Problem 9 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 1-12, find the volume of the solid generated when the region \(R\) bounded by the given curves is revolved about the indicated axis. Do this by performing the following steps. (a) Sketch the region \(R\). (b) Show a typical rectangular slice properly labeled. (c) Write a formula for the approximate volume of the shell generated by this slice. (d) Set up the corresponding integral. (e) Evaluate this integral. \(x=y^{2}, y=1, x=0 ;\) about the \(x\)-axis

Step-by-Step Solution

Answer

The volume of the solid is \( \frac{\pi}{2} \).

Step 1: Sketch the Region

The region \(R\) is defined by the curves \(x = y^2\), \(y = 1\), and \(x = 0\). The parabola \(x = y^2\) opens to the right, \(y = 1\) is a horizontal line, and \(x = 0\) is the \(y\)-axis. To sketch this region, draw the parabola starting from the origin (0,0) to (1,1) where the line \(y = 1\) intersects it. The region is bounded by the parabola from \(y=0\) to \(y=1\), and to the left by \(x = 0\).

Step 2: Sketch the Typical Slice

The typical slice is a horizontal rectangular strip parallel to the \(x\)-axis. It is at height \(y\) and has a small thickness of \(dy\). The width of the strip is the distance between \(x = 0\) and \(x = y^2\), which is \(y^2\).

Step 3: Volume of the Shell

When this horizontal slice is revolved around the \(x\)-axis, it forms a cylindrical shell. The radius of the shell is \(y\), and its height is \(y^2\). The thickness of the shell is \(dy\). The approximate volume of the shell is given by \(dV = 2\pi (\text{radius})(\text{height})(\text{thickness}) = 2\pi y(y^2)dy\).

Step 4: Set Up the Integral

To find the total volume of the solid, integrate the volume of the shells from \(y = 0\) to \(y = 1\): \[V = \int_{0}^{1} 2\pi y(y^2) \, dy.\] Simplifying the integrand, we have \(2\pi y^3\).

Step 5: Evaluate the Integral

Evaluate the integral to find the volume. \[ V = 2\pi \int_{0}^{1} y^3 \, dy \] First, compute the antiderivative of \(y^3\) which is \((1/4)y^4\). Then evaluate from 0 to 1: \[ V = 2\pi \left[ \frac{1}{4} y^4 \right]_{0}^{1} = 2\pi \left( \frac{1}{4} \cdot 1^4 - \frac{1}{4} \cdot 0^4 \right) = 2\pi \left( \frac{1}{4} \right) = \frac{\pi}{2}. \]

Step 6: Conclusion

The volume of the solid generated by revolving the region around the \(x\)-axis is \(\frac{\pi}{2}\).

Key Concepts

Volume of Revolution

The volume of revolution is a concept in calculus that helps determine the volume of a solid formed when a region in the plane is revolved around a line (axis). Imagine a region between curves or straight lines that you rotate around the x-axis or y-axis. This spinning creates a 3D object, much like spinning a flat drawing on paper to get a vase.
\(\)
In our exercise, we take a region defined by specific curves and revolve it to form a solid. The curves given are the equation of a parabola and straight lines that bound a specific area. By revolving this bounded region about the x-axis, we establish a volume that can be calculated precisely using integration.

• This concept is crucial as it extends the idea of finding areas under curves to discovering volumes.
• It bridges geometry and calculus by using integrals to work out 3D figures.

Understanding the setup of the region and identifying the axis of revolution are crucial first steps in solving these types of problems.

Integrals

Integrals are a core concept in calculus used to calculate areas, volumes, central points, and many other useful things. In problems like our given exercise, they help us find the volume of a solid by pointing out the total accumulation of infinitesimal elements (slices or shells).
\(\)
For calculating volumes of revolved regions, the integral takes into account an infinite sum of small volumes:

• An integral aggregates these small pieces (tiny slices or shells) to give us the total volume.
• We perform this task by setting up the integral and solving it using standard techniques.

The step-by-step exercise demonstrates how to set up and evaluate an integral for the given region. Understanding how to simplify the integrand and execute the integral accurately is key to solving these problems.

Cylindrical Shells

The cylindrical shells method is an ingenious way to compute the volume of revolution. Instead of considering disks and washers, this method considers 'shells'— hollow cylinders—formed when a tiny slice of the region revolves around an axis.
\(\)
In our specific task, a horizontal slice creates a cylindrical shell when rotated about the x-axis. Here's how it works:

• The radius of each shell varies depending on the position of the slice in the original region.
• The height of the shell corresponds to the width of the slice.
• Each shell's volume is determined by multiplying its circumference, height, and thickness, which integrates to give the full volume.

This method is particularly effective when the region is awkwardly shaped or bounded by complex curves, making the disk method cumbersome. Recognizing when to use cylindrical shells helps solve challenging calculus problems efficiently.

Parabolas

Parabolas are smooth, symmetrical curves that appear frequently in calculus problems due to their simple algebraic expression: a quadratic equation. In the exercise at hand, the parabola is given by the function \(x = y^2\).
\(\)
Here's why parabolas are significant in this context:

• Parabolas are used to outline the boundary of the region that is revolved to form a solid.
• Understanding their geometry helps identify the limits for integration when setting up the volume calculations.
• The shape properties of parabolas simplify the integral calculations due to their predictable symmetry.

In calculus, working with parabolas involves recognizing their vertex, focus, and directrix, helping us sketch the region accurately and set boundaries for volume integration.