Problem 10 - Thomas Calculus in SI Units

Question

Find the lateral (side) surface area of the cone generated by revolving the line segment \(y=x / 2,0 \leq x \leq 4,\) about the \(x\) -axis. Check your answer with the geometry formula Lateral surface area \(=\frac{1}{2} \times\) base circumference \(\times\) slant height.

Step-by-Step Solution

Answer

Lateral surface area = \(4\pi \sqrt{5}\) square units.

Step 1: Understand the problem

We need to find the lateral surface area of a cone formed by revolving the line segment given by the equation \(y = \frac{x}{2}\) from \(x = 0\) to \(x = 4\) around the \(x\)-axis. This involves using calculus to integrate, as well as verifying using the geometry formula for lateral surface area of a cone.

Step 2: Set up the integral for lateral surface area

To find the lateral surface area of the cone, use the formula for the surface area of revolution: \[ A = \int_{a}^{b} 2\pi y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \] Here, \(y = \frac{x}{2}\), so \(\frac{dy}{dx} = \frac{1}{2}\). Substitute this into the formula.

Step 3: Substitute and simplify the integral

Substitute \(y = \frac{x}{2}\) and \(\frac{dy}{dx} = \frac{1}{2}\) into the integral: \[ A = \int_{0}^{4} 2\pi \left(\frac{x}{2}\right) \sqrt{1 + \left(\frac{1}{2}\right)^2} \, dx \] Simplify to \[ A = \pi \int_{0}^{4} x \sqrt{1 + \frac{1}{4}} \, dx \] Which becomes \[ A = \pi \int_{0}^{4} x \sqrt{\frac{5}{4}} \, dx \] \[ A = \pi \sqrt{\frac{5}{4}} \int_{0}^{4} x \, dx \]

Step 4: Calculate the integral

Calculate the integral: \[ \int_{0}^{4} x \, dx = \left. \frac{x^2}{2} \right|_{0}^{4} = \frac{4^2}{2} - \frac{0^2}{2} = 8 \] Thus, the area \( A = \pi \sqrt{\frac{5}{4}} \times 8 = 4\pi \sqrt{5} \) square units.

Step 5: Verify using the geometry formula

According to the geometry formula, the lateral surface area of a cone is \(\frac{1}{2} \times\) base circumference \(\times\) slant height. Here, base radius \(r = 2\) and height \(h = 4\). The slant height \(l\) is calculated as: \[ l = \sqrt{r^2 + h^2} = \sqrt{2^2 + 4^2} = \sqrt{20} = 2\sqrt{5} \] Base circumference \(= 2\pi \times r = 4\pi\). Therefore, lateral surface area \(= \frac{1}{2} \times 4\pi \times 2\sqrt{5} = 4\pi \sqrt{5} \) square units.

Key Concepts

Surface Area of Revolution

The concept of the surface area of revolution is fascinating because it allows us to calculate surfaces formed by rotating curves around an axis. In this context, imagine revolving a straight line to create a 3D cone. The line segment given by the equation \( y = \frac{x}{2} \) from \( x = 0 \) to \( x = 4 \) is rotated around the \( x \)-axis to generate a cone.

When we deal with such a scenario, we use the formula:

• \( A = \int_{a}^{b} 2\pi y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \)

This formula is crucial because it incorporates both the curve's function \( y \) and its derivative \( \frac{dy}{dx} \), which helps account for the slope and shape of the generating curve. Understanding how each element of the integral contributes to calculating an accurate surface area is key. In this problem, the derivative \( \frac{dy}{dx} \) is constant since \( y = \frac{x}{2} \), simplifying calculations.

Integration in Calculus

Integration in calculus plays a significant role in calculating the lateral surface area of a cone formed by a line segment revolution. Once we set up the proper integral, we perform integration to find the exact surface area. Let's break it down:

For a cone formed by revolving the line \( y = \frac{x}{2} \), our main task is to tackle the integral:

• \( \pi \int_{0}^{4} x \sqrt{\frac{5}{4}} \, dx \)

Upon simplifying, this becomes:

• \( \pi \sqrt{\frac{5}{4}} \int_{0}^{4} x \, dx \)

Solving this integral requires knowledge of basic integration rules. Here the integral of \( x \) with respect to \( x \) is \( \frac{x^2}{2} \), calculated from bounds 0 to 4. Plug in the limits, and you find the value of the integral as 8. Finally, multiplying it by \( \pi \sqrt{\frac{5}{4}} \) gives the surface area \( 4\pi \sqrt{5} \) square units. Each step of integration builds a framework for solving complex geometrical surface areas.

Geometry Formula Verification

Verification through geometry formula is a vital step in ensuring our calculated values align with known geometric principles. In particular, for a cone, the lateral surface area can be verified with:

• Geometry Formula: \( \frac{1}{2} \times \text{base circumference} \times \text{slant height} \)

To break it down:

• Base radius \( r = 2 \)
• Height \( h = 4 \)
• Slant height \( l = \sqrt{r^2 + h^2} = 2\sqrt{5} \)
• Base circumference \( 2\pi r = 4\pi \)

Using these, calculate: \( \frac{1}{2} \times 4\pi \times 2\sqrt{5} = 4\pi \sqrt{5} \) square units. This confirms that both the calculus-based derivation and the geometric formula yield the same result. Such cross-verification reinforces the reliability of using calculus for geometric problem-solving.