Question

Consider the region \(R\) bounded by the right branch of the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) 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: Find the volume of the solid generated when the region R, bounded by the right branch of the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) and the vertical line through the right focus, is revolved about the x and y-axis. Answer: The volume of the solid generated when the region R is revolved about the x-axis is: \(V_x=\pi b^{2} \left(\frac{a(1+\frac{b^{2}}{a^{2}})^{\frac{3}{2}}}{3}-\frac{a^{2}(1+\frac{b^{2}}{a^{2}})}{2}-\frac{a}{3}\right)\) The volume of the solid generated when the region R is revolved about the y-axis is: \(V_y = \pi\left(a^{3}\left(\frac{b}{b^{2}}+ \frac{1}{3b^{2}}\left(1+\frac{b^{2}}{b^{2}}\right)^{\frac{3}{2}}-\sqrt{1+\frac{b^{2}}{a^{2}}}b\right)\right)\)

Step-by-step solution

Identify the equations of the region R

The region R is bounded by the right branch of the hyperbola \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) and the vertical line through the right focus. Let's first find the equation of the vertical line, which goes through the right focus. The right focus of the hyperbola is given by \((a\sqrt{1+e^{2}},0)\). Here, \(e=\frac{c}{a}\) is the eccentricity and \(c=\sqrt{a^{2}+b^{2}}\). So the x-coordinate of the right focus is \(a\sqrt{1+\frac{a^{2}+b^{2}}{a^{2}}} = a\sqrt{1+\frac{b^{2}}{a^{2}}}\). The equation of the vertical line through the right focus is \(x=a\sqrt{1+\frac{b^{2}}{a^{2}}}\) Now we have the two equations that define the region R: 1. \(\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1\) 2. \(x=a\sqrt{1+\frac{b^{2}}{a^{2}}}\)

Find the volume when R is revolved about the x-axis

Let's first find the volume when the region R is revolved about the x-axis. We will use the disk method. The area of a disk at x is given by \(A(x) = \pi [f(x)]^{2}\), where f(x) gives the radius. We need to find an expression for the radius in terms of x, which can be acquired by isolating y from the hyperbola equation. Since the region R is on the right branch of the hyperbola and above the x-axis, it is required to consider the positive solution for y: \(y=b\sqrt{\frac{x^{2}}{a^{2}}-1}\). The element of the volume is \(dV = A(x) dx = \pi [b\sqrt{\frac{x^{2}}{a^{2}}-1}]^{2} dx\). Now we will integrate this expression to find the volume. The limits of integration are from \(x=a\) (the right vertex of the hyperbola) to \(x=a\sqrt{1+\frac{b^{2}}{a^{2}}}\) (the vertical line through the right focus). \(V_x = \int_{a}^{a\sqrt{1+\frac{b^{2}}{a^{2}}}} \pi b^{2} \left(\frac{x^{2}}{a^{2}}-1\right) dx\)

Evaluate the integral and find the volume

Now we just need to evaluate the integral to find the volume when R is revolved about the x-axis. \(V_x = \pi b^{2} \left[\frac{x^{3}}{3a^{2}}-\frac{x^{2}}{2}\right]_{a}^{a\sqrt{1+\frac{b^{2}}{a^{2}}}}\) Plug in the limits of integration: \(V_x =\pi b^{2} \left(\frac{a^{3}(1+\frac{b^{2}}{a^{2}})^{\frac{3}{2}}}{3a^{2}}-\frac{a^{2}(1+\frac{b^{2}}{a^{2}})}{2}-\frac{a}{3}\right)\) Finally, we have the volume when R is revolved about the x-axis: \(V_x=\pi b^{2} \left(\frac{a(1+\frac{b^{2}}{a^{2}})^{\frac{3}{2}}}{3}-\frac{a^{2}(1+\frac{b^{2}}{a^{2}})}{2}-\frac{a}{3}\right)\)

Find the volume when R is revolved about the y-axis

Let's now find the volume when the region R is revolved about the y-axis. We will use the washer method this time. We want to find the inner and outer radii, which are the distances from any point on either curve to the y-axis. From the hyperbolic equation, the outer radius is given by \(R(y) = x\), and for the vertical line, the inner radius is given by \(r(y) = a\sqrt{1+\frac{b^{2}}{a^{2}}}\). To find x, we need to isolate x from the hyperbola equation by taking the positive solution, since the region is on the right branch of the hyperbola: \(x=a\sqrt{1+\frac{y^{2}}{b^{2}}}\). The element of the volume is: \(dV = \pi[(R(y))^2-(r(y))^2]dy = \pi\left[(a\sqrt{1+\frac{y^{2}}{b^{2}}})^{2}-(a\sqrt{1+\frac{b^{2}}{a^{2}}})^{2}\right]dy\) Now we will integrate this expression to find the volume. The limits of integration are from \(y=0\) to \(y=b\). \(V_y = \int_{0}^{b} \pi\left[(a\sqrt{1+\frac{y^{2}}{b^{2}}})^{2}-(a\sqrt{1+\frac{b^{2}}{a^{2}}})^{2}\right]dy\)

