Question

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.)

Short answer

The surface area of the bulb and the amount of glass needed to make the bulb can be calculated by identifying the correct formula for the surface area of a solid of revolution, differentiating the given function, substituting into the formula and evaluating the integral. Finally, multiply the derived surface area by the thickness of the glass to approximate the amount of glass needed.

Step-by-step solution

Formula Identification

To start, identify the correct formula to use. The surface area \(S\) of a solid of revolution obtained by rotating a curve \(y=f(x)\), \(a \leq x \leq b\) around the x-axis can be given by \(S=2\pi \int_{a}^{b} y \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} dx\).\n

Derive the function

Differentiate \(y=\frac{1}{3} x^{1 / 2}-x^{3 / 2}\) to get \(\frac{dy}{dx}=\frac{1}{6} x^{-1 / 2}-\frac{3}{2} x^{1 / 2}\). \nThen, square the derivative and add 1, to get \(1+\left(\frac{d y}{d x}\right)^{2}=1+\left(\frac{1}{6} x^{-1 / 2}-\frac{3}{2} x^{1 / 2}\right)^{2}\).

Substitute into the formula

Next, substitute \(y\), \(\frac{d y}{d x}\), \(a\), and \(b\) into the formula. This gives: \(S=2\pi \int_{0}^{1 / 3}\left(\frac{1}{3}x^{1 / 2} - x^{3 / 2}\right) \sqrt{1+\left(\frac{1}{6} x^{-1 / 2}-\frac{3}{2} x^{1 / 2}\right)^{2}} dx\) which is the integral to be computed.

Calculate the Integral

Evaluate this integral to find the surface area of the bulb in square feet.

Approximate the amount of glass needed

Multiply the derived surface area by the thickness of the glass (converted to feet) to approximate the amount of glass needed to make the bulb.

Key concepts

Calculus

Calculus is a branch of mathematics focused on limits, functions, derivatives, integrals, and infinite series. It's divided into two main branches: differential calculus, which deals with the rate at which quantities change, and integral calculus, which deals with the accumulation of quantities and the areas under and between curves. In the context of finding the surface area of a solid of revolution, calculus plays a vital role, as it provides the tools necessary to calculate complex areas and volumes that cannot be easily measured using basic geometry.
Using calculus, we can solve real-world problems like determining the amount of material needed for a three-dimensional object, which is exactly what we're doing in the case of this ornamental light bulb. Calculus allows us to transition from the precise definition to the integral calculation needed to find a solid's surface area.

Solids of Revolution

Solids of revolution are three-dimensional shapes created by rotating a two-dimensional curve about an axis. The resulting solid resembles a pottery piece on a wheel, hence the term 'revolution.' These solids are prevalent in real life, resembling objects such as vases, bowls, and light bulbs, like the one in our exercise.
Calculating the surface area of such an object requires understanding how the shape is generated and employing integral calculus to sum up all the infinitesimally small areas that make up the surface. In education, this method is used to demonstrate a practical application of calculus and to show students how abstract mathematical concepts can be applied to tangible engineering and design problems.

Integration

Integration is a fundamental concept of calculus and a process of finding the integral of a function. An integral, represented by the integral sign \( \int \), essentially calculates the total size or value, such as areas under curves, volumes of solids, and other physical quantities. Integration comes in two primary forms: indefinite integrals, which represent families of functions, and definite integrals, which calculate the net area between the curve and the x-axis over an interval.
In the context of finding the surface area of a solid of revolution, we use definite integration over the interval from \( a \) to \( b \) to precisely calculate the area of the curved surface.

Definite Integrals

Definite integrals have a wide range of applications in physics, engineering, and other sciences, and they are essential for calculating areas, volumes, and other quantities that have clear start and end points. The integral sign with the upper and lower bounds represents the total accumulation of a quantity, such as the surface area of a solid, over the specified interval \( [a, b] \).
When calculating the surface area of the light bulb in our exercise, we need to set up and evaluate a definite integral from \( 0 \) to \( \frac{1}{3} \) to cover the full shape of the bulb. This process symbolizes the practical application of advanced mathematical techniques to design and engineering challenges, emphasizing the importance of a strong foundation in calculus for technical professions.