Question

A solid is described. Orient the solid along the \(x\) -axis such that a cross- sectional area function \(A(x)\) can be obtained, then apply Theorem 7.2 .1 to find the volume of the solid. A skew right circular cone with height of 10 and base radius of 5.

Short answer

The volume of the skew cone is \(\frac{250\pi}{3}\) cubic units.

Step-by-step solution

Understand the Geometrical Configuration

A skew right circular cone is a cone whose apex is not aligned right above the center of its base, making it an asymmetrical cone. Given the height of 10 and a base radius of 5, understand that the cross-sections perpendicular to the height (x-axis) will be circular with varying radii.

Set Up the Cross-sectional Area Function

The cross-sectional area function, \(A(x)\), describes the area of the circle when a slice is taken at a distance \(x\) from the apex along the x-axis. As the cone is skew, the radius linearly changes from 0 at the apex to 5 at height 10. The radius at a distance \(x\) is computed as \(5 - \frac{5}{10}x = 5 - 0.5x\). Hence, the area function is \(A(x) = \pi (5 - 0.5x)^2\).

Apply Theorem 7.2.1

Theorem 7.2.1, known as the Disk Method, is used to calculate the volume of a solid of revolution. The volume \(V\) of the solid by integrating along the x-axis is given by \(V = \int_0^{10} A(x) \, dx = \int_0^{10} \pi(5 - 0.5x)^2 \, dx\).

Simplify and Integrate

Simplify the integrand: \((5 - 0.5x)^2 = 25 - 5x + 0.25x^2\). The integral becomes \(\int_0^{10} \pi(25 - 5x + 0.25x^2) \, dx\). Simplifying \(\int_0^{10} \pi(25 - 5x + 0.25x^2) \, dx = \pi \left[ 25x - \frac{5}{2}x^2 + \frac{0.25}{3}x^3 \right]_0^{10}\).

Compute the Definite Integral

Calculate the definite integral: Substitute \(x = 10\) and \(x = 0\) into the antiderivative obtained:\[V = \pi \left[ 25(10) - \frac{5}{2}(10)^2 + \frac{0.25}{3}(10)^3 \right] - \pi \left[ 0 \right]\]This simplifies to \[V = \pi \left( 250 - 250 + \frac{250}{3} \right)\]. Evaluate to find \(V = \frac{250\pi}{3}\).

Final Calculation and Volume

The final volume of the skew cone is computed in step 5 as \(V = \frac{250\pi}{3}\) cubic units. This takes into account the variability in the radius along the cone's height.

Key concepts

Disk Method

The Disk Method is a vital technique in calculus used to find the volume of a solid of revolution. It involves the concept of slicing the solid into thin disks, each having a small thickness along the axis of revolution.
These disks can be thought of as tiny circular slices of the solid. When you stack them together, they approximate the entire volume.
The radius of each disk is determined by a function, often referred to as the cross-sectional area function. By integrating the area of these disks from one end of the solid to the other, you can find the total volume.

• Conceptualize the solid as made of many thin slices.
• The disks are perpendicular to the axis of revolution.
• Volume is the sum of all these infinitesimally thin disks.

Integration

Integration is fundamental in calculating the volume of solids using the disk method.
Essentially, integration helps you sum up an infinite number of infinitely small quantities.
In the context of volumetric calculations, these quantities are the tiny disks we've mentioned previously. To solve these problems:

• Transform the problem into a mathematical expression featuring integration.
• Set up limits according to the solid's height or other relevant dimensions.
• Perform integration to calculate the total volume.

The integration process transforms theoretical concepts into real numerical solutions. This makes it a powerful tool for engineers and scientists.

Cross-sectional Area Function

The cross-sectional area function, often denoted as \(A(x)\), plays a critical role in finding the volume of a solid. It represents the area of a slice at any point \(x\) along the given axis.
For solids like cones or cylinders, this function often results in a formula involving pi because the shapes of the slices are circular.
When setting up \(A(x)\):

• Identify how the shape's radius changes along the axis.
• Express this change as a function of \(x\).
• Compute the area using the formula for the area of a circle, \(A(x) = \pi r(x)^2\).

Understanding and correctly setting up \(A(x)\) are crucial for applying the disk method.

Definite Integral

The definite integral is the bridge between the theoretical concept of integration and the actual numerical volume.
To evaluate a definite integral, you calculate the antiderivative and then apply the limits which represent the start and end points of integration.
This gives the volume of the solid:

• Find the antiderivative of the integrand, which includes the area function \(A(x)\).
• Use limits of integration based on the solid's physical dimensions.
• Subtract the value of the antiderivative at the lower limit from its value at the upper limit to find the volume.

The process of evaluating definite integrals connects theoretical calculus with real-world volumetric calculation.