Problem 65 - Calculus of a Single Variable

Question

Approximating Arc Length or Surface Area In Exercises \(63-66,\) set up the definite integral for finding the indicated arc length or surface area. Then use the integration capabilities of a graphing utility to approximate the are length or surface area. (You will learn how to evaluate this type of integral in Section \(8.8 . )\) Astroid Find the area of the surface formed by revolving the portion in the first quadrant of the graph of \(x^{2 / 3}+y^{2 / 3}=4,0 \leq y \leq 8,\) about the \(y\) -axis.

Step-by-Step Solution

Answer

The surface area of the solid formed by revolving the given curve about the y-axis can be calculated by setting up and approximating the definite integral \( S = 2 \pi \int_{0}^{8} (4 - y^{2 / 3})^{3 / 2} \sqrt{1+\left(- \frac{3y{^{-1/3}}(4 - y^{2 / 3})^{1 / 2}}{2}\right)^2} dy\).

Step 1: Recognize the type of problem

This is a standard calculus problem where we are supposed to find the surface area of a solid of revolution. We are given the equation of the curve, \(x^{2 / 3}+y^{2 / 3}=4\), and asked to find the surface area formed when this curve is revolved about the y-axis between the limits \(0 \leq y \leq 8\).

Step 2: Re-arrange the equation for x

Rearrange the given equation of the curve to solve for x. After rearranging, the equation becomes \(x = (4 - y^{2 / 3})^{3 / 2}\).

Step 3: Find the integral expression for the surface area

The formula for the surface area of a solid of revolution where \(y\) is also a variable is given by \( S = 2 \pi \int_{c}^{d} x \sqrt{1+(\frac{dx}{dy})^{2}} dy\). Calculate the derivative of \(x\) with respect to \(y\), which is \(\frac{dx}{dy}\). After calculation, we get \(\frac{dx}{dy} = - \frac{3y{^{-1/3}}(4 - y^{2 / 3})^{1 / 2}}{2}\). Now substitute \(x\) and \(\frac{dx}{dy}\) into the equation for \( S\). The result is an integral that represents the surface area.

Step 4: Set up the integral

Using the formula derived above, the definite integral to find the surface area revolves the function about the y axis between the limits y=0 to y=8, is calculated. The integral to be evaluated now is \( S = 2 \pi \int_{0}^{8} (4 - y^{2 / 3})^{3 / 2} \sqrt{1+\left(- \frac{3y{^{-1/3}}(4 - y^{2 / 3})^{1 / 2}}{2}\right)^2} dy\).

Step 5: Calculate approximation

Now, use a graphing utility to approximate the value of the integral. The graphing calculator will give a numerical approximation for the surface area. This is expected as the exercise does not yet reach the point of evaluating the integral.

Key Concepts

Surface Area

The surface area of a solid figure is essentially the total area covering its surface. In calculus, when dealing with shapes formed by revolving a curve around an axis, we refer to this as the surface area of a solid of revolution. The key components to understanding this concept are the curve being revolved, the axis of revolution, and the limits (or bounds) within which the surface area is to be calculated.
The general formula for finding the surface area of a solid of revolution is:

• For revolution around the x-axis: \[S = 2 \pi \int_{a}^{b} y\sqrt{1+\left(\frac{dy}{dx}\right)^2} \, dx\]
• For revolution around the y-axis: \[S = 2 \pi \int_{c}^{d} x\sqrt{1+\left(\frac{dx}{dy}\right)^2} \, dy\]

The equation depends on both the function being revolved and its derivative. By calculating the definite integral, we find the surface area of the solid created by the given boundary of the curve.

Solid of Revolution

A solid of revolution is a three-dimensional object obtained by rotating a two-dimensional curve around a straight line (the axis of revolution).
In this task, the curve defined by the equation \(x^{2/3} + y^{2/3} = 4\) is revolved around the y-axis to form the solid. Only the first quadrant of the graph is considered, which implies restrictions on x and y: \(0 \leq y \leq 8\).
The solid of revolution's surface area is calculated by integrating the function with respect to the distance unpacked along the body as it rotates around the axis. When the revolved curve is different from the axis of revolution, it forms a complicated surface, requiring calculus to solve.

Definite Integral

The definite integral is a fundamental concept in calculus that represents the accumulation of quantities. It provides a precise method for calculating the total area under a curve within a specific range, which is crucial for solving problems involving surface area or arc length.
In practical applications, like finding the surface area of a solid of revolution, the definite integral is limited by upper and lower boundaries. These are denoted in the integral notation as the limits of integration, say from \(c\) to \(d\):

• \(\int_{c}^{d} f(y) \, dy\)

These bounds ensure that only the section of the curve within these limits is considered in the calculation. Approximations using graphing utilities or numerical methods can handle complex integrals that are sometimes hard to evaluate analytically.

Arc Length

The arc length in calculus refers to the measure of the distance along a curve. It is calculated using an integral and is essential for determining the length of a curve between two specified points.
In functions revolved about an axis to form a solid of revolution, the arc length formula can guide how the curve's different parts contribute to the overall shape and surface area of the resultant solid.
To find the arc length of a curve represented by a function \(f(x)\) from \(x=a\) to \(x=b\), you use the formula:

• \[L = \int_{a}^{b} \sqrt{1 + \left(\frac{df}{dx}\right)^2} \, dx\]

This integral utilizes the curve’s derivative to consider how steepness affects the total length. In the context of calculating surface areas, understanding the arc length can sometimes be a precursor to setting up your integral for the revolution.