Volume of the Remaining Sphere After Drilling a Cylindrical Hole

In this article, we explore the fascinating problem of calculating the volume of the remaining part of a sphere after a cylindrical hole of radius r 2 is drilled through its axis, with the sphere having a radius R 3.

Calculating the volume of such a complex shape involves a combination of geometric principles and integration techniques. Let's break down the steps to solve this problem in a clear and detailed manner.

Step-by-Step Solution

1. Calculate the Volume of the Sphere

First, we calculate the volume of the original sphere using the formula:

$$V_s frac{4}{3}pi R^3$$

Given that the radius R 3, we can substitute this into the formula:

$$V_s frac{4}{3}pi 3^3 frac{4}{3}pi 27 36pi$$

2. Calculate the Volume of the Cylindrical Hole

The volume of a cylinder is given by:

$$V_c pi r^2 h$$

Here, r 2 is the radius of the cylindrical hole. To find h, the height of the cylinder, we use the Pythagorean theorem in the cross-section of the sphere. The height can be calculated as:

$$h 2 sqrt{R^2 - r^2} 2 sqrt{3^2 - 2^2} 2 sqrt{9 - 4} 2 sqrt{5}$$

Substituting r 2 and h 2 sqrt{5} into the cylinder volume formula:

$$V_c pi 2^2 2 sqrt{5} 8pi sqrt{5}$$

3. Calculate the Volume of the Remaining Part of the Sphere

The volume of the remaining part of the sphere, V_r, is the volume of the sphere minus the volume of the cylindrical hole:

$$V_r V_s - V_c 36pi - 8pi sqrt{5}$$

This gives us the volume of the remaining part of the sphere.

Alternative Method: Integration

A more elegant approach involves integrating the volume directly. Let's say the sphere radius is R and the hole radius is r. We arrange coordinates with the origin at the center and the hole along the x-axis. A slice of the volume perpendicular to the x-axis at x is a ring with inner radius r and the square of the outer radius equal to R^2 - x^2. The area of the ring is a disk of the outer radius minus a disk of the inner radius. We want to integrate this along the x-axis:

$$V int_{-a}^{a} pi (R^2 - x^2 - r^2) , dx$$

The limits correspond with where the hole meets the surface of the sphere, so a sqrt{R^2 - r^2}. Integrating we get:

$$V pi R^2 - r^2 x - frac{x^3}{3} Big|_{-a}^{a} pi 2R^2 - r^2 sqrt{R^2 - r^2}^{1.5} - (R^2 - r^2) sqrt{R^2 - r^2}^{1.5} frac{4}{3}pi R^2 - r^2 sqrt{R^2 - r^2}^{1.5}$$

For the given R 3 and r 2, we get that:

$$V frac{20pi sqrt{5}}{3}$$

This formula checks out when r 0, reducing to the volume of the sphere.

Conclusion

The problem of calculating the remaining volume of a sphere after drilling a cylindrical hole through its axis is a classic example of how geometric principles can be combined with integration to solve complex problems. The final result for our specific case is given by:

$$36pi - 8pi sqrt{5}$$

or equivalently:

$$frac{20pi sqrt{5}}{3}$$