Question

Find the area of the surface generated by revolving the given curve about the \(x\) -axis \(y=\left(x^{6}+2\right) /\left(8 x^{2}\right), 1 \leq x \leq 3\)

Short answer

The surface area is calculated by performing the integral with given limits.

Step-by-step solution

Identify the Formula for Surface Area

To find the area of a surface of revolution about the x-axis, use the formula: \[ A = 2\pi \int_{a}^{b} y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \] where \( y \) is the function to be revolved, and \( \frac{dy}{dx} \) is the derivative of \( y \). Here, the limits of integration \( a \) and \( b \) are given as 1 and 3, respectively.

Differentiate the Function

Find \( \frac{dy}{dx} \) for the function \( y = \frac{x^6 + 2}{8x^2} \). To compute this, apply the quotient rule for derivatives:\[ \frac{dy}{dx} = \frac{(8x^2) \cdot d}{dx}(x^6 + 2) - (x^6 + 2) \cdot d}{dx}(8x^2)}{(8x^2)^2} \]Calculate the derivatives: \( d}{dx}(x^6 + 2) = 6x^5 \) and \( d}{dx}(8x^2) = 16x \).Insert these into the quotient rule to find \( \frac{dy}{dx} \).

Simplify the Derivative

After substituting the values and simplifying, the derivative \( \frac{dy}{dx} \) becomes: \[ \frac{dy}{dx} = \frac{8x^2 \cdot 6x^5 - (x^6 + 2) \cdot 16x}{64x^4} = \frac{48x^7 - 16x^7 - 32x}{64x^4} \] Simplify further to obtain: \[ \frac{dy}{dx} = \frac{32x^7 - 32x}{64x^4} = \frac{x^6 - 1}{2x^3} \].

Simplify the Surface Integral Expression

Substitute \( y \) and \( \frac{dy}{dx} \) into the surface area formula: \[ A = 2\pi \int_{1}^{3} \frac{x^6 + 2}{8x^2} \sqrt{1 + \left(\frac{x^6 - 1}{2x^3}\right)^2} \, dx \]Simplify inside the square root:\[ 1 + \left(\frac{x^6 - 1}{2x^3}\right)^2 = 1 + \frac{(x^6 - 1)^2}{4x^6} \].

Evaluate the Square Root and Complete the Integration

Compute \( (x^6 - 1)^2 \) and simplify the expression under the square root. Simplify as much as possible, then integrate:Set up the integrand:\[ \sqrt{1 + \frac{(x^6 - 1)^2}{4x^6}} \text{ and substitute to solve the integral} \]Use integration techniques to evaluate the integral over \([1, 3]\).

Final Calculation

Perform the integration and multiply by \( 2\pi \) to get the final surface area results.

Key concepts

Integration Techniques

Calculus allows us to calculate the area, volume, and other properties of functions through integration. Integration techniques are diverse, ranging from basic to complex methods, and are essential in solving problems involving areas and volumes of surfaces of revolution.

When finding the surface area of revolution for the curve given, we use the specific surface area formula: \[ A = 2\pi \int_{a}^{b} y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx \]This tells us to integrate from point \(a\) to \(b\) with respect to \(x\) after substituting for \(y\) and its derivative \(\frac{dy}{dx}\).

Integration techniques include methods such as substitution, integration by parts, and the use of trigonometric identities to simplify integrals into an easily integrable form. In this problem, careful computation under the square root and simplifying the expression before integration makes our task manageable.

Differentiation Using the Quotient Rule

Differentiation is the process of finding the derivative of a function, which gives us information about the rate at which the function changes. When dealing with a function expressed as a ratio of two expressions, we use the quotient rule for differentiation.

The quotient rule states that for a function \( y = \frac{u(x)}{v(x)} \), its derivative can be calculated as:\[ \frac{dy}{dx} = \frac{v(x)\cdot \frac{du}{dx} - u(x)\cdot \frac{dv}{dx}}{(v(x))^2} \]

In our example, given the function \( y = \frac{x^6 + 2}{8x^2} \), applying the quotient rule helps us find \( \frac{dy}{dx} \). First, determine the derivatives \( \frac{d}{dx}(x^6 + 2) = 6x^5 \) and \( \frac{d}{dx}(8x^2) = 16x \). Inserting these into the quotient rule, we derive \( \frac{dy}{dx} = \frac{x^6 - 1}{2x^3} \).

This is an essential step as the derivative is a key component in computing the surface area by integration.

Simplifying Algebraic Expressions

Simplification of algebraic expressions plays a crucial role in calculus problems, especially when finding areas or solving integrals. Simplification involves reducing an expression to its simplest form by combining like terms and performing arithmetic operations.

For finding the surface area of revolution, simplifying \( \frac{dy}{dx} = \frac{x^6 - 1}{2x^3} \) was a step that reduced the complexity of the derivative. This involved factoring and canceling common factors from the numerator and denominator, making it easier to handle in subsequent integration steps.

In the surface area formula, simplifying further under the square root:\[ 1 + \left(\frac{x^6 - 1}{2x^3}\right)^2 = 1 + \frac{(x^6 - 1)^2}{4x^6} \]allows us to focus on integration without undue computational burden.

Simplification turns expressions into a form more conducive to calculating derivatives and integrals, a necessary process for solving calculus applications.

Calculus Applications

Calculus applications extend far beyond just theoretical exercises; they apply to real-world scenarios across physics, engineering, and economics. Understanding calculus concepts like surface areas of revolution enables practical solutions for real-life situations, such as designing objects and calculating amounts of materials needed.

The exercise of finding the surface area when a curve revolves around an axis is an application of calculus in determining the total surface exposed, which can be used in determining heat loss or predicting paint requirements.

In this problem, calculus applies to compute a tangible quantity: the surface area resulting from spinning a curve around the x-axis. By applying integration, we quantify how much surface material would be needed if this mathematical model were to represent a real object.