Problem 18 - Thomas Calculus

Question

Find the areas of the surfaces generated by revolving the curves in Exercises \(13-23\) about the indicated axes. If you have a grapher, you may want to graph these curves to see what they look like. $$x=(1 / 3) y^{3 / 2}-y^{1 / 2}, \quad 1 \leq y \leq 3 ; \quad y -axis$$

Step-by-Step Solution

Answer

The surface area is given using the formula for revolution about the y-axis with integration bounds 1 to 3.

Step 1: Understand the Formula for Surface Area of Revolution

When a curve is revolved around the y-axis, the formula for the surface area is given by \[ A = 2\pi \int_{c}^{d} x \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \ dy \]where \(x\) is the curve in terms of \(y\), \(\frac{dx}{dy}\) is the derivative of \(x\) with respect to \(y\), and \([c, d]\) is the interval over which the function is defined.

Step 2: Compute the Derivative of x with Respect to y

Given \(x = \frac{1}{3}y^{3/2} - y^{1/2}\), we need to find \(\frac{dx}{dy}\). Start by differentiating the expression:\[\frac{dx}{dy} = \frac{d}{dy} \left(\frac{1}{3}y^{3/2} - y^{1/2}\right) = \frac{1}{2}y^{1/2} - \frac{1}{2}y^{-1/2}\].

Key Concepts

Integral Calculus

Integral calculus is a fundamental branch of mathematics focused on finding areas, volumes, and accumulations over a particular interval. It is essential in computing the surface area of a curve when revolved around an axis. This problem specifically utilizes integral calculus in the form of definite integrals to calculate the surface area.
The formula for the surface area of revolution, when revolving around the y-axis, utilizes the integral: \[A = 2\pi \int_{c}^{d} x \sqrt{1 + \left(\frac{dx}{dy}\right)^2} \, dy\]This formula calculates the surface area by integrating over an interval \(c, \, d\), which represents the bounds of the curve along the axis of revolution. To apply this, you first need to express the function in terms of the variable over which you are integrating, here, that is in terms of \(y\).

Derivative of Functions

The derivative of a function measures how the function changes as its input changes. It is vitally important in finding the slope of a function, and in our context, it is crucial for computing surface areas.
To solve the problem presented, you need to differentiate the given function of \(x = \frac{1}{3}y^{3/2} - y^{1/2}\) with respect to \(y\). This derivative \(\frac{dx}{dy}\) reflects the rate of change of x as y changes.Differentiating, you find:\[\frac{dx}{dy} = \frac{d}{dy} \left(\frac{1}{3}y^{3/2} - y^{1/2}\right) = \frac{1}{2}y^{1/2} - \frac{1}{2}y^{-1/2}\]This derivative is then used in the integral to account for the curve's slope, playing a critical role in the computation of the surface area.

Curve Sketching

Curve sketching is the graphical representation of a function. It helps visualize the behavior and characteristics of a curve. Although not always required in mathematical computations, it helps to understand the shape and main features of the function before solving complex problems like surface area of revolution.
For this exercise, if you have access to graphing tools, sketching the curve \(x=\frac{1}{3}y^{3/2}-y^{1/2}\) over the interval \(1 \leq y \leq 3\) can be incredibly helpful. This sketch lets you see the curve’s relationship with the y-axis, which is crucial since it is being revolved around this axis. Seeing the curve helps validate the correctness of your setup before jumping into calculations.

Axis of Revolution

The axis of revolution is the line around which a two-dimensional shape or curve is revolved to generate a three-dimensional solid. Identifying this axis is critical as it directly influences the calculation of the surface area of the resultant solid.
In this case, the curve defined by \(x=\frac{1}{3}y^{3/2}-y^{1/2}\) is revolved around the y-axis to create a surface area. Therefore, all calculations must consider that the function is described in terms of \(y\). Knowing the correct axis ensures you apply the formula for surface area of revolution properly, making it possible to correctly integrate values over the specified interval \([1, 3]\). This precise setup is foundational in solving revolving surface problems accurately.