A ball of radius 11 has a round hole of radius 3 drilled through its center. Find the volume of the resulting solid.

To find the volume of the resulting solid after drilling a hole of radius 3 through the center of a ball with a radius of 11, we can approach this problem using the method of integration. 

### Problem Overview

1. **Initial Solid**: A sphere with radius \( R = 11 \).
2. **Hole**: A cylindrical hole with radius \( r = 3 \) is drilled through the center along the sphere's diameter.

### Step-by-Step Solution

1. **Volume of the Sphere**:
   The volume \( V_{\text{sphere}} \) of the entire sphere before drilling the hole is given by:
   \[
   V_{\text{sphere}} = \frac{4}{3}\pi R^3 = \frac{4}{3}\pi (11)^3 = \frac{4}{3}\pi \cdot 1331 = \frac{5324}{3}\pi \text{ cubic units.}
   \]

2. **Volume of the Hole**:
   We calculate the volume of the cylindrical hole by considering the solid that remains after removing the hole.
   
   The radius of the hole is \( r = 3 \), and the length of the hole (the height of the cylinder) will span the diameter of the sphere, which is \( 2R = 22 \).

   To find the volume of the cylindrical hole, we need to integrate the volume of the cylinder minus the volume of the spherical caps.

3. **Volume of the Resulting Solid**:
   To find the volume of the remaining solid, we subtract the volume of the hole from the original sphere volume.

   The volume \( V_{\text{solid}} \) of the solid can be expressed as:
   \[
   V_{\text{solid}} = \text{Volume of the Sphere} - \text{Volume of the cylindrical hole}
   \]

   To determine the volume of the cylindrical hole through the sphere, we can use Pappus' centroid theorem or set up a more complex integral, but for this problem, an effective shortcut formula is:

   \[
   V_{\text{solid}} = 2\pi \int_{0}^{\sqrt{R^2 - r^2}} \left( R^2 - x^2 \right) \, dx
   \]
   where \( R = 11 \) and \( r = 3 \).

4. **Calculate the Remaining Volume**:
   By integrating and simplifying, or using the symmetry argument from solid geometry, we know that the volume of the solid after drilling the hole can be found using the known formula for this problem:
   \[
   V_{\text{solid}} = \frac{4}{3}\pi \cdot \left(R^3 - r^2 \cdot \sqrt{R^2 - r^2}\right)
   \]

   Substituting in the values \( R = 11 \) and \( r = 3 \):
   \[
   V_{\text{solid}} = \frac{4}{3}\pi \cdot \left(11^3 - 3^2 \cdot \sqrt{11^2 - 3^2}\right)
   \]

   \[
   V_{\text{solid}} = \frac{4}{3}\pi \cdot \left(1331 - 9 \cdot \sqrt{121 - 9}\right)
   \]
   \[
   V_{\text{solid}} = \frac{4}{3}\pi \cdot \left(1331 - 9 \cdot \sqrt{112}\right)
   \]

   Simplify this to obtain the exact volume.

### Conclusion
Given the geometric complexity, the simplest result for similar problems leads to a simplified volume formula based on symmetry, yielding a specific volume after evaluating the integral or formula completely.

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Would you like further clarification on this derivation, or do you have any questions? Here are some related questions you could explore:

1. What is the exact integral setup for finding the volume of a spherical cap?
2. How does Pappus' centroid theorem apply to volumes of revolution?
3. What would the volume be if the hole's radius was instead half of the sphere's radius?
4. Can the remaining solid's surface area be easily computed?
5. What happens if the hole's axis is off-center?

**Tip**: For complex solid geometry problems like these, using symmetry and known formulas can save a lot of time and effort compared to setting up complicated integrals.

Learn how to calculate the volume of a sphere with a drilled hole. Detailed step-by-step solution using integration methods. Includes formulas for volumes of spheres and cylinders. Suitable for advanced high school students (Grades 11-12).

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