Problem 30 - Thomas Calculus

Question

For some regions, both the washer and shell methods work well for the solid generated by revolving the region about the coordinate axes, but this is not always the case. When a region is revolved about the \(y\) -axis, for example, and washers are used, we must integrate with respect to \(y .\) It may not be possible, however, to express the integrand in terms of \(y .\) In such a case, the shell method allows us to integrate with respect to \(x\) instead. Exercises 29 and 30 provide some insight. $$ \begin{array}{l}{\text { Compute the volume of the solid generated by revolving the triangular }} \\ {\text { region bounded by the lines } 2 y=x+4, y=x, \text { and } x=0} \\ {\text { about }}\end{array} $$ $$ \begin{array}{l}{\text { a. the } x \text { -axis using the washer method. }} \\\ {\text { b. the } y \text { -axis using the shell method. }} \\ {\text { c. the line } x=4 \text { using the shell method. }} \\ {\text { d. the line } y=8 \text { using the washer method. }}\end{array} $$

Step-by-Step Solution

Answer

a. \(8\pi\), b. \(\frac{32\pi}{3}\), c. \(32\pi\), d. \(64\pi\).

Step 1: Determine region boundaries and intersections

The equations of the lines are given as \( 2y = x + 4 \) or \( y = \frac{x}{2} + 2 \), \( y = x \), and \( x = 0 \). Find the intersections of these lines to determine the vertices of the triangular region. Solving \( y = \frac{x}{2} + 2 \) and \( y = x \) gives \( x = 4 \). Thus, the vertices are A: \( (0, 0) \), B: \( (4, 4) \), and C: \( (0, 2) \).

Step 2a: Volume about the x-axis using the washer method

When revolving around the \( x \)-axis, use the washer method with outer radius \( R(y) = \frac{y - 2}{1/2} \) (solving \( y = \frac{x}{2} + 2 \) for \( x \)) and inner radius \( r(y) = y \). The volume is \[ V = \pi \int_{0}^{2} \left(R(y)^2 - r(y)^2\right)\, dy = \pi \int_{0}^{2} \left( \left(2(y-2)\right)^2 - y^2 \right) \ dy. \] Compute the integral to find \( V \).

Step 3a: Compute integral for x-axis

The integrand becomes \( \left(2(y-2)\right)^2 - y^2 = 4(y-2)^2 - y^2 = 4(y^2 - 4y + 4) - y^2 = 4y^2 - 16y + 16 - y^2 = 3y^2 - 16y + 16 \). Integrating: \[ V = \pi \left[ \frac{3y^3}{3} - 16 \frac{y^2}{2} + 16y \right]_0^2 = \pi \left[ 8 - 32 + 32 \right]_0^2 = 8\pi. \]

Step 2b: Volume about the y-axis using the shell method

For a shell method revolving around the \( y \)-axis, use \( x \) as variable, height of shell \( h(x) = \frac{x}{2} + 2 - x = -\frac{x}{2} + 2 \), and radius \( r(x) = x \). The volume is \[ V = 2\pi \int_{0}^{4} x \left(-\frac{x}{2} + 2 \right) \, dx = 2\pi \int_{0}^{4} \left(-\frac{x^2}{2} + 2x \right)\, dx. \] Integrate to find the volume.

Step 3b: Compute integral for y-axis

Simplify and integrate: \( 2\pi \left[ -\frac{x^3}{6} + x^2 \right]_0^4 = 2\pi \left[ -\frac{64}{6} + 16 \right] = 2\pi \left[ \frac{-32}{3} + 16 \right] = 2\pi \cdot \frac{16}{3} = \frac{32\pi}{3} \).

Step 2c: Volume about the line x=4 using shell method

For rotation around \( x=4 \), set shell height the same as for the \( y \)-axis, but adjust the radius as \( r(x) = 4-x \). The volume formula is \( V = 2\pi \int_{0}^{4} (4-x)\left( \frac{-x}{2} + 2 \right) \, dx \). Compute the integral.

Step 3c: Compute integral for line x=4

After simplifying, integrate: \( 2\pi \int_{0}^{4} (4-x)\left( \frac{-x}{2} + 2 \right) \, dx = 2\pi \left[ -\frac{x^3}{6} + 2x^2 - 2x^2 + 8x \right]_0^4 = 2\pi \cdot 16 = 32\pi \).

