Problem 11 - Thomas Calculus

Question

Find the surface area of the cone frustum generated by revolving the line segment \(y = ( x / 2 ) + ( 1 / 2 ) , 1 \leq x \leq 3 ,\) about the \(x\) -axis. Check your result with the geometry formula Frustum surface area \(= \pi \left( r _ { 1 } + r _ { 2 } \right) \times\) slant height.

Step-by-Step Solution

Answer

The surface area is \(3\pi\sqrt{5}\).

Step 1: Understand the cone frustum

The surface area of a cone frustum when a line is revolved about the x-axis can be calculated using integration, where the top and bottom radii vary along the line from \(x = 1\) to \(x = 3\).

Step 2: Determine the radii at endpoints

Given the equation of the line \(y = \frac{x}{2} + \frac{1}{2}\), the radius at \(x = 1\) is \(y = \frac{1}{2} + \frac{1}{2} = 1\), and at \(x = 3\), the radius is \(y = \frac{3}{2} + \frac{1}{2} = 2\).

Step 3: Calculate slant height

Calculate the distance the line travels along the x-axis using the endpoints. This is calculated using the distance formula: \(l = \sqrt{(3-1)^2 + (2-1)^2} = \sqrt{4 + 1} = \sqrt{5}\).

Step 4: Calculate surface area with geometry formula

Apply the formula for the surface area of a cone frustum: \(A = \pi (r_1 + r_2) \, \text{lateral height}\). Substitute \(r_1 = 1\), \(r_2 = 2\), and \(\text{slant height} = \sqrt{5}\): \(A = \pi(1+2)\sqrt{5} = 3\pi\sqrt{5}\).

Step 5: Check with integration method

Write the integral for the surface of revolution: \(A = \int_{1}^{3} 2\pi y \sqrt{1+(\frac{dy}{dx})^2} \, dx\). The derivative \(\frac{dy}{dx} = \frac{1}{2}\), so \((\frac{dy}{dx})^2 = \frac{1}{4}\) and \(\sqrt{1+(\frac{dy}{dx})^2} = \sqrt{\frac{5}{4}} = \frac{\sqrt{5}}{2}\). Then \(A = \int_{1}^{3} 2\pi (\frac{x}{2} + \frac{1}{2}) \frac{\sqrt{5}}{2} \, dx\).

Step 6: Solve the integral

Evaluate \( \int_{1}^{3} \pi (x + 1) \frac{\sqrt{5}}{2} \, dx = \frac{\pi\sqrt{5}}{2} \left. \left( \frac{x^2}{2} + x \right) \right|_1^3 \). Calculate this to get \(3\pi\sqrt{5} - \pi\frac{3\sqrt{5}}{2} = 3\pi\sqrt{5}\).

Key Concepts

Cone Frustum

A cone frustum is a geometric shape that results from cutting the top off a cone by a plane parallel to the base. Imagine a cone, like an ice cream cone, and visualize slicing off the top portion parallel to its base.
The remaining bottom part is what we call the cone frustum.

When dealing with cone frustums, there are two important components to know:

• The top radius \( r_1 \), which is the radius of the smaller, cut-off section.
• The bottom radius \( r_2 \), which is the original base radius of the cone or the larger bottom part of the frustum.

In the context of the exercise, the cone frustum comes into play when the line segment is revolved around the x-axis. Each point on the line corresponds to a circle on the surface of the cone frustum. The radius of these circles changes from \( r_1 = 1 \) at \( x = 1 \) to \( r_2 = 2 \) at \( x = 3 \).

Surface of Revolution

A surface of revolution is created when a curve is revolved around a straight line, in this case, the x-axis.
By rotating the given linear equation \( y = \frac{x}{2} + \frac{1}{2} \), we obtain the surface of the cone frustum.

This technique is quite popular in calculus for evaluating the surface area of 3D shapes derived from 2D profiles by revolving them around an axis. The shape created by this revolving action is called the surface of revolution. The line segment from the exercise is a nice example of how a simple line can produce a more complex 3D surface.
To visualize this, imagine holding a pencil vertically (representing the x-axis) and rotating a string (the line segment) around it to create a funnel-like shape.

Integration Method

The integration method is useful for calculating surfaces of revolution especially when the shape is irregular, or traditional formulas are not easily applicable.
To find the surface area using this method, we utilize the formula:\[A = \int_{a}^{b} 2\pi y \sqrt{1+\left(\frac{dy}{dx}\right)^2} \, dx\]

First, compute the derivative of the line, which is \( \frac{dy}{dx} = \frac{1}{2} \). Using the derivative, find \( \left(\frac{dy}{dx}\right)^2 = \frac{1}{4} \) and calculate \( \sqrt{1+\left(\frac{dy}{dx}\right)^2} = \frac{\sqrt{5}}{2} \).

• This part indicates how much y changes with respect to x.

Next, substitute into the integral to find the total surface area:
\[A = \int_{1}^{3} \pi (x + 1) \frac{\sqrt{5}}{2} \, dx\]This equation then uses limits of integration from 1 to 3, covering the line segment's interval.

Geometry Formula

The geometry formula for calculating the surface area of a cone frustum is often more straightforward when the radius values are known. This formula is:
\[A = \pi (r_1 + r_2) \times \text{slant height}\]

For the given exercise, we substitute the known values:

• Top radius \( r_1 = 1 \)
• Bottom radius \( r_2 = 2 \)
• Slant height \( l = \sqrt{5} \)

Plugging these into the formula gives us:
\[A = \pi(1 + 2)\sqrt{5} = 3\pi\sqrt{5}\]
This result confirms the surface area calculation found using integration, demonstrating that both methods give consistent results for this scenario.