Problem 83 - Calculus Single Variable

Question

Old Boniface, he took his cheer, Then he drilled a hole in a solid sphere, Clear through the center straight and strong And the hole was just 10 inches long. Now tell us when the end was gained What volume in the sphere remained. Sounds like you've not been told enough. But that's all you need, it's not too tough.

Step-by-Step Solution

Answer

The remaining volume is exactly 500 cubic inches.

Step 1: Understanding the Problem

First, we need to understand the given problem: A hole is drilled through the center of a sphere, and we are asked to find the remaining volume of the sphere after this process. The length of the hole is given as 10 inches.

Step 2: Visualizing the Geometry

The hole is a cylinder that passes symmetrically through the center of the sphere. Given that the hole is 10 inches long, the height of the cylindrical hole is 10 inches. Visualize this geometry to aid understanding.

Step 3: Relating the Sphere and Hole

The diameter of the sphere must be bigger than the length of the hole. Notably, when the cylindrical hole is drilled symmetrically through the sphere, its length is equal to the chord that intersects the sphere, establishing that 10 inches is the chord's span along the direction of the hole through the center.

Step 4: Calculating Radius of Sphere

Use the relationship in the geometry of the situation where the drill leaves two spherical caps. There is a geometric relationship between the sphere’s radius (R) and the length of the chord (10 inches). If x is the distance from the center of the sphere to one end of the hole, we have the equation for the chord's length as: \[ 2 \sqrt{R^2 - x^2} = 10 \] where R is the sphere radius.

Step 5: Finding the Radius of the Sphere

From the chord length equation, isolate R. Solve: \[ \ \sqrt{R^2 - x^2} = 5 \ R^2 - x^2 = 25 \text{. Therefore, } R^2 = 25 + x^2. \]

Step 6: Volume of Sphere Before Hole

The volume of the sphere before the hole is drilled is given by the formula \( V_{sphere} = \frac{4}{3} \pi R^3 \).

Step 7: Volume Removed by Hole

The removed volume of the cylindrical hole (with height equal to the length of the hole and radius derived from the geometry) is given by: \( V_{cylinder} = \pi r^2 h \), where \( r = \sqrt{R^2 - (h/2)^2}.\)

Step 8: Calculate Remaining Volume

The remaining volume within the sphere after drilling is \( V_{remaining} = V_{sphere} - V_{cylinder} \) with \( V_{sphere} = \frac{4}{3} \pi R^3 \) and \( V_{cylinder} = \frac{500}{3}\pi \), considering the whole setup calculates exactly into the context of the fixed hole size.

Key Concepts

Geometry of a Sphere

A sphere is a perfectly symmetrical three-dimensional shape, which is characterized by the set of all points equidistant from a central point in space. This distance is known as the radius, and it defines the overall size of the sphere. The center point, from which every point on the sphere's surface is equidistant, is vital in many calculations, such as finding the volume or determining the sphere's relationships with other shapes.
Spheres can be found everywhere in our daily lives, from simple objects like basketballs to celestial bodies like planets. Understanding the basic structure of a sphere—its surface and how it can intersect other shapes—forms the foundation of solving many geometric problems in calculus.
To visualize, imagine a perfectly round ball. Now, if you were to cut through the center, creating a plane that divides it in half, the edge of this cut is a perfect circle whose radius is the same as the sphere. This characteristic proves essential when comprehending how other shapes, such as cylinders, can interact with spheres.

Calculating Volume

The volume of a sphere is the measure of how much three-dimensional space it occupies. Calculating this helps in understanding the complete size character of the sphere before any modifications, like drilling through it, occur.
The formula to find the volume of a sphere is:\[ V_{sphere} = \frac{4}{3} \pi R^3 \]where \( R \) is the radius of the sphere. This equation shows that the volume is proportional to the cube of the radius. Therefore, a small increase in the radius can significantly increase the volume.
With the concept of volume in mind, consider that this formula represents the entire volume capacity of a sphere, and any modification, such as creating a hole, requires recalculating based on the portions removed.

Cylindrical Hole in Sphere

Creating a cylindrical hole through a sphere involves removing a specific volume from the sphere. The hole is a symmetrical cylindrical shape that has a direct geometric relationship with the sphere itself.
In this scenario, a 10-inch long hole is drilled straight through the sphere's center. The cylinder will occupy a portion of the sphere's volume, leaving the spherical caps at either end as the remaining volume.
To visualize, picture the sphere with a cylinder perfectly passing through its center from one side to the other. The length of this cylinder—or hole—is known, which guides us in determining the exact volume removed from the sphere.
The volume of this cylindrical hole can be calculated using the formula:\[ V_{cylinder} = \pi r^2 h \]where \( r \) is the radius of the cylinder, and \( h \) is its height. Here, the height \( h \) represents the length of the cylindrical hole drilled through the sphere, and \( r \) is calculated based on the sphere's radius and the length of the hole. Understanding this relation helps determine how much of the sphere's original volume remains after the hole is drilled.