Problem 41 - Thomas Calculus

Question

A bead is formed from a sphere of radius 5 by drilling through a diameter of the sphere with a drill bit of radius 3. $$ \begin{array}{l}{\text { a. Find the volume of the bead. }} \\ {\text { b. Find the volume of the removed portion of the sphere. }}\end{array} $$

Step-by-Step Solution

Answer

a. Volume of the bead: \( \frac{106}{3} \pi \) cubic units. b. Volume of removed portion: \( \frac{394}{3} \pi \) cubic units.

Step 1: Calculate the Volume of the Sphere

The volume of a sphere is given by the formula \( V = \frac{4}{3} \pi r^3 \), where \( r \) is the radius of the sphere. Here, the radius is 5, so we calculate: \[V = \frac{4}{3} \pi (5)^3 = \frac{4}{3} \pi \times 125 = \frac{500}{3} \pi.\]

Step 2: Calculate the Volume of the Cylinder Drilled Out

The cylinder is drilled through the sphere, and its volume is given by the formula \( V = \pi r^2 h \), where \( r = 3 \) is the radius and \( h = 2 \times 5 = 10 \) is the height (equal to the diameter of the sphere). Thus, the volume of the cylinder is: \[V = \pi (3)^2 (10) = 90\pi.\]

Step 3: Calculate the Volume of the Spherical Caps

Since drilling the cylinder through the sphere removes two spherical caps, we need to calculate their total volume. Each cap volume is given by \( V = \frac{1}{6} \pi h (3a^2 + h^2) \) where \( h = 5 - 3 = 2 \) is the height of the cap and \( a = 3 \). Calculating the volume for one cap: \[V_{cap} = \frac{1}{6} \pi (2) (3(3)^2 + (2)^2) = \frac{1}{6} \pi (2)(27 + 4) = \frac{62}{3} \pi.\]Thus, for two caps: \[2 \times \frac{62}{3} \pi = \frac{124}{3} \pi.\]

Step 4: Calculate the Volume of the Removed Portion

The removed portion includes both the cylinder and the two spherical caps. Therefore, the total removed volume is the sum of the cylinder volume and the volume of the two caps: \[V_{removed} = 90\pi + \frac{124}{3} \pi = \frac{270}{3} \pi + \frac{124}{3} \pi = \frac{394}{3} \pi.\]

Step 5: Calculate the Volume of the Bead

The volume of the bead is the volume of the original sphere minus the volume of the removed portion: \[V_{bead} = \frac{500}{3} \pi - \frac{394}{3} \pi = \frac{106}{3} \pi.\]

Key Concepts

Sphere

A sphere is a perfectly symmetrical three-dimensional shape, where every point on its surface is equidistant from its center. It's the shape you'd see in a basketball or a perfectly round soap bubble. The radius (\( r \)) of a sphere is the distance from the center to any point on its surface.

To find the volume of a sphere, we use the formula: \[V = \frac{4}{3} \pi r^3\]This equation helps us calculate how much space is inside the sphere. For example, with a radius of 5, like the sphere in our exercise, the calculation becomes:

• First, calculate \( r^3 = 5^3 = 125 \)
• Then, apply the formula: \( V = \frac{4}{3} \pi \times 125 = \frac{500}{3} \pi \)

This will give you the volume of our original sphere before any modifications.

Cylinder

A cylinder is a 3D shape with two parallel circular bases connected by a curved surface. Think of it as the shape of a soup can or a drum. The volume of a cylinder is calculated by considering the area of its base and its height.

We use the formula: \[V = \pi r^2 h\]where \( r \) is the radius of the circle base and \( h \) is the height. In the sphere drilling exercise, the cylinder drilled out has a radius of 3 and a height of 10 (the diameter of the sphere since it goes right through the center).

The volume of this drilled cylinder can be calculated by:

• First, calculate the base area: \( \pi (3)^2 = 9\pi \)
• Then, multiply by the height: \( 9\pi \times 10 = 90\pi \)

The result gives you the volume of the cylindrical portion removed from the sphere.

Spherical caps

Spherical caps are essentially "lids" from a sphere when you slice it. Visually, consider cutting off the top of an orange—what's left is a spherical cap. When you drill a cylinder through a sphere, two caps are created, one at each junction of the cylinder with the sphere.

The formula to find the volume of a spherical cap is: \[V = \frac{1}{6} \pi h (3a^2 + h^2)\]where \( h \) is the height of the cap and \( a \) is the base radius. In our scenario:

• The height \( h \) is calculated as \( 5 - 3 = 2 \)
• The radius \( a \) is 3
• One cap volume: \( \frac{1}{6} \pi \times 2 \times (3(3)^2 + 2^2) = \frac{62}{3} \pi \)

Since there are two such caps, multiply by 2 to get \( \frac{124}{3} \pi \). This volume is part of the removed portion of the sphere.

Geometry problem solving

Solving geometry problems often requires breaking down complex shapes into simpler parts and using formulas to find volumes or areas. This exercise showcased such an approach with a sphere, cylinder, and spherical caps.

Here are important steps in geometry problem solving, using the example above:

• Determine the individual shapes involved—identify any combinations of basic shapes.
• Calculate the volume (or area) of each basic shape using the correct formula.
• For a drilled object (or combined shapes), calculate subtracted or added volumes for changes like cuts or holes.
• Check your work, ensuring all steps add up to the original total.

In the exercise presented, we calculated volumes step-by-step: the original sphere, the drilled cylinder, and the caps. Subtracting the drilled volume from the original gave the volume of what remains. Clear and organized calculations help solve such problems accurately and efficiently.