Problem 64 - 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 \(8.8 . )\) Bulb Design An ornamental light bulb is designed by revolving the graph of $$y=\frac{1}{3} x^{1 / 2}-x^{3 / 2}, \quad 0 \leq x \leq \frac{1}{3}$$ about the \(x\) -axis, where \(x\) and \(y\) are measured in feet (see figure). Find the surface area of the bulb and use the result to approximate the amount of glass needed to make the bulb. (Assume that the glass is 0.015 inch thick.)

Step-by-Step Solution

Answer

The integral can be resolved using the graphing utility. To approximate the glass needed, multiply the surface area by the thickness of the glass (0.015 inches).

Step 1: Formula for the Area

Start with the formula for finding the surface area of a volume of revolution around the x-axis, which is \(A = 2\pi \int_{a}^{b} y \sqrt{1 + (y')^2} dx\) where \(y = \frac{1}{3}x^{1/2} - x^{3/2}\), \(y'\) is the derivative of \(y\) with respect to \(x\), and \(a\) and \(b\) are 0 and \(\frac{1}{3}\) respectively.

Step 2: Finding the derivative

Before proceeding with the calculation, first, find the derivative of the function \(y = \frac{1}{3}x^{1/2} - x^{3/2}\). The derivative is \(y' = \frac{1}{6}x^{-1/2} - \frac{3}{2}x^{1/2}\).

Step 3: Substituting Values

Substitute the values of \(y\), \(y'\), \(a\), and \(b\) into the formula and we get \[A = 2\pi \int_{0}^{1/3} (\frac{1}{3}x^{1/2} - x^{3/2}) \sqrt{1+[(\frac{1}{6}x^{-1/2} - \frac{3}{2}x^{1/2})]^2} dx\].

Step 4: Evaluation of the Integral

At this point, it is instructed to use a graphing utility to approximate the definite integral. So, evaluate the integral with this graphing utility.

Step 5: Approximation for the Glass

Lastly, approximate the amount of glass needed, but multiplying the surface area of the bulb by the thickness of the glass.

Key Concepts

Arc Length

The concept of arc length in calculus is essential when dealing with curves on a graph. In simple terms, arc length refers to the distance along a curve between two points. To calculate the arc length of a curve, we typically use an integral formula:
\[L = \int_{a}^{b} \sqrt{1 + \left(f'(x)\right)^2} \, dx\]
Where:

• \( L \) is the arc length.
• \( f'(x) \) is the derivative of the function you're working with.
• \( a \) and \( b \) are the interval limits defined on the x-axis.

The process generally involves finding the derivative of the function to determine how the curve behaves. You then apply the integration formula above. It's important to use graphing utilities when the integral is too complex for manual computation.
In practice, calculating arc length helps in various applications, like designing aesthetically pleasing paths, or in engineering, where precise lengths are necessary for construction.

Surface Area

Surface area plays a crucial role, especially when dealing with volumes of revolution. When a curve on a graph is revolved around an axis, it creates a 3D surface. Calculating the surface area of such a shape involves using calculus to measure the geometric surface.
The basic formula for finding the surface area of a surface of revolution around the x-axis is:
\[A = 2\pi \int_{a}^{b} y \, \sqrt{1 + (y')^2} \, dx\]
In this formula:

• \( A \) represents the surface area.
• \( y \) is the function you're revolving.
• \( y' \) is the derivative of \( y \) with respect to \( x \).
• \( a \) and \( b \) are the bounds along the x-axis.

To find the surface area, you first differentiate \( y \), substitute it back into the formula, and then compute the integral. Often, graphing utilities are used to approximate because manual calculations can be difficult. Problems like finding the surface area help understand how to apply calculus to real-world objects, such as bulbs, and calculate the necessary material.

Definite Integral

A definite integral is a core concept in calculus used to calculate the area under a curve between two points on the x-axis. It is represented by:
\[\int_{a}^{b} f(x) \, dx\]
Where:

• \( a \) and \( b \) are the lower and upper bounds of integration.
• \( f(x) \) is the function being integrated.

The definite integral gives us a numerical value which represents the accumulation of quantities, which could be areas, volumes, or other small amounts summed into a whole.
To solve a definite integral, you may require anti-differentiation, and sometimes graphing utilities become necessary to handle complex integrals. Understanding definite integrals is vital when dealing with scenarios that involve finding total values over a range, such as calculating total materials needed, like glass in an ornamental bulb.

Volume of Revolution

When you rotate a curve around an axis, a 3D shape is created, and the volume of this shape can be calculated using the concept of volume of revolution. This is vital in engineering and manufacturing where shapes are designed for functional purposes. The formula for the volume of revolution when rotating a curve around the x-axis is:
\[V = \pi \int_{a}^{b} [f(x)]^2 \, dx\]
Where:

• \( V \) stands for the volume.
• \( f(x) \) is the function that defines the curve.
• \( a \) and \( b \) are the interval limits.

The process involves squaring the function, multiplying by \( \pi \), and integrating to find the volume. This technique is often used for objects like vases, bottles, or bulbs—essentially any object that can be formed by rotating a profile around an axis. By applying these concepts, you can design objects efficiently, ensuring the right amount of material is used and the design is structurally sound.