Question

Use the shell method to find the volume of the following solids. The solid formed when a hole of radius 3 is drilled symmetrically along the axis of a right circular cone of radius 6 and height 9

Short answer

Answer: The volume of the solid after drilling the hole is 918π cubic units.

Step-by-step solution

Formula for the volume of a cone

The formula for finding the volume of a cone is given as: V = (1/3) * π * r^2 * h, where V is the volume, π is the mathematical constant pi (approximately 3.14159), r is the radius of the base of the cone, and h is the height. In our case, the radius r is 6, and the height h is 9.

Calculate the volume of the cone

Now, using the formula, we can find the volume of the cone: V = (1/3) * π * (6^2) * 9 V = (1/3) * π * 36 * 9 V = π * 3 * 36 * 9 V = 972π So, the volume of the cone is 972π cubic units. #Part 2: Finding the volume of the hole using the shell method#

Set up the integral for the shell method

The shell method formula is given as: V_hole = 2π * ∫[a, b] r(x) * h(x) dx, where r(x) is the radius of the cylindrical shell and h(x) is the height. For this exercise, r(x) = x, and h(x) is the remaining height at position x. The height of the shell cylinder can be found using similar triangles: 9/6 = h(x)/(x + 3). Thus, h(x) = (9/6)*(x+3). The bounds of integration for x are [0, 3].

Calculate the integral for the shell method

Now, we can set up our integral for the shell method: V_hole = 2π * ∫[0, 3] x * (9/6)*(x + 3) dx V_hole = 2π * ∫[0, 3] (3/2)x(x + 3) dx Now, proceed with integrating the expression: V_hole = 2π [(1/2)(3/2)x^2 * (x + 3)] (from 0 to 3) Calculate the antiderivative at the limits of integration: V_hole = 2π [(1/2)(3/2)(3^2) * (6) - (1/2)(3/2)(0^2) * (3)] V_hole = 2π [27 - 0] V_hole = 54π So, the volume of the hole is 54π cubic units. #Part 3: Finding the volume of the solid after drilling the hole#

Subtract the volume of the hole from the volume of the cone

Now, to find the volume of the solid after drilling the hole, we subtract the volume of the hole from the volume of the cone: V_solid = V_cone - V_hole V_solid = 972π - 54π V_solid = 918π So, the volume of the solid after drilling the hole is 918π cubic units.

Key concepts

Volume of a Cone

The volume of a cone can be determined using a straightforward formula, which captures the cone's three-dimensional nature. This formula is given by:

• \[ V = \frac{1}{3} \pi r^2 h \]
• Where:
  • \( V \) - Volume of the cone
  • \( \pi \) - Pi, approximately 3.14159
  • \( r \) - The radius of the base
  • \( h \) - The height of the cone

To use this formula, you simply plug in the values for the radius and height. In this example, the cone has a base radius of 6 units and a height of 9 units. By substituting these values, we find the volume of the cone to be \(972\pi\) cubic units. Remember that cones are partially defined by their circular base, making \(\pi\) an essential component of the formula.

Integration

Integration is a fundamental concept in calculus used to find areas, volumes, central points, and many useful things. In this context, integration helps us calculate volumes, particularly when dealing with irregular shapes such as those formed by rotation.

• The process typically involves setting up an integral that sums an infinite number of infinitesimally small quantities (like slices or shells), essentially accumulating the total volume.
• The formula for integration to find volumes can vary, but it often involves setting up bounds of integration and determining the function to integrate based on the geometry of the problem.
• In the shell method, for example, the integration accounts for varying radii along the axis of rotation.

With integration, we can precisely compute not only regular geometric shapes but also complex ones, enhancing our ability to analyze drilled, rotated, or otherwise manipulated solids.

Cylindrical Shells

Cylindrical shells are a powerful concept in volume calculation. They provide a way to visualize and compute volumes of solids of revolution, particularly when the axis of rotation is different from the object's axis of symmetry. The method involves:

• Considering a series of thin cylindrical "shells" to approximate the volume.
• Each shell is like a thin circular band, with infinitesimal thickness and a formula for volume:
  • \[ V_{\text{shell}} = 2\pi \text{(average radius)} \times \text{(height)} \times \text{(thickness)} \]
• Setting up an integral to sum the contributions from all these shells across the defined bounds.

In the exercise, the shell method is used to determine the volume of a hole drilled through the cone, where the integration considers a radius function \(r(x)\) and a height function \(h(x)\), over the limits which define the drill.

Drilled Solids

Drilled solids are fascinating because they involve calculating the volume of objects with voids—holes that often run through the object's central axis. Here's how this works:

• First, determine the volume of the solid without any holes. For example, calculate the full cone's volume first.
• Then, use methods like the shell method to calculate the volume of the hole separately. This is essential because holes can have complex, non-uniform volumes.
• Finally, subtract the volume of the hole from the total volume of the original object. This gives the volume of the remaining solid.

For our cone example, a hole of radius 3 is drilled along the axis. By calculating both the cone's full volume and the hole's volume, and then performing a simple subtraction, we find that the volume of the remaining solid is \(918\pi\) cubic units, showcasing the object's structural modifications.