Question

A round hole of radius \(a\) is drilled through the center of a solid sphere of radius \(b\) (assume that \(b>a\) ). Find the volume of the solid that remains.

Short answer

The remaining solid volume is \(\frac{4}{3} \pi b^3 - 2\pi a^2 \sqrt{b^2-a^2}\).

Step-by-step solution

Understand the Problem

We need to find the volume of a sphere with a cylindrical hole drilled through its center. The sphere has a radius of \(b\), and the cylindrical hole has a radius of \(a\) and extends through the entire sphere, from one side to the other.

Set Up the Integral

Let's use cylindrical coordinates to evaluate the volume of the solid. The remaining volume is equal to the volume of the sphere minus the volume of the drill-out cylinder. The equation of the sphere is \(x^2 + y^2 + z^2 = b^2\). When drilling, the cylinder is defined by \(x^2 + y^2 \leq a^2\) and because the sphere is symmetric around the z-axis, this extends from \(-z\) to \(z\).

Find the Limits of the Cylinder

To find the vertical extent of the cylinder within the sphere, consider the boundary where the cylinder ends. The equation is \(z^2 = b^2 - x^2 - y^2\). Therefore, the limits for \(z\) are \(-\sqrt{b^2 - a^2}\) to \(\sqrt{b^2 - a^2}\), since the base of the cylinder is defined by \(x^2 + y^2 = a^2\).

Calculate Volume of the Cylinder

The cylinder's volume is given by the integral \[V_{cyl} = \int_{-\sqrt{b^2-a^2}}^{\sqrt{b^2-a^2}} \int_0^{2\pi} \int_0^a r \, dr \, d\theta \, dz\]. Evaluate this triple integral to get \(V_{cyl} = \pi a^2 (2\sqrt{b^2-a^2})\).

Calculate Volume of the Sphere

The volume of the sphere is \(V_{sphere} = \frac{4}{3} \pi b^3\).

Find the Remaining Solid Volume

The remaining volume after subtracting the volume of the cylinder is: \[V_{remaining} = V_{sphere} - V_{cyl} = \frac{4}{3} \pi b^3 - \pi a^2 (2\sqrt{b^2-a^2})\].

Simplify the Expression

Simplify the expression for an exact or approximate numerical answer if needed. However at this stage, it can be left as the calculated formula.

Key concepts

Cylindrical coordinates

Cylindrical coordinates are a system of coordinates that extend the two-dimensional polar coordinates with a third coordinate, analogous to the height in Cartesian coordinates. This system is particularly useful in problems involving circular symmetry, like our problem here with a sphere and a cylindrical hole drilled through it. In cylindrical coordinates, a point in space is represented by

• a radial distance from the z-axis, denoted as \(r\)
• an angle \(\theta\) in the xy-plane from the positive x-axis,
• and a height \(z\) parallel to the z-axis.

Use cylindrical coordinates whenever a system has cylindrical shapes or symmetry around an axis. For our problem, this means using a set of polar-like coordinates extended with height, which simplifies dealing with the hollow cylindrical shape inside the sphere.

Volume of a sphere

The volume of a sphere is calculated using the formula: \[ V_{sphere} = \frac{4}{3} \pi b^3 \]where \(b\) is the sphere's radius. This formula gives the total volume of space occupied by a solid sphere without any holes. In our problem, the full volume of the sphere is important to establish the starting point for how much material was initially present before any was removed by drilling the hole. Remember:

• The sphere is a perfect three-dimensional object, with uniform distance from the center to any point on its surface.
• Calculating its volume is straightforward due to its symmetry.

Understanding the sphere's volume helps in understanding the problem as you need this original volume to perform corrections or subtractions, like removing the cylinder's volume from it.

Triple integral

In our solution to finding the volume of the cylindrical hole within the sphere, we utilize a triple integral. This mathematical concept is used to compute volumes by integrating over three dimensions — height, circumference, and radius. The volume of the cylindrical hole is given by the triple integral: \[ V_{cyl} = \int_{-\sqrt{b^2-a^2}}^{\sqrt{b^2-a^2}} \int_0^{2\pi} \int_0^a r \, dr \, d\theta \, dz \]Let's break it down:

• The integration limits for \(z\) are from \(-\sqrt{b^2 - a^2}\) to \(\sqrt{b^2 - a^2}\), determining the height of the cylinder through the sphere.
• The limits for \(\theta\) are from \(0\) to \(2\pi\), covering a full circle around the z-axis (complete revolution).
• And \(r\), from \(0\) to \(a\), spanning from the center of the cylinder, outward to the radius of the hole.

By solving this triple integral, you find the total volume of the cylindrical hole that has been removed from the sphere.

Volume subtraction

Volume subtraction involves removing the volume of one shape from another to find the remaining volume. In our case, it's the volume of the sphere originally minus the volume of the cylindrical hole drilled through it. This step is crucial as we want to know how much of the solid remains after the drilling.

• Start with the total volume of the sphere: \(V_{sphere} = \frac{4}{3} \pi b^3\).
• Subtract the volume of the cylinder: \(V_{cyl} = \pi a^2 (2\sqrt{b^2-a^2})\).

The calculation is straightforward after both volumes are known:\[ V_{remaining} = V_{sphere} - V_{cyl} = \frac{4}{3} \pi b^3 - \pi a^2 (2\sqrt{b^2-a^2}) \]This gives the volume of the remaining solid, showing how much material is left after removing the cylindrical portion.