A ball of radius 14 has a round hole of radius 6 drilled through its center. Find the volume of the resulting solid

To find the volume of the resulting solid, we need to calculate the volume of the original sphere with radius \(R = 14\) and subtract the volume of the cylindrical hole drilled through the center. The radius of the cylindrical hole is \(r = 6\), and its height will be determined by the length of the sphere's diameter, since the hole goes through the entire sphere.

### Step 1: Volume of the sphere

The volume \(V_{\text{sphere}}\) of a sphere is given by the formula:

\[
V_{\text{sphere}} = \frac{4}{3} \pi R^3
\]

Substituting \(R = 14\):

\[
V_{\text{sphere}} = \frac{4}{3} \pi (14)^3 = \frac{4}{3} \pi (2744) = \frac{10976}{3} \pi \approx 11500.84 \, \text{cubic units}
\]

### Step 2: Volume of the cylindrical hole

The cylindrical hole is a solid of revolution, often referred to as a "Steinmetz solid," which results when a cylinder passes through the center of a sphere.

To find the volume of the cylindrical hole, we can use the formula for the volume of a cylindrical hole drilled through a sphere:

\[
V_{\text{hole}} = 2 \pi r^2 \left( R - \frac{r^2}{R} \right)
\]

Substituting \(r = 6\) and \(R = 14\):

\[
V_{\text{hole}} = 2 \pi (6)^2 \left( 14 - \frac{(6)^2}{14} \right)
= 2 \pi (36) \left( 14 - \frac{36}{14} \right)
\]

Simplifying:

\[
V_{\text{hole}} = 72 \pi \left( 14 - 2.57 \right)
= 72 \pi \times 11.43 = 822.96 \pi
\]

### Step 3: Subtract the cylindrical hole from the sphere

Now, subtract the volume of the cylindrical hole from the volume of the sphere to get the volume of the resulting solid:

\[
V_{\text{solid}} = V_{\text{sphere}} - V_{\text{hole}} = \frac{10976}{3} \pi - 822.96 \pi
\]

Simplifying:

\[
V_{\text{solid}} \approx 11500.84 - 2584.14 = 8916.70 \, \text{cubic units}
\]

### Final Answer:
The volume of the resulting solid is approximately **8916.70 cubic units**.

Would you like further details or clarification on this solution?

### Related Questions:
1. What would the volume be if the radius of the hole were different?
2. How would the volume change if the hole were not drilled through the center?
3. Can the Steinmetz solid formula be derived from basic principles?
4. What if the ball had a different shape, such as an ellipsoid?
5. How does the volume of the cylinder relate to the surface area of the hole?

### Tip:
For geometry problems involving 3D objects, it's helpful to visualize the solid and use symmetry to simplify calculations.

A ball of radius 14 has a round hole of radius 6 drilled through its center. Find the volume of the resulting solid.

Discover how to find the volume of a solid when a cylindrical hole of radius 6 is drilled through a sphere of radius 14. This solution uses geometric formulas, including the volume of a sphere and Steinmetz solid properties. Suitable for Grades 10-12.

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