Problem 15 - Thomas Calculus in SI Units

Question

Find the areas of the surfaces generated by revolving the curves about the indicated axes. If you have a grapher, you may want to graph these curves to see what they look like. $$y=\sqrt{2 x-x^{2}}, \quad 0.5 \leq x \leq 1.5 ; \quad x \text { -axis }$$

Step-by-Step Solution

Answer

The surface area is found by evaluating the integral of the given function's revolving surface, needing computation techniques for exact value.

Step 1: Understand the problem

We are asked to find the area of a surface generated by revolving a curve around the x-axis. The curve is defined by the function \( y = \sqrt{2x - x^2} \) and the interval \( 0.5 \leq x \leq 1.5 \).

Step 2: Set up the formula for the surface area of revolution

The formula to calculate the surface area \( S \) of a curve \( y = f(x) \) when revolved around the x-axis from \( x = a \) to \( x = b \) is given by: \[ S = \int_{a}^{b} 2\pi y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \: dx \]

Step 3: Differentiate the function

Find the derivative \( \frac{dy}{dx} \) of the function \( y = \sqrt{2x - x^2} \). Use the chain rule: \[ \frac{dy}{dx} = \frac{1}{2\sqrt{2x - x^2}} \cdot (2 - 2x) = \frac{1 - x}{\sqrt{2x - x^2}} \]

Step 4: Substitute into the surface area formula

Substitute \( y = \sqrt{2x - x^2} \) and \( \frac{dy}{dx} = \frac{1-x}{\sqrt{2x - x^2}} \) into the surface area formula: \[ S = \int_{0.5}^{1.5} 2\pi \sqrt{2x - x^2} \sqrt{1 + \left(\frac{1-x}{\sqrt{2x - x^2}}\right)^2} \: dx \]

Step 5: Simplify the integral

Simplify the expression under the square root: \[ \sqrt{1 + \left(\frac{1-x}{\sqrt{2x - x^2}}\right)^2} = \sqrt{\frac{2x - x^2 + (1-x)^2}{2x - x^2}} = \sqrt{\frac{2 - 2x + x^2}{2x - x^2}} \]This simplifies further to \( \sqrt{\frac{2(1-x) + x^2}{2x - x^2}} \).

Step 6: Evaluate the integral

Now the problem is to evaluate the integral: \[ \int_{0.5}^{1.5} 2\pi \sqrt{2x - x^2} \cdot \sqrt{\frac{2 - 2x + x^2}{2x - x^2}} \: dx \]The integration can be done by first simplifying inside and then using a computer algebra system or numerical methods for exact solutions, as this integral may not have a simple closed form.

Key Concepts

Surface Area of Revolution

The concept of the surface area of revolution is an exciting area in calculus. It allows us to calculate the area of a three-dimensional surface formed by revolving a two-dimensional curve around a line (usually one of the coordinate axes). In this exercise, the curve defined by the function \( y = \sqrt{2x - x^2} \) is being revolved around the x-axis. This creates a symmetrical surface, often resembling a bowl or vase.

To find this surface area mathematically, we rely on an integral formula. For a curve \( y = f(x) \), the surface area \( S \) resulting from revolving the curve about the x-axis from \( x = a \) to \( x = b \) is calculated using the formula:

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

This formula accounts for both the distance around the curve (circumference in a differential strip) and the slope of the curve to provide a precise surface area calculation.

Understanding this concept involves visualizing how the curve rotates and imagining the three-dimensional shape it produces. It's a great way to tie together geometry, trigonometry, and calculus.

Integration Techniques

Integration techniques are crucial for solving problems like calculating the surface area of revolution. When we set up an integral for this purpose, we often need to manipulate and simplify expressions to ensure we can solve them effectively. These techniques can include methods like substitution, integration by parts, or using technology when a solution can't be easily found analytically.

In this particular exercise, simplifying the expression inside the integral was necessary before attempting to solve it. The expression \( \sqrt{1 + \left(\frac{1-x}{\sqrt{2x - x^2}}\right)^2} \) was simplified using algebraic manipulation to make the integral more approachable.

• This involves expanding and reducing the expression \( \sqrt{\frac{2x - x^2 + (1-x)^2}{2x - x^2}} = \sqrt{\frac{2 - 2x + x^2}{2x - x^2}} \).

Though the integration requires numerical methods or a computer algebra system due to its complexity, the simplification steps make it possible by reducing potential errors and clarifying what must be integrated.

Differentiation

Differentiation is a fundamental concept in calculus, involving finding a function's derivative. The derivative gives us the rate at which a function changes at any given point and is essential for many applications, including the calculation of surface areas of revolution.

For this exercise, we started with a function \( y = \sqrt{2x - x^2} \). To apply the surface area formula, knowing \( \frac{dy}{dx} \) was necessary.

• Using the chain rule, the derivative was determined as \( \frac{dy}{dx} = \frac{1-x}{\sqrt{2x - x^2}} \).

The chain rule is particularly useful here as it allows differentiation of composite functions, which are common in real-world problems.

Understanding how to differentiate accurately ensures that we can apply integral calculus correctly, especially where the slope of the curve influences the area calculation like in this surface area formula.

Definite Integral

The definite integral is a key tool in calculus that computes the accumulation of quantities and is particularly powerful for finding the total value of a function across a specific interval. In this exercise, the definite integral helps find the exact surface area for the curve \( y = \sqrt{2x - x^2} \) as it revolves around the x-axis from \( x = 0.5 \) to \( x = 1.5 \).

The process involves setting up and evaluating the integral:

• \( \int_{0.5}^{1.5} 2\pi \sqrt{2x - x^2} \sqrt{\frac{2 - 2x + x^2}{2x - x^2}} \, dx \)

Given that the integral lacks a straightforward antiderivative, numerical methods or technology are often employed for evaluation.

The definite integral not only finds lengths, areas, and volumes but also gives us insight into how variables accumulate across an interval, providing precise results necessary in scientific and engineering applications.