Problem 37 - Essential Calculus

Question

In Exercises \(35-38\), set up and evaluate the definite integral for the area of the surface generated by revolving the curve about the \(x\) -axis. $$ y=\frac{x^{3}}{6}+\frac{1}{2 x}, \quad 1 \leq x \leq 2 $$

Step-by-Step Solution

Answer

The area of the surface generated by revolving the given curve about the x-axis is given by the definite integral \(A = 2\pi \int_{1}^{2} \left(\frac{x^{3}}{6} + \frac{1}{2x}\right) \sqrt{1 + \frac{x^{8} - 2x^{4} + 1}{4x^{4}}} dx\). The exact value of this integral can be calculated using standard integration techniques or a numerical method.

Step 1: Differentiate the function

Differentiate \(y = \frac{x^{3}}{6} + \frac{1}{2x}\) with respect to \(x\). This gives \(y' = \frac{x^{2}}{2} - \frac{1}{2x^{2}}\).

Step 2: Simplify the derivative

Simplify \(y'\) by finding a common denominator and combining the fractions. This gives \(y' = \frac{x^{4} - 1}{2x^{2}}\).

Step 3: Find Square of derivative

Square \(y'\) to find \((y')^{2}\). This equals to \(\left(\frac{x^{4} - 1}{2x^{2}}\right)^{2} = \frac{x^{8} - 2x^{4} + 1}{4x^{4}}\).

Step 4: Find value of function y(x)

Substitute \(x = 1\) and \(x = 2\) in \(y = \frac{x^{3}}{6} + \frac{1}{2x}\) to find \(y(1) = \frac{1}{6} + \frac{1}{2} = \frac{2}{3}\) and \(y(2) = \frac{8}{6} + \frac{1}{4} = \frac{4}{3}\).

Step 5: Compute the integral

Substitute \(y, y', a, b\) into the surface area formula and evaluate the integral: \(A = 2\pi \int_{1}^{2} \left(\frac{x^{3}}{6} + \frac{1}{2x}\right) \sqrt{1 + \frac{x^{8} - 2x^{4} + 1}{4x^{4}}} dx \). This can be calculated using standard integration techniques or a numerical method.

Key Concepts

Definite Integral

A definite integral is a mathematical concept that represents the area under a curve between two specified points on the x-axis. In the context of surface area of revolution, the definite integral helps us calculate the total surface area formed by revolving a curve around an axis.
To evaluate the definite integral, we need the two limits of integration, which are often given as the lower and upper bounds of the x-values. In our exercise, these bounds are from 1 to 2.

The definite integral is computed by taking the antiderivative (or integral) of the function, and then evaluating this antiderivative at the upper and lower limits. The result gives us the accumulated value, or the surface area in this context, over the specified interval.

Curve Revolution About X-axis

When we revolve a curve around the x-axis, we create a 3D shape. This concept is crucial in understanding how to calculate the surface area of revolution, which is the total surface area of this 3D shape.
The process involves revolving a 2D curve around the x-axis, effectively sweeping the curve in a circular motion. In our exercise, the function given is \( y = \frac{x^3}{6} + \frac{1}{2x} \), and as each point on this curve rotates around the x-axis, it forms a new shape.

The surface area can be calculated through a formula that involves integrating the function multiplied by \(2\pi\) times the distance from the axis of revolution. This method captures the circular path of the curve, and the definite integral helps accumulate these circular paths into a measurable surface area.

Differentiation

Differentiation is a fundamental calculus concept used to determine how a function changes at any given point. It involves finding the derivative of a function, which represents the rate of change or slope at any point along the curve.
In our solution, differentiating the function \( y = \frac{x^3}{6} + \frac{1}{2x} \) resulted in \( y' = \frac{x^2}{2} - \frac{1}{2x^2} \).

This derivative is essential when calculating the surface area of revolution, as it contributes to the formula for the arc length of the curve segments being revolved. This, in turn, influences the shape and extent of the surface area calculated. The derivative, when squared and compounded with 1, forms part of what is integrated to find the surface area.

Integration Techniques

Integration techniques are various methods used to solve integrals. Some standard methods include substitution, integration by parts, and numerical integration for more complex expressions.
In our given problem, the expression to be integrated involves the function and its derivative squared inside a square root, making it a bit complex.

For solving these integrals, especially when they involve products or complex fractions, substitution can simplify the equation. Additionally, numerical methods such as Simpson's Rule or the Trapezoidal Rule can provide approximate solutions if an analytical solution is too cumbersome.

• Substitution: Used to simplify the integral by changing variables.
• Numerical Integration: Approximates the value of integrals that are difficult to evaluate analytically.

These techniques are indispensable when dealing with integral calculus, particularly for calculating areas and volumes related to curves on rotated axes.