Problem 46 - Thomas Calculus

Question

Ellipsoid For what value of \(c\) is the volume of the ellipsoid \(x^{2}+(y / 2)^{2}+(z / c)^{2}=1\) equal to 8\(\pi ?\)

Step-by-Step Solution

Answer

The value of \( c \) is 3.

Step 1: Understand the Standard Ellipsoid Formula

An ellipsoid is represented by the equation \( \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 \). The given equation \( x^2 + \left( \frac{y}{2} \right)^2 + \left( \frac{z}{c} \right)^2 = 1 \) can be rewritten in the standard form as \( \frac{x^2}{1} + \frac{y^2}{4} + \frac{z^2}{c^2} = 1 \), with \( a = 1 \), \( b = 2 \), and \( c \) unknown.

Step 2: Identify the Volume Formula for an Ellipsoid

The volume \( V \) of an ellipsoid is given by the formula \( V = \frac{4}{3} \pi a b c \). This is crucial for determining the value of \( c \) that will satisfy the given conditions.

Step 3: Substitute Known Values into the Volume Formula

Since we know \( a = 1 \), \( b = 2 \), and the desired volume is 8\(\pi\), substitute these into the volume formula: \( 8\pi = \frac{4}{3} \pi \times 1 \times 2 \times c \).

Step 4: Solve for \(c\)

Simplify and solve the equation from the previous step: \( 8\pi = \frac{8}{3} \pi c \). Divide both sides by \( \pi \), leaving \( 8 = \frac{8}{3} c \). Multiply both sides by \( \frac{3}{8} \) to isolate \( c \), resulting in \( c = 3 \).

Key Concepts

Ellipsoid Formula

An ellipsoid represents a three-dimensional shape resembling a stretched sphere. It is defined by its three semi-axes lengths. The equation to describe an ellipsoid is typically written as:

• \( \frac{x^2}{a^2} + \frac{y^2}{b^2} + \frac{z^2}{c^2} = 1 \)

This equation indicates that:- \( a \), \( b \), and \( c \) represent the lengths of the semi-axes along the x, y, and z directions, respectively.The given problem involves transforming the equation of an ellipsoid into this standard form. Originally expressed as \( x^2 + \left( \frac{y}{2} \right)^2 + \left( \frac{z}{c} \right)^2 = 1 \), this becomes \( \frac{x^2}{1} + \frac{y^2}{4} + \frac{z^2}{c^2} = 1 \). Understanding this conversion is crucial because it makes the ellipsoid's geometric properties, like its volume, easier to calculate.

Volume Calculation

The volume of an ellipsoid is a measure of the amount of space it occupies in three dimensions. Calculating this volume relies on understanding the ellipsoid's semi-axes. The formula used to find the volume of an ellipsoid is:

• \( V = \frac{4}{3} \pi a b c \)

In this formula:- \( a \), \( b \), and \( c \) represent the semi-axes.- \( \pi \) is a mathematical constant approximately equal to 3.14159.The problem asks for the value of \( c \) when the volume is given as \( 8\pi \). We know \( a = 1 \), \( b = 2 \), and using the volume formula, we set up the equation: \( 8\pi = \frac{4}{3} \pi \times 1 \times 2 \times c \). Solving this helps determine the specific geometry of the ellipsoid described in the exercise.

Geometric Shapes

Ellipsoids are part of the family of geometric shapes which include more familiar forms like spheres and cylinders. They provide an elegant example of how geometry extends into three dimensions. Some characteristics of ellipsoids include:

• Symmetry along their three principal axes.
• Defined by lengths known as semi-axes, which dictate their stretch along each axis.
• Varying curvature that can be seen as a generalization of a circle or sphere.

Understanding the geometric properties of ellipsoids allows us to apply mathematical formulas more effectively. It contextualizes problems like the given exercise, which involves finding parameters such as volume.

Calculus Problem-Solving

Solving calculus-based problems often involves understanding and manipulating complex mathematical forms. The given problem illustrates this through:

• Transforming an equation to better understand its parameters.
• Applying the volume formula of an ellipsoid, which is an integral part of calculus studies.
• Simplifying and solving equations by logical reasoning and basic mathematical operations.

In the provided exercise, precise problem-solving techniques allow us to determine the correct value of \( c \) when calculating the volume of an ellipsoid. Step-by-step analysis showcases how complex calculus concepts can become manageable when broken down correctly. This example underscores the value of systematic approaches in applying calculus to geometry.