Problem 48 - Thomas Calculus

Question

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

Step-by-Step Solution

Answer

Volume of the solid is given by \(V = \pi ab \left( h + \frac{h^3}{3c^2} \right)\). Expressed in areas: \(V = \frac{h}{2}(A_0 + A_h) = \frac{h}{6}(A_0 + 4A_m + A_h)\).

Step 1: Setting up the volume integral

To find the volume bounded by the hyperboloid and the planes, consider the equation of the hyperboloid, which, rewritten for fixed z, resembles an ellipse: \[ \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 + \frac{z^2}{c^2}. \]This equation describes ellipses stacked between the planes \(z=0\) and \(z=h\). The volume \(V\) can be found by integrating the area of these ellipses across \(z\) from 0 to \(h\).

Step 2: Calculating the Area of the Ellipses

The general area \(A(z)\) of an ellipse is given by the formula:\[ A(z) = \pi \cdot \left(a^2(1 + \frac{z^2}{c^2})\right)^{1/2} \cdot \left(b^2(1+\frac{z^2}{c^2})\right)^{1/2}.\]Simplify it to,\[ A(z) = \pi ab(1 + \frac{z^2}{c^2}). \]

Step 3: Integration setup

To find the volume, integrate the area function \(A(z)\) from \(z=0\) to \(z=h\):\[ V = \int_{0}^{h} \pi ab \left(1 + \frac{z^2}{c^2}\right) \; dz. \]

Step 4: Solving the Integral

Evaluate the integral:\[ V = \pi ab \int_{0}^{h} \left(1 + \frac{z^2}{c^2}\right) \; dz \]\[ = \pi ab \left[ z + \frac{z^3}{3c^2} \right]_{0}^{h} \]\[ = \pi ab \left( h + \frac{h^3}{3c^2} \right). \]

Step 5: Expressing in terms of $A_0$ and $A_h$

At \(z=0\), \(A_0 = \pi ab\), and at \(z=h\), \(A_h = \pi ab(1 + \frac{h^2}{c^2})\). We can express volume \(V\) as:\[ V = \frac{h}{2}(A_0 + A_h). \]

Step 6: Alternative formula with $A_m$

Show that the volume can be seen in terms of \(A_m\):The area at \(z=h/2,A_m = \pi ab(1+\frac{(h/2)^2}{c^2}) = \pi ab(1+\frac{h^2}{4c^2}).\)This gives the new volume formula:\[ V = \frac{h}{6}(A_0 + 4A_m + A_h). \]

Key Concepts

Hyperboloid

A hyperboloid is a type of quadratic surface which can be encountered in various applications, including architecture and physics. It is defined by a specific equation: \( \frac{x^2}{a^2} + \frac{y^2}{b^2} - \frac{z^2}{c^2} = 1 \). This characterizes the shape as being open along the \(z\)-axis, forming a double-curved surface that resembles a v-shaped bowl.

In this particular exercise, we are dealing with a hyperboloid of one sheet. It is important to note that the hyperboloid intersects different planes along \(z\), creating distinct elliptical cross-sections. When viewed along the \(z\)-axis, the hyperboloid generates a series of ellipses, which are critical for calculating the volume of the solid.

Understanding how to work with hyperboloids is crucial for solving problems that involve contours and volumes bounded by specific surfaces. You can encounter such problems in courses focusing on vector calculus and advanced geometry, where the ability to analyze three-dimensional forms is essential.

Volume Integral

Integrating over a volume is a fundamental technique in calculus that allows us to find the space enclosed by a surface. For a hyperboloid bounded by the planes \(z = 0\) and \(z = h\), the volume is computed by integrating the area of its elliptical cross-sections.

To calculate this, we use a definite integral:

• Set up the integral of the area function of the ellipse.
• The area function for an ellipse cut on the hyperboloid is \(A(z) = \pi ab(1 + \frac{z^2}{c^2})\).
• Integrate this function from \(z = 0\) to \(z = h\).

The result is:\[ V = \pi ab \int_{0}^{h} (1 + \frac{z^2}{c^2}) \, dz = \pi ab \left( h + \frac{h^3}{3c^2} \right)\]

This integral provides a convenient way to sum up the infinitesimally small volumes of stacked ellipses between two planes. Understanding this concept is key to solving more complex geometric problems involving arbitrary shapes and limits.

Ellipses

Ellipses are fundamental shapes in geometry, and they appear as sections of three-dimensional objects like hyperboloids. In this context, they play a crucial role in calculating the volume of the region under consideration.

This exercise involves finding the cross-sectional area of an ellipse on the hyperboloid at any particular height \(z\). The formula used is: \(A(z) = \pi ab (1+\frac{z^2}{c^2})\). Here’s how different parameters affect the size of the ellipse:

• \(a\) and \(b\) represent the semi-major and semi-minor axes of the ellipse at \(z=0\).
• As \(z\) changes, the ellipse's dimensions change, affected by the term \(1+\frac{z^2}{c^2}\).

At specific slices, like \(z=0\), the original ellipse has an area \(A_0\), and at \(z=h\), the area is \(A_h\). At halfway \(z=h/2\), an interim area \(A_m\) is present.

These measurements assist in computing the total volume, offering insight into how simple curves like ellipses contribute to forming complex three-dimensional volumes.