Problem 23 - 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. \(x=\left(y^{4} / 4\right)+1 /\left(8 y^{2}\right), \quad 1 \leq y \leq 2 ; \quad x\) -axis \(\quad\) (Hint: \(\quad\) Express \(d s=\sqrt{d x^{2}+d y^{2}}\) in terms of \(d y,\) and evaluate the integral \(S=\int 2 \pi y d s\) with appropriate limits.)

Step-by-Step Solution

Answer

The surface area is calculated through the integral: \( S = \int_{1}^{2} 2 \pi y \sqrt{1 + \left( y^3 - \frac{1}{4y^3} \right)^2} \ dy \).

Step 1: Derive Expression for dx/dy

Start by finding the derivative \( \frac{dx}{dy} \) of the given function \( x = \frac{y^4}{4} + \frac{1}{8y^2} \). Use the power rule to differentiate each term, resulting in \( \frac{dx}{dy} = y^3 - \frac{1}{4y^3} \).

Step 2: Setup ds in Terms of dy

We know that \( ds = \sqrt{(dx)^2 + (dy)^2} \). Substitute \( dx = \frac{dx}{dy} \cdot dy \) to express \( ds \) as \( ds = \sqrt{1 + \left( \frac{dx}{dy} \right)^2} dy \).

Step 3: Express the Surface Area Integral

Given the integral equation for surface area is \( S = \int 2 \pi y \ ds \), substitute our \( ds \) expression to get \( S = \int 2 \pi y \sqrt{1 + \left( y^3 - \frac{1}{4y^3} \right)^2} \ dy \).

Step 4: Simplify and Evaluate the Integral

Simplify the expression inside the square root and then set up the integral with the limits of \( y: 1 \) to \( 2 \). It becomes \( S = \int_{1}^{2} 2 \pi y \sqrt{1 + \left( y^3 - \frac{1}{4y^3} \right)^2} \ dy \). Compute the integral through substitution or numerical methods if needed.

Key Concepts

Calculus

Calculus is a branch of mathematics that helps us understand changes, motion, and behaviors of functions. It is divided into two main parts: differential calculus and integral calculus. Differentiate this with simple examples can include understanding how a car's speedometer reading is the rate of change of position, this involves differentiation. Meanwhile, finding the distance traveled over a period from speedometer readings involves integration. This wonderful duality of calculating rates and sums forms the core of calculus.

In the exercise, calculus enables us to find the surface area of a shape created by rotating a curve around an axis, harnessing the power of integrals to capture the surface's entire area. Leveraging these calculus tools provides a structured approach to tackling problems about continuously changing phenomena.

Definite Integral

A definite integral helps find the total accumulation of quantities, such as area or volume, over a defined range. It is represented as \( \int_a^b f(x) \, dx \), where \( a \) and \( b \) denote the limits of integration. When calculating a definite integral, one finds the net area beneath the curve \( f(x) \) between points \( a \) and \( b \).

In the given exercise, the definite integral \( \int_{1}^{2} 2 \pi y \sqrt{1 + \left( y^3 - \frac{1}{4y^3} \right)^2} \, dy \) is essential for determining the surface area of the revolution. Here, the integration boundaries \( y = 1 \) and \( y = 2 \) set the initial and final values across which the process of integration sums up incremental area values around the axis, allowing for a complete calculation of the surface area.

Differentiation

Differentiation is a foundational concept of calculus that deals with finding the rate at which a quantity changes. When you differentiate a function, you're identifying its derivative. This tells us how quickly the function's value is changing at any given point.

In this exercise, differentiation is applied to determine \( \frac{dx}{dy} \) for the given curve \( x = \frac{y^4}{4} + \frac{1}{8y^2} \). Using the power rule, we compute the derivative \( \frac{dx}{dy} = y^3 - \frac{1}{4y^3} \). Such deriving steps are essential before moving towards integrating, as the derivative informs the slope and helps configure the expression for \( ds \), which is crucial in calculating surface area. This rate of change forms an integral part of acquiring the necessary \( ds \) formula for our final integration step.

Surface of Revolution

The surface of revolution emerges when a curve is revolved around an axis, creating a three-dimensional shape. To find the area of such a surface, calculus leverages integration techniques to sum the infinitesimal slices of area into a coherent total.

In the problem addressed, revolving the curve \( x = \frac{y^4}{4} + \frac{1}{8y^2} \) about the x-axis generates a surface of revolution. To find the surface area, we use the formula \( S = \int 2 \pi y \ ds \). Here, \( ds \) involves an expression derived from \( dx/dy \) and \( dy \), providing the component necessary to compute the entire curve's contribution to the surface area. Solving this integral from \( y = 1 \) to \( y = 2 \) completes the calculation, giving us the total surface area for this revolution. This application showcases the power of calculus in crafting meaningful geometric interpretations from functions.