A ball of radius 10 has a round hole of radius 9 drilled through its center. Find the volume of the resulting solid.

The problem is to find the volume of the resulting solid after a hole with a radius of 9 is drilled through the center of a ball with a radius of 10. This problem involves subtracting the volume of the cylindrical hole from the volume of the sphere.

### Step 1: Volume of the Sphere

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

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

where \( R \) is the radius of the sphere. In this case, \( R = 10 \), so:

\[
V_{\text{sphere}} = \frac{4}{3} \pi (10)^3 = \frac{4}{3} \pi (1000) = \frac{4000}{3} \pi
\]

### Step 2: Shape of the Hole

The hole drilled through the sphere is a cylinder with a radius of 9. However, this is a special case where the hole passes completely through the sphere, making it a cylindrical hole that truncates part of the sphere. The volume of such a figure is not simply a cylindrical volume subtraction but rather needs to be calculated using the method of **solid of revolution**.

### Step 3: Set up the integral for the volume of the remaining solid

We need to calculate the volume of the remaining solid by rotating the cross-section of the sphere with the cylindrical hole about the z-axis. The equation of the sphere is:

\[
x^2 + y^2 + z^2 = R^2
\]

where \( R = 10 \). For the cross-section in the xy-plane, the equation becomes:

\[
x^2 + z^2 = 100
\]

We want to find the volume of the solid that remains after the cylindrical hole, which has a radius of 9, is removed. This corresponds to considering the volume from \( x = -9 \) to \( x = 9 \) and using symmetry to find the volume.

We calculate the volume by revolving this around the z-axis:

\[
V = 2 \pi \int_0^9 (100 - x^2) \, dx
\]

### Step 4: Solve the integral

Now, we compute the integral:

\[
V = 2 \pi \int_0^9 (100 - x^2) \, dx
\]

First, split the integral:

\[
V = 2 \pi \left( \int_0^9 100 \, dx - \int_0^9 x^2 \, dx \right)
\]

Evaluate each part:

\[
\int_0^9 100 \, dx = 100x \Big|_0^9 = 900
\]

\[
\int_0^9 x^2 \, dx = \frac{x^3}{3} \Big|_0^9 = \frac{9^3}{3} = 243
\]

So, the total volume is:

\[
V = 2 \pi \left( 900 - 243 \right) = 2 \pi \times 657 = 1314 \pi
\]

### Final Answer:

The volume of the resulting solid is:

\[
1314 \pi \text{ cubic units}
\]

Would you like further details or clarifications?

Here are 5 related questions to extend the topic:
1. How would the volume change if the radius of the hole were 8 instead of 9?
2. How is the volume of the cylindrical hole different from a straight cylinder through the center?
3. What happens to the volume if the sphere's radius increases to 12?
4. Can you generalize the formula for any spherical radius \( R \) and cylindrical hole radius \( r \)?
5. What mathematical tools are typically used to solve problems involving solids of revolution?

**Tip:** Always sketch the shape of the solid and its cross-sections to visualize the problem before integrating.

Explore how to find the volume of a sphere of radius 10 with a cylindrical hole of radius 9 drilled through its center. This problem involves geometry and calculus concepts, including the solid of revolution and Cavalieri's Principle, suitable for students in Grades 11-12.

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