Problem 80 - Thomas Calculus

Question

a. Find the volume of the solid bounded by the hyperboloid $$ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}-\frac{z^{2}}{c^{2}}=1 $$ and the planes \(z=0\) and \(z=h, h>0\) b. Express your answer in part (a) in terms of \(h\) and the areas \(A_{0}\) and \(A_{h}\) of the regions cut by the hyperboloid from the planes \(z=0\) and \(z=h.\) c. Show that the volume in part (a) is also given by the formula $$ V=\frac{h}{6}\left(A_{0}+4 A_{m}+A_{2}\right) $$ where \(A_{m}\) is the area of the region cut by the hyperboloid from the plane \(z=h / 2\)

Step-by-Step Solution

Answer

The volume using integration is \( V = \pi ab \left( h + \frac{h^3}{3c^2} \right) \) and by Simpson's rule \( V = \frac{h}{6} (A_0 + 4A_m + A_h) \).

Step 1: Set up equations for cross-sections

The hyperboloid is given by \( \frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1 \). The cross-section at a constant \( z \) is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 + \frac{z^2}{c^2} \), which is an ellipse.

Step 2: Express cross-sectional area as a function of z

The area of the ellipse at height \( z \) is \( A(z) = \pi a_z b_z \), where \( a_z = a \sqrt{1 + \frac{z^2}{c^2}} \) and \( b_z = b \sqrt{1 + \frac{z^2}{c^2}} \). Thus, \( A(z) = \pi ab (1 + \frac{z^2}{c^2}) \).

Step 3: Calculate volume using integration

The volume \( V \) of the solid from \( z=0 \) to \( z=h \) is given by the integral \( V = \int_0^h A(z) \, dz = \int_0^h \pi ab \left(1 + \frac{z^2}{c^2}\right) \, dz \).

Step 4: Solve the integral

Integrating, we find \( V = \pi ab \left[ z + \frac{z^3}{3c^2} \right]_0^h = \pi ab \left( h + \frac{h^3}{3c^2} \right) \).

Step 5: Express volume in terms of areas

The area \( A_0 \) at \( z=0 \) is \( \pi ab \). The area \( A_h \) at \( z=h \) is \( \pi ab \left( 1 + \frac{h^2}{c^2} \right) \). Thus, \( V = \frac{h}{2} (A_0 + A_h) \).

Step 6: Calculate area at z=h/2 for Simpson's rule

The area \( A_m \) at \( z = \frac{h}{2} \) is \( \pi ab \left( 1 + \frac{h^2}{4c^2} \right) \).

Step 7: Apply Simpson's rule for volume

By Simpson's rule, \( V = \frac{h}{6} (A_0 + 4A_m + A_h) = \frac{h}{6} \left( \pi ab + 4 \pi ab \left( 1 + \frac{h^2}{4c^2} \right) + \pi ab \left( 1 + \frac{h^2}{c^2} \right) \right) \).

Key Concepts

Hyperboloid

A hyperboloid is a fascinating geometric shape that appears often in mathematics and engineering. It can look like a "double" hourglass or like a "one-sheet" surrounding, depending on its type. In this exercise, we focus on a type of hyperboloid highlighted by the formula \( \frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1 \).
This equation represents a hyperboloid of one sheet, a surface that extends infinitely in the space.
The core idea is to find the volume of this hyperboloid limited between two planes, specifically at \( z = 0 \) and \( z = h \). These planes effectively "cut off" the ends of the hyperboloid.

Integration

Integration allows us to calculate the volume of complex shapes. The idea is to break down the three-dimensional shape into an infinite number of thin cross-sectional areas.
Each cross-section contributes a small volume element, and by adding all these elements together using integration, we find the total volume.
In this exercise, the target is to integrate the area of cross-sections from start, \( z=0 \), to finish, \( z=h \), to determine the complete volume of the bounded hyperboloid.
With the hyperboloid equation given, we know each cross-section is an ellipse. By integrating over the cross-sectional areas, we can find how much space the solid occupies between the two planes.

Cross-Sectional Area

The concept of cross-sectional areas is fundamental in calculating volumes of solids. For the hyperboloid, each vertical slice of the shape is an ellipse.
When the hyperboloid intersects with a horizontal plane at a given height \( z \), it forms an ellipse defined by the equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 + \frac{z^2}{c^2} \).
The area of this ellipse can be determined by the formula \( A(z) = \pi a_z b_z \), where

• \( a_z = a \sqrt{1 + \frac{z^2}{c^2}} \)
• \( b_z = b \sqrt{1 + \frac{z^2}{c^2}} \)

Finding these areas at various \( z \) heights is crucial to using integration to ascertain the total volume.

Simpson's Rule

Simpson's Rule is a method to approximate definite integrals, especially useful when dealing with data-derived functions or when obtaining an exact integral is complex.
Typically used for calculating areas or volumes, it gives a more accurate approximation by considering the shape and pattern of the function between endpoints.
In this exercise, Simpson's Rule helps by using the areas of three distinct cross-sections: one at \( z = 0 \), one at \( z = h/2 \), and one at \( z = h \).
This rule then formulates the volume as \( V = \frac{h}{6}(A_0 + 4A_m + A_h) \), effectively averaging discrepancies among possible inaccuracies and leveraging approximations to achieve a precise result.