Question

Consider the region \(R\) bounded by the right branch of the hyperbola \(x^{2} / a^{2}-y^{2} / b^{2}=1\) and the vertical line through the right focus. a. What is the volume of the solid that is generated when \(R\) is revolved about the \(x\) -axis? b. What is the volume of the solid that is generated when \(R\) is revolved about the \(y\) -axis?

Short answer

Question: Compute the volumes of the solids generated when the region R, bounded by the right branch of the hyperbola \(x^{2}/a^{2}-y^{2}/b^{2}=1\) and the vertical line through the right focus, is revolved about the x-axis and the y-axis. Answer: The volume of the solid generated when the region R is revolved about the x-axis is \(V = \pi a^2 c + \pi a^2\frac{c^3}{3b^2}\), and the volume of the solid generated when the region R is revolved about the y-axis is \(V = \pi a b^2 \left(\frac{2}{3}({1 + \frac{c^2}{b^2}})^{\frac{3}{2}} - \frac{2}{3}\right)\), where \(c = \sqrt{a^2 + b^2}\).

Step-by-step solution

Integrate using the disk method

Since we are revolving around the x-axis, we will use the disk method. First, we need to express the equation of the right branch of the hyperbola in terms of y: \(x = a\sqrt{1 + \frac{y^2}{b^2}}\) Now, we will set up the integral to find the volume. Since we are revolving the region about the x-axis, we need to integrate with respect to x. The radius of the disk will be the distance from the curve to the x-axis, given by \(y\). Therefore, the volume is given by: \(V = \pi\int_{0}^{c} (a\sqrt{1 + \frac{y^2}{b^2}})^2 dy\) Step 2: Solve the integral

Solve the integral

Now, we need to solve the integral: \(V = \pi\int_{0}^{c} (a^2 + a^2\frac{y^2}{b^2}) dy\) Separate the terms and integrate: \(V = \pi\int_{0}^{c} a^2 dy + \pi\int_{0}^{c} a^2\frac{y^2}{b^2} dy\) \(V = \pi a^2[c] + \pi a^2\frac{1}{b^2}\frac{c^3}{3}\) \(V = \pi a^2 c + \pi a^2\frac{c^3}{3b^2}\) Now, let's find the volume of the solid generated when the region R is revolved about the y-axis. Step 3: Set up the integral

Integrate using the shell method

Since we are revolving around the y-axis, we will use the shell method. We already have the equation for the right branch of the hyperbola in terms of y. The height of the shell will be the distance from the curve to the y-axis, given by \(x\). Therefore, the volume is given by: \(V = 2\pi\int_{0}^{b\sqrt{a^2+b^2}} {y} (a\sqrt{1 + \frac{y^2}{b^2}}) dy\) Step 4: Solve the integral

Solve the integral

Now, we need to solve the integral: \(V = 2\pi a \int_{0}^{b\sqrt{a^2+b^2}} y\sqrt{1 + \frac{y^2}{b^2}} dy\) This integral is more challenging to compute. We can use the substitution method: let \(u = 1 + \frac{y^2}{b^2}\), then \(du = \frac{2y}{b^2} dy\). Now the integral becomes: \(V = 2\pi a b^2 \int_{1}^{1 + \frac{c^2}{b^2}} \sqrt{u} \frac{1}{2} du\) \(V = \pi a b^2 \int_{1}^{1 + \frac{c^2}{b^2}} u^{\frac{1}{2}} du\) Now, integrate: \(V = \pi a b^2 \left[\frac{2}{3} u^{\frac{3}{2}}\right]_{1}^{1 + \frac{c^2}{b^2}}\) \(V = \pi a b^2 \left(\frac{2}{3}({1 + \frac{c^2}{b^2}})^{\frac{3}{2}} - \frac{2}{3}\right)\) So, the volume of the solid generated when the region R is revolved about the x-axis is \(V = \pi a^2 c + \pi a^2\frac{c^3}{3b^2}\), and the volume of the solid generated when the region R is revolved about the y-axis is \(V = \pi a b^2 \left(\frac{2}{3}({1 + \frac{c^2}{b^2}})^{\frac{3}{2}} - \frac{2}{3}\right)\).

Key concepts

Disk Method

The Disk Method is a technique used to find the volume of a solid of revolution. When a region is revolved around the x-axis, we use circular disks to approximate the volume of the resulting solid. Each disk has its circular face perpendicular to the axis of rotation, with its radius typically defined by the distance from the curve to the axis.

To calculate the volume, we slice the solid into numerous thin disks. The volume of a single disk can be approximated by the formula for the volume of a cylinder, \( V = \pi r^2 h \), where \( r \) is the radius and \( h \) is the height (or thickness of the disk). The total volume is calculated by integrating these disks along the axis of revolution.

For example, when revolving the right branch of a hyperbola around the x-axis, the radius of each disk is determined by the value of \( y \) at that x-coordinate. The integral is set up as:\[V = \pi \int_{a}^{b} [f(x)]^2 dx\]where \( f(x) \) represents the equation of the curve. This results in a single continuous operation that neatly sums the volume of the disks from start to end.

Shell Method

The Shell Method is an alternative approach used when revolving a region around an axis, such as the y-axis. In this method, we consider cylindrical shells rather than disks. The axis of rotation determines the radius of each shell, and the height of the shell corresponds to the function's value as a function of the other variable.

To calculate the volume using the Shell Method, imagine peeling away layers of the solid to form thin cylindrical shells. For each shell, the formula for the volume is given by \( V = 2\pi rh \), where \( r \) is the radius (distance to the axis of revolution) and \( h \) is the height of the shell.

This technique is especially useful when dealing with revolution around the y-axis. The integral formed by these shells is:\[V = 2\pi \int_{c}^{d} x \, f(x) \, dx\]where \( f(x) \) represents the height of the shell. This method can often simplify the process in cases where the Disk Method would result in more complicated integrals.

Hyperbola

A hyperbola is a type of conic section, or curve, that appears in many mathematical contexts. It is defined by the equation\[\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\]for its standard position. The graph forms two separate curves known as branches, each mirror images across the central axis of the hyperbola.

The hyperbola can be described by its asymptotes, which are lines that the curve approaches but never touches. They form a crucial part of the hyperbola's geometric properties. The hyperbola features two foci, and the distance between these points defines many of its properties, including its eccentricity.

• For the right branch of a hyperbola revolving about an axis, the shape can significantly dictate how the volume is calculated.
• Understanding the geometry of the hyperbola can help set up the limits and integrals correctly when using methods such as Disk or Shell.

In problems involving integration, transforming the hyperbola equation into useful integral bounds and curves is a key step.

Integration Techniques

Integration techniques are essential in finding the volumes of solids of revolution. Solving integrals often involves several methods such as substitution, partial fractions, or integration by parts, depending on the complexity of the function.

For instance, in the exercise involving the right branch of a hyperbola:

• The Disk Method used an integration of squared terms, which required distributing and integrating each term separately.
• When using the Shell Method, substitution was crucial, as in the case of transforming the variable to simplify the square root terms.

These techniques help manage and simplify complex integrals, enabling us to find the exact volume enclosed by the revolving region. The choice of method can greatly affect the simplicity of the solution, thus making it necessary to understand the dynamics of the given integrals and potential substitutions or rearrangements needed to reach the answer efficiently.