To find the volume of the solid resulting from drilling a cylindrical hole through a sphere, we can use the method of cylindrical shells or symmetry. However, there's a specific formula for this type of problem, often referred to as the volume of a spherical segment with a cylindrical hole:

Step 1: Visualize the problem

Imagine a sphere with radius $R = 10$ and a cylindrical hole drilled through its center with radius $r = 3$. The length of the hole is equal to the diameter of the sphere, and the height of the cylinder will extend across the entire diameter of the sphere.

Step 2: Set up the integral

The volume of the spherical segment with a cylindrical hole can be computed by subtracting the volume of the cylindrical hole from the volume of the spherical cap it removes.

However, we can also compute this using the following standard formula for a spherical segment with height $h$ and hole radius $r$ (if $r < R$):

$V = \frac{4}{3} \pi R^3 - \pi r^2 (R - h)$

Where $h$ is the distance from the center of the sphere to the edge of the cylindrical hole.

Step 3: Calculate the height $h$

The height $h$ of the spherical cap is related to the sphere’s radius $R$ and the hole’s radius $r$ by:

$h = \sqrt{R^2 - r^2}$

Substitute the values $R = 10$ and $r = 3$:

$h = \sqrt{10^2 - 3^2} = \sqrt{100 - 9} = \sqrt{91}$

Step 4: Compute the volume of the resulting solid

The volume of the cylindrical hole itself can be determined by subtracting the volume of the solid portion of the sphere from the volume of the entire sphere:

$V_{\text{solid}} = \frac{4}{3} \pi R^3 - \pi r^2 \cdot 2h$

First, calculate the total volume of the sphere:

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

Now, compute the volume of the cylindrical hole:

$V_{\text{cylinder}} = \pi r^2 \cdot 2h = \pi (3)^2 \cdot 2 \times \sqrt{91} = 9\pi \cdot 2\sqrt{91} = 18\pi \sqrt{91}$

Final Volume of the Resulting Solid:

Subtract the volume of the cylindrical hole from the volume of the sphere:

$V_{\text{resulting}} = \frac{4000}{3} \pi - 18 \pi \sqrt{91}$

Thus, the volume of the resulting solid is:

$\boxed{V = \frac{4000}{3} \pi - 18\pi \sqrt{91} \text{ cubic units}}$

Would you like to go over any specific step in detail?

Related Questions:

1. How would the volume change if the hole’s radius was increased?
2. What is the surface area of the resulting solid after the hole is drilled?
3. Can this method be extended to ellipsoids instead of spheres?
4. How does the volume change if the hole is not drilled through the center?
5. What happens to the volume if the sphere is sliced instead of having a hole drilled?

Tip: Always double-check your setup when dealing with three-dimensional geometry problems, especially when applying integrals or standard formulas!