Step 2d: Volume about line y=8 using the washer method

For revolutions about \( y = 8 \), outer radius \( R(x) = 8 - x \) and inner radius \( r(x) = 8 - (\frac{x}{2} + 2) \). The volume expression: \[ V = \pi \int_{0}^{4} \left( (8-x)^2 - (6-\frac{x}{2})^2 \right) dx. \] Simplify and compute.

Step 3d: Compute integral for line y=8

Simplifying the expression gives \( (8-x)^2 - (6-\frac{x}{2})^2 \) results in: \( 64 - 16x + x^2 - (36 - 6x + \frac{x^2}{4}) \). Calculate and integrate: \[ \pi \int_{0}^{4} \left( \frac{3x^2}{4} - 10x + 28 \right) dx = \pi \left[ \frac{x^3}{4} - 5x^2 + 28x \right]_0^4 = \pi \times 64 = 64\pi. \]

Key Concepts

shell method

The shell method is a powerful technique for finding the volume of solids of revolution, particularly useful when dealing with regions that are revolved around vertical lines like the y-axis. In the shell method, we imagine slicing the solid into thin cylindrical "shells" rather than "washers." Each shell is essentially a hollow tube, where the volume can be computed by integrating the circumferences of these shells multiplied by their height.

To implement this method, consider the variable of integration as perpendicular to the axis of rotation. For a rotation about the y-axis, the variable will typically be \(x\). The formula is:

• Volume, \( V = 2\pi \int_{a}^{b} (radius) imes (height) imes (thickness) \ dx \)

Here, the radius is the distance from the y-axis to a point on the shell, which is simply \(x\). The height, derived from the function expressing the top minus the bottom of the region, is the function value at that point. The thickness is an infinitesimal increase in the x direction, \(dx\).

By employing this method, integration becomes straightforward, as functions of \(x\) are often simpler to work with, especially when the integration limits naturally align with the problem's domain.

volume of rotational solids

Understanding the volume of rotational solids is crucial in various fields, including engineering and physics. These volumes are created by rotating a given region around an axis, forming a 3D object. The process of deriving these volumes often involves two fundamental methods: the washer method and the shell method.

• The washer method is akin to stacking circular disks (washers) with holes, useful when revolving around a horizontal axis like the x-axis. It involves identifying an outer and an inner radius based on the limits of the region.
• The shell method involves considering cylindrical shells, useful for vertical rotations like around the y-axis or any vertical line.

Both methods leverage definite integrals to sum up these infinitesimal volumes created by revolution. The choice between the two often depends on the orientation of the area being revolved, and whether it's easier to express the functions in terms of \(x\) or \(y\). Thus, selecting the appropriate method can significantly simplify calculations.

integration techniques

When calculating volumes of complex shapes, integration techniques become indispensable. Selection of the correct integration method helps in simplifying the problem and in reaching the correct solution efficiently.

There are several integration techniques at your disposal:

• Substitution: Simplifies integrals by changing variables, often used to make expressions easier to handle.
• Integration by Parts: Useful for products of functions, applies the integration formula \( \int u \, dv = uv - \int v \, du \).
• Partial Fraction Decomposition: Breaks rational functions into simpler fractions that are easier to integrate.

$Applying\; one\; or\; more\; of\; these\; techniques\; often\; leads\; to\; successfully\; solving\; integrals\; that\; define\; the\; volumes\; of\; solids.\; Integration\; underpins\; the\; entire\; process,\; transforming\; algebraic\; expressions\; into\; measurable\; volumes.$

coordinate geometry

Coordinate geometry, also known as analytic geometry, plays a crucial role in the process of finding volumes of rotating solids. By expressing geometric properties algebraically, it allows for precise calculations using coordinates and equations.

Here's how coordinate geometry aids in these problems:

• Defining Regions: Equations of boundaries (like lines or curves) precisely describe regions to be revolved.
• Finding Intersections: Solving sets of equations to identify intersection points helps establish limits and integration bounds.
• Calculating Distances: Formulas derived from coordinate geometry, such as distance between points or heights, feed directly into the calculations for radii and heights in volume formulas.

Effectively using coordinate geometry is a key step in setting up the problem correctly. Mastery of equations and transformations in this domain underline every accurate calculation in the finding of volumes.