Problem 7 - Thomas Calculus

Question

Hole in sphere \(A\) circular cylindrical hole is bored through a solid sphere, the axis of the hole being a diameter of the sphere. The volume of the remaining solid is $$ V=2 \int_{0}^{2 \pi} \int_{0}^{\sqrt{3}} \int_{1}^{\sqrt{4-z^{2}}} r d r d z d \theta $$ a. Find the radius of the hole and the radius of the sphere. b. Evaluate the integral.

Step-by-Step Solution

Answer

The hole's radius is 1, the sphere's radius is 2, and the volume is \(4\pi\sqrt{3}\).

Step 1: Understanding the Geometry

Firstly, let's understand the geometry of the problem. We have a sphere with a cylindrical hole bored through it. The axis of the cylinder is aligned with one of the sphere's diameters, and the cylindrical hole has a total height of 2 because it ranges from -1 to 1 in terms of the distance along the z-axis, indicating that the diameter of this cylindrical hole is 2. Thus, the radius of this hole is 1. Since the limits of integration suggest symmetry around the z-axis, the radius of the sphere must be 2 as it encompasses the entire range from -2 to 2 in the z-axis.

Step 2: Setting up the Integral

The volume of the remaining solid is expressed as a triple integral in cylindrical coordinates. The integrand is given as: \[V = 2 \int_{0}^{2 \pi} \int_{0}^{\sqrt{3}} \int_{1}^{\sqrt{4-z^{2}}} r \, d r \, d z \, d \theta\]In this setup, \(r\) represents the radial distance in cylindrical coordinates from the z-axis, bounded by 1 and \(\sqrt{4-z^2}\). The angle \(\theta\) ranges from 0 to \(2\pi\), and \(z\) ranges from 0 to \(\sqrt{3}\). This defines a cylindrical segment inside the sphere.

Step 3: Integrate with Respect to r

We first integrate with respect to \(r\):\[\int_{1}^{\sqrt{4-z^{2}}} r \, dr = \left[ \frac{1}{2} r^2 \right]_{1}^{\sqrt{4-z^{2}}} = \frac{1}{2} (4 - z^2) - \frac{1}{2}\]This simplifies to: \[\frac{1}{2} (3 - z^2)\]

Step 4: Integrate with Respect to z

Now, integrate the result with respect to \(z\):\[\int_{0}^{\sqrt{3}} \left(\frac{1}{2} (3 - z^2)\right) \, dz = \frac{1}{2}\left( \int_{0}^{\sqrt{3}} 3 \, dz - \int_{0}^{\sqrt{3}} z^2 \, dz \right)\]Compute these integrals separately:\[\int_{0}^{\sqrt{3}} 3 \, dz = 3z \bigg|_{0}^{\sqrt{3}} = 3\sqrt{3}\]\[\int_{0}^{\sqrt{3}} z^2 \, dz = \frac{1}{3}z^3 \bigg|_{0}^{\sqrt{3}} = \frac{1}{3}(\sqrt{3})^3 = \frac{1}{3} \cdot 3\sqrt{3} = \sqrt{3}\]Substituting these results gives us:\[\frac{1}{2} (3\sqrt{3} - \sqrt{3}) = \frac{1}{2} \cdot 2\sqrt{3} = \sqrt{3}\]

Step 5: Integrate with Respect to θ

Finally, integrate with respect to \(\theta\):\[2 \cdot \int_{0}^{2\pi} \sqrt{3} \, d\theta = 2\sqrt{3} \cdot \theta \bigg|_{0}^{2\pi} = 2\sqrt{3} \cdot 2\pi = 4\pi\sqrt{3}\]

Step 6: Conclusion

The volume of the remaining solid after the cylindrical hole is bored through the sphere is \( 4\pi\sqrt{3} \). The radius of the hole is 1, and the radius of the sphere is 2.

Key Concepts

Volume of Solids

When calculating the volume of solids, especially those with complex shapes like a sphere with a cylindrical hole, it's crucial to break down the solid into more manageable parts. This typically involves using techniques from calculus, such as integration, to find the volume of the solid by adding up infinitesimally small pieces.

• Spheres: A complete sphere has a simple volume formula, \( V = \frac{4}{3} \pi r^3 \), where \( r \) is the radius. But when a hole is bored through it, the problem becomes more complex.
• Cylinders: A cylinder's volume is \( V = \pi r^2 h \), where \( r \) is the radius and \( h \) is the height. This straightforward calculation changes when part of the cylinder is missing, like when a hole runs through it.

To determine the volume of the solid remaining after boring a hole, we need to evaluate the free volume using integration methods. By visualizing the problem and understanding the shapes involved, we can set up integrals that account for all parts of the object.

Cylindrical Coordinates

The cylindrical coordinate system is particularly useful when dealing with problems involving symmetry around an axis, such as a cylindrical hole within a sphere. In this system, each point in space is described using three parameters: the radial distance \( r \), the angle \( \theta \), and the height \( z \).

• Radial Distance (\( r \)): It measures the distance from the point to the z-axis. In our problem, it changes as we move away from the center of the sphere, constrained by the radius of the cylinder.
• Angle (\( \theta \)): It signifies the counterclockwise angle from the positive x-axis on the xy-plane, and varies between 0 and \( 2\pi \) to cover the full circle.
• Height (\( z \)): It describes how "high" or "low" a point is relative to the xy-plane. In this scenario, the z varies from 0 to \( \sqrt{3} \), providing partial symmetry.

Cylindrical coordinates make it easier to integrate over circular paths and cylindrical surfaces by transforming complex integrals into more straightforward calculations.

Triple Integrals

Triple integrals allow us to calculate the volume of three-dimensional regions in space. When combined with cylindrical coordinates, they offer a powerful way to solve problems involving shapes with rotational symmetry.

• Setup: To evaluate a triple integral, \( \int \int \int f(x, y, z) \, dx \, dy \, dz \), the region of integration must be clearly defined, considering the bounds of each variable.
• Integration Order: The sequence of integration (such as \( dr, dz, d\theta \)) can significantly affect the complexity of the problem, and is chosen based on the symmetry and limits of the region.
• Application: In the exercise, the volume of the solid with a cylindrical hole is obtained by integrating in the order of \( r, z, \theta \), where the bounds are influenced by the geometry of the sphere and cylinder.

Using triple integrals, we can calculate how much material is left in the sphere after the hole is removed, taking full advantage of the symmetrical properties to simplify the computations.