Evaluate the integral and find the volume

Now we just need to evaluate the integral to find the volume when R is revolved about the y-axis. \(V_y = \pi \left[\frac{a^{3}y}{b^{2}}\sqrt{1+\frac{y^{2}}{b^{2}}} + \frac{a^{3}}{3b^{2}}\left(1+\frac{y^{2}}{b^{2}}\right)^{\frac{3}{2}}-a^{3}\sqrt{1+\frac{b^{2}}{a^{2}}}y\right]_0^b\) Plug in the limits of integration: \(V_y = \pi\left(\frac{a^{3}b}{b^{2}}\sqrt{1+\frac{b^{2}}{b^{2}}} + \frac{a^{3}}{3b^{2}}\left(1+\frac{b^{2}}{b^{2}}\right)^{\frac{3}{2}}-a^{3}\sqrt{1+\frac{b^{2}}{a^{2}}}b\right)\) Finally, we have the volume when R is revolved about the y-axis: \(V_y = \pi\left(a^{3}\left(\frac{b}{b^{2}}+ \frac{1}{3b^{2}}\left(1+\frac{b^{2}}{b^{2}}\right)^{\frac{3}{2}}-\sqrt{1+\frac{b^{2}}{a^{2}}}b\right)\right)\) To summarize, the volume of the solid generated when the region R is revolved about the x-axis is: \(V_x=\pi b^{2} \left(\frac{a(1+\frac{b^{2}}{a^{2}})^{\frac{3}{2}}}{3}-\frac{a^{2}(1+\frac{b^{2}}{a^{2}})}{2}-\frac{a}{3}\right)\) The volume of the solid generated when the region R is revolved about the y-axis is: \(V_y = \pi\left(a^{3}\left(\frac{b}{b^{2}}+ \frac{1}{3b^{2}}\left(1+\frac{b^{2}}{b^{2}}\right)^{\frac{3}{2}}-\sqrt{1+\frac{b^{2}}{a^{2}}}b\right)\right)\)

Key concepts

Disk Method

The disk method is an efficient and straightforward way to calculate the volume of a solid formed by revolving a shape around the x-axis. Think of a stack of pancakes: each pancake represents a disk, and together, they form the solid. To find the volume of such a solid, you slice it into thin disks perpendicular to the axis of rotation.

The radius of each disk is found using the distance from the axis to the curve. For the given problem, with the region under a hyperbola, the radius at any point x is expressed as the y-coordinate of the curve, which is the height of the disk.

• The formula for the area of each disk is: \[ A(x) = \pi [f(x)]^2 \] where \( f(x) \) is the radius.
• The volume of a small slice of the solid is the product of the disk's area and its thickness, \( dx \): \[ dV = A(x) \, dx = \pi [f(x)]^2 \, dx \]
• To find the total volume, integrate the volume of these disks from one endpoint of the region to the other.

In the exercise, this method was used to revolve the area between the hyperbola and a vertical line around the x-axis.

Washer Method

The washer method is used when the solid of revolution has a hole in the middle, resembling a washer or donut. This is particularly useful when dealing with regions that are revolved around the y-axis, as seen in the given problem.

The key here is to identify both inner and outer radii, which determine the thickness of the washer. This is like imagining a series of hollow disks stacked together.

• Find the outer radius \(R(y)\), which is the distance from the axis to the outer curve. Here, \( R(y) = x = a\sqrt{1 + \frac{y^2}{b^2}} \).
• Determine the inner radius \(r(y)\), given by the distance to the vertical boundary, \( r(y) = a\sqrt{1 + \frac{b^2}{a^2}} \).
• The volume of the washer is: \[ dV = \pi (R(y)^2 - r(y)^2) \, dy \]
• To find the total volume, integrate this expression over the specified limits.

This method was used to find volumes generated by revolving around the y-axis in the exercise.

Hyperbola

A hyperbola is a type of conic section formed by intersecting a double cone with a plane. In this exercise, the hyperbola is given by the equation \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), representing its standard form.

There are a few important aspects of hyperbolas that are critical for solving problems related to solids of revolution:

• The branches of a hyperbola curve away from a central point, forming two separate curves.
• Each hyperbola has foci, points located along the axis of the hyperbola. They are crucial in defining the vertical line through which the region is bounded in the given exercise.
• Understanding the distances from these foci helps determine the bounding region for any area under consideration.

The focus used in the problem helps define a boundary, creating a region that is revolved around the axes to find volumes.

Integration of Volumes

Integration plays a crucial role in finding volumes of solids of revolution. It involves accumulating infinitesimally small volume elements, using integrals to sum them across the desired bounds.

There are a couple of integration techniques applied based on the context of revolutions and the methods used (disk or washer).

• For the **disk method**, the volume is integrated with respect to x, summed over the bounds defined by the region on the x-axis.
• For the **washer method**, the integration is with respect to y, allowing calculation of volumes where there are inner and outer radii.This involves determining the function values that define these radii based on how the region interacts with the y-axis.
• The integral is usually set up in the form of: \[ V = \int (\text{{cross-sectional area}}) \, d(variable) \]This integral gathers all individual contributions from each infinitesimal slice or washer.

Without precise integration, the accumulated volume could not be determined accurately. It's the fundamental tool that allows us to transform geometric problems into solvable equations.