant. In this exercise, the given hyperbola is \( y^2 - x^2 = 1 \).

• In Cartesian coordinates, a hyperbola consists of two separate curves called branches.
• Unlike circles or ellipses, hyperbolas open indefinitely.
• The graph of a hyperbola can be asymmetric, intersecting the axes at specific points depending on its equation.

To solve such problems, it's essential to identify the region bounded by the hyperbola and any other given lines or curves. The region of interest is typically bound by the intersection points of these curves.

Definite Integrals

Definite integrals are used to calculate the exact area under the curve of a function, between two specified points. They also play a critical role in finding the volume, area, and other properties of geometric shapes in calculus.

• The notation for a definite integral is \( \int_a^b f(x) \, dx \), where a and b represent the lower and upper limits, respectively.
• Definite integrals consider the entire area between the curve and the x-axis from x = a to x = b.

In the context of the problem, the integral \( \pi \int_{-\sqrt{3}}^{\sqrt{3}} (3 - x^2) \, dx \) is evaluated to find the total volume. The process of computing a definite integral involves integrating the function algebraically and then substituting the boundary values to find the net area or volume between those bounds.

Calculus

Calculus is a branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. It has two main branches: differential calculus and integral calculus.

• Differential calculus involves finding the rate of change of quantities, serving as the foundation for solving problems related to slopes, velocities, and optimizations.
• Integral calculus deals with finding the accumulation of quantities, such as areas under curves and volumes of solids.

This problem utilizes principles from integral calculus to find the volume of a hyperbolic region when rotated about an axis. Using definite integrals helps us not only determine volumes but also allows us to handle complex curves and compute areas that would otherwise be arduous with basic geometric formulas. Understanding these calculus concepts is crucial for solving advanced mathematical problems.