Problem 12 - Thomas Calculus

Question

Find the volume of the largest right circular cone that can be inscribed in a sphere of radius \(3 .\)

Step-by-Step Solution

Answer

The volume is \(2\pi \sqrt{3}\).

Step 1: Understand the Problem

We need to find the volume of the largest right circular cone that can be inscribed inside a sphere with a radius of 3 units. This means the height and radius of the cone must be determined based on the geometry of the sphere.

Step 2: Visualize the Geometry

Visualize the sphere. Placing the sphere in a coordinate system with its center at the origin, the general equation for the sphere's surface is \(x^2 + y^2 + z^2 = 9\). The cone's vertex is at the top of the sphere, its base is parallel to the xy-plane, with height \(h\) and base radius \(r\).

Step 3: Relate Cone's Dimensions to Sphere

If the cone has height \(h\), then since it is inside the sphere, from the vertex to the center of its base is along the z-axis (let's assume this distance is \(x\)). Using Pythagorean Theorem in the right triangle inside the sphere with one vertex at the sphere's center and the hypotenuse along its radius, the distance \(x\) satisfies \(x + r = 3\). Adding the opposite base height \(h\) gives \(h^2 + r^2 = 9\) (from the sphere equation and cone's orientation).

Step 4: Express 'r' in Terms of 'h'

From the relation in Step 3, express \(r\): \(r = 3 - x = \sqrt{9 - h^2}\). Therefore, \(r = \sqrt{9 - h^2}\).

Step 5: Volume of Cone

The volume \(V\) of a cone is given by the formula \( V = \frac{1}{3}\pi r^2 h \), substituting \(r = \sqrt{9-h^2}\) into this gives: \( V = \frac{1}{3}\pi (9-h^2)h \).

Step 6: Differentiate and Optimize Volume

Differentiate the volume function \( V(h) = \frac{\pi}{3} (9h - h^3) \) with respect to \(h\): \( V'(h) = \frac{\pi}{3} (9 - 3h^2)\). Set this derivative equal to zero to find critical points: \(9 - 3h^2 = 0\). Solving this provides \( h = \sqrt{3} \) or \(h = -\sqrt{3}\); since \(h\) must be positive, \(h = \sqrt{3}\).

Step 7: Calculate Radius and Volume

Substitute \(h = \sqrt{3}\) back into \(r = \sqrt{9-h^2}\): \( r = \sqrt{9 - 3} = \sqrt{6}\). The volume: \( V = \frac{1}{3}\pi (\sqrt{6})^2 \sqrt{3} = \frac{1}{3}\pi \times 6 \times \sqrt{3} = 2\pi \sqrt{3}\).

Step 8: Conclude Largest Volume Calculation

The volume of the largest right circular cone that can be inscribed in the sphere is \(2\pi \sqrt{3}\).

Key Concepts

Geometry

Geometry is at the heart of understanding the relationship between different shapes and forms in space. In this exercise, the main shapes are a sphere and a right circular cone. A sphere is a perfectly round geometrical shape, where every point on its surface is equidistant from its center.
The equation for a sphere with a radius of 3, centered at the origin in a coordinate system, is expressed as:

• \( x^2 + y^2 + z^2 = 9 \)

A right circular cone, on the other hand, is a three-dimensional shape with a circular base and a pointed top (vertex) that is directly above the center of the base when the cone is upright.
To solve the problem, it's essential to visualize how the cone fits inside the sphere. The center of the sphere aligns with the base of the cone, while the apex touches the sphere's inner surface.

Optimization

Optimization in calculus involves finding the maximum or minimum value of a function within a given domain. In this context, it's about determining the maximum volume that a cone can have while still fitting inside a sphere with a radius of 3.
The process starts by expressing the variable relationships that confine the cone within the sphere. In step-by-step terms, we first derived relationships between the cone's dimensions using geometric constraints.
To find the maximum volume, we then took the derivative of the volume function with respect to the cone's height, resulting in a critical point that indicates the optimal height for maximum volume when set to zero.
The calculated height \( h = \sqrt{3} \) is verified as a maximum by checking the sign changes or using the second derivative test if necessary.

Right Circular Cone

The right circular cone is the featured element in this optimization problem. Its characteristics include a height \( h \), a base radius \( r \), and a slant height forming a straight line from the edge of the base to the apex at the top.
In this exercise, the cone's vertex is on the top of the sphere and it is inscribed, meaning it touches the sphere at precisely one point along its surface.
The relationship between the cone's dimensions \( h \) and \( r \) is determined by its position within the sphere, leading to the relation:

• \( h^2 + r^2 = 9 \)

This indicates the cone's constant presence entirely within the sphere, ensuring no part exceeds the confines of the sphere.

Volume Calculation

Calculating the volume of a cone is central to solving this optimization problem. The basic formula for the volume of a cone is given by:

• \( V = \frac{1}{3} \pi r^2 h \)

In this case, using geometric constraints, we determined that \( r = \sqrt{9-h^2} \), which allows us to express the volume in terms of a single variable, \( h \).
The resulting volume function becomes:

• \( V(h) = \frac{1}{3} \pi (9-h^2)h \)

Differentiating this function and finding its maximum value ensures we are getting the right volume for the cone that fits snugly inside the sphere.
This process combines geometry principles with calculus techniques to solve real-world problems involving spatial constraints and optimal dimensions.