Problem 15 - Thomas Calculus

Question

Find the areas of the surfaces generated by revolving the curves in Exercises \(13-22\) about the indicated axes. If you have a grapher, you may want to graph these curves to see what they look like. \(y=\sqrt{2 x-x^{2}}, \quad 0.5 \leq x \leq 1.5 ; \quad x\) -axis

Step-by-Step Solution

Answer

The surface area is \( 2\pi \).

Step 1: Understanding the Surface Area Formula

To find the surface area generated by revolving a curve around the x-axis, we use the formula:\[ A = 2\pi \int_{a}^{b} y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \]where \( y \) is the function we are revolving, and \( \frac{dy}{dx} \) is its derivative.

Step 2: Calculate the Derivative

We need to find \( \frac{dy}{dx} \) for \( y = \sqrt{2x - x^2} \). Let's use the chain rule. Let \( u = 2x - x^2 \), so \( y = \sqrt{u} \) implies \( \frac{dy}{dx} = \frac{1}{2\sqrt{u}} \cdot \frac{du}{dx} \). Calculate \( \frac{du}{dx} = 2 - 2x \).So, \( \frac{dy}{dx} = \frac{2 - 2x}{2\sqrt{2x - x^2}} = \frac{1 - x}{\sqrt{2x - x^2}} \).

Step 3: Set Up the Integral for the Surface Area

Substitute the function and its derivative into the surface area formula. We have:\[ A = 2\pi \int_{0.5}^{1.5} \sqrt{2x - x^2} \sqrt{1 + \left(\frac{1-x}{\sqrt{2x - x^2}}\right)^2} \, dx \]Simplify the expression inside the integral before solving.

Step 4: Simplify the Expression Inside the Integral

The expression under the square root becomes:\[ 1 + \left(\frac{1-x}{\sqrt{2x - x^2}}\right)^2 = \frac{(2x - x^2) + (1-x)^2}{2x - x^2} \]\[ = \frac{2x - x^2 + 1 - 2x + x^2}{2x - x^2} = \frac{1}{2x - x^2} \]This simplifies the integral to:\[ A = 2\pi \int_{0.5}^{1.5} \sqrt{2x - x^2} \cdot \frac{1}{\sqrt{2x - x^2}} \, dx \]which reduces to:\[ A = 2\pi \int_{0.5}^{1.5} 1 \, dx \].

Step 5: Evaluate the Integral

The integral \( \int_{0.5}^{1.5} 1 \, dx \) is straightforward to evaluate. It equals:\[ \left[ x \right]_{0.5}^{1.5} = 1.5 - 0.5 = 1 \].Thus, the surface area is:\[ A = 2\pi \times 1 = 2\pi \].

Step 6: Conclusion

The surface area of the surface generated by revolving \( y = \sqrt{2x - x^2} \) around the x-axis from \( x = 0.5 \) to \( x = 1.5 \) is \( 2\pi \).

Key Concepts

Calculus

Calculus is a branch of mathematics that deals with changes and motion. It is essential for understanding various types of physical quantities such as speed and acceleration.
There are two main branches of calculus:

• Differential Calculus: This focuses on rates of change and slopes of curves.
• Integral Calculus: This primarily deals with total size or value, such as volumes and areas.

Calculus plays a critical role in analyzing functions and their behaviors. Finding the surface area of revolution is one such application. This involves rotating a curve around a specific axis to form a 3-dimensional object. Understanding the nature of the curve and the process of rotation is crucial for accurate calculations.
It’s also important to use derivatives from calculus to express such rotations in mathematical terms, paving the way for integral calculus to take over and compute total surfaces.

Integral Calculus

Integral calculus involves the process of finding integrals, which are used to determine areas, volumes, and other cumulative quantities.
In terms of a surface area of revolution, integral calculus is used to sum up infinitely small pieces of the surface to get the entire area. The formula for the surface area of a function revolved around the x-axis is:
\[ A = 2\pi \, \int_{a}^{b} y \, \sqrt{1 + \left( \frac{dy}{dx} \right)^2} \, dx \]
Here, the function, represented by \( y \), is revolved to form a surface. The integral sign \( \int \) indicates the accumulation of all surface pieces between limits \( a \) and \( b \). The term \( 2\pi \) accounts for the circular nature of the revolution, and the derivative \( \frac{dy}{dx} \) provides information about the curve's slope that adjusts its twisting and turning.
Integral calculus is often challenging, requiring a good grasp of integration techniques and sometimes simplifications like algebraic manipulations. It is an invaluable tool in handling complex geometries in mathematics and physics.

Chain Rule

The chain rule is a fundamental technique in differential calculus used for finding the derivative of composite functions.
If you have a function \( y = f(u) \) and \( u = g(x) \), the chain rule allows you to find \( \frac{dy}{dx} \) by computing:
\[ \frac{dy}{dx} = \frac{dy}{du} \times \frac{du}{dx} \]
In the context of the example given, the function \( y = \sqrt{2x - x^2} \) can be managed using the chain rule by setting \( u = 2x - x^2 \). This transforms \( y \) into \( \sqrt{u} \). To find the derivative, you first compute \( \frac{dy}{du} = \frac{1}{2\sqrt{u}} \), and \( \frac{du}{dx} = 2 - 2x \).
Using the chain rule, you multiply these derivatives to find \( \frac{dy}{dx} \). This approach is essential whenever functions are composed within each other, maintaining smooth calculation progress in larger calculus problems.

Curve Sketching

Curve sketching is a technique used to gain insights into the behavior and shape of functions. It involves drawing the graph of a function based on its formula and known properties.
A good sketch should capture important features of the curve like:

• Intercepts: Points where the curve crosses the axes.
• Asymptotes: Lines that the curve approaches but doesn't touch.
• Critical points: Locations where the slope is zero, indicating possible peaks, troughs, or inflection points.
• Intervals: Sections on the x-axis where the function is increasing or decreasing.

When dealing with revolving curves, a proper sketch will help visualize the volume or surface area being calculated.
For example, sketching the curve \( y = \sqrt{2x - x^2} \) helps identify its symmetry and boundaries. Knowing that it revolves around the x-axis, recognizing its highest and lowest points assists in calculating areas more accurately.
Curve sketching acts as a precursor to analytical methods, providing intuitive insights into how calculus principles apply to real-world scenarios.