Problem 14 - Calculus Early Transcendentals: Pearson New International Edition

Question

In Problems 11-16, sketch the region \(R\) bounded by the graphs of the given equations and show a typical horizontal slice. Find the volume of the solid generated by revolving \(R\) about the \(y\)-axis. $$ x=y^{2 / 3}, y=27, x=0 $$

Step-by-Step Solution

Answer

The volume is \( \frac{6561\pi}{7} \) cubic units.

Step 1: Graph the Region R

First, we need to sketch the region \( R \) that is bounded by the given equations. The equation \( x=y^{2/3} \) represents a curve. The lines \( y=27 \) and \( x=0 \) bound the region vertically and horizontally, respectively. When \( x=0 \), \( y=0 \). The curve \( x=y^{2/3} \) crosses the line \( y=27 \) at the point \( (27^{2/3}, 27) \). Graph this to visualize the bounded region \( R \).

Step 2: Illustrate a Typical Horizontal Slice

A typical horizontal slice of the region is a horizontal line segment with a small thickness \( \, dy \, \) at a height \( y \). The slice extends from \( x=0 \) to the curve, where \( x=y^{2/3} \). This will be used to set up the volume integral.

Step 3: Set Up the Integral for Volume Using the Disk Method

Since we are revolving around the \( y \)-axis, we use the disk method. The volume of a typical slice (a disk) is \( \pi \times \text{{(radius)}}^2 \times \text{{(thickness)}} = \pi (y^{2/3})^2 dy \). Set up the integral from \( y=0 \) to \( y=27 \):\[ V = \int_{0}^{27} \pi \left(y^{2/3}\right)^2 \, dy \]

Step 4: Simplify the Integral Expression

Simplify the integral:\[ \int_{0}^{27} \pi \cdot y^{4/3} \, dy \]This is because \( (y^{2/3})^2 = y^{4/3} \).

Step 5: Perform the Integration

Integrate the expression with respect to \( y \):\[ V = \pi \left[ \frac{y^{4/3+1}}{4/3+1} \right]_{0}^{27} \]\[ V = \pi \left[ \frac{y^{7/3}}{7/3} \right]_{0}^{27} \]\[ V = \frac{3\pi}{7} \, [y^{7/3}]_{0}^{27} \]

Step 6: Evaluate the Definite Integral

Now, evaluate the definite integral:\[ V = \frac{3\pi}{7} \left(27^{7/3} - 0^{7/3}\right) \]Solving \( 27^{7/3} = (3^3)^{7/3} = 3^7 \). Therefore, \( 3^7 = 2187 \).So, \[ V = \frac{3\pi}{7} \times 2187 = \frac{6561\pi}{7} \]

Step 7: Write the Final Volume

The final volume of the solid obtained by revolving the region \( R \) around the \( y \)-axis is:\[ V = \frac{6561\pi}{7} \] cubic units.

Key Concepts

Disk Method

The Disk Method is a technique used to find the volume of a solid of revolution. It is particularly useful when we revolve a region around an axis perpendicular to the axis of the function. In this exercise, we're revolving around the y-axis.

When using the Disk Method, imagine slicing the solid into thin, flat disks. Each disk is formed by rotating a small slice of the region.- The volume of each disk is calculated using the formula: \( V = \pi \times (\text{radius})^2 \times \text{thickness} \).

- The radius is determined by the distance from the axis of rotation to the edge of the region; for our exercise, it's given by the function \( y^{2/3} \).

- The thickness is an infinitesimally small change along the axis of rotation, represented as \( dy \) since we're revolving around the y-axis.

To find the total volume, you add up (integrate) the volumes of all the disks from start to end of the region, which brings us to the next key concept: definite integration.

Definite Integration

Definite integration is a powerful mathematical tool used for finding the total accumulated value, like area or volume, of a continuously changing quantity.

In calculating the volume of a solid of revolution using the Disk Method, the process of definite integration accumulates the volume of each infinitesimally small disk. For our problem:
- We're interested in summing the volumes of disks for values of \( y \) ranging from 0 to 27.

- The integral we set up is \( \int_{0}^{27} \pi \left(y^{2/3}\right)^2 \, dy \), representing the sum of all disk volumes as they are rotated around the y-axis.
In practical terms, definite integration embraces:- Evaluating the integral by integrating with respect to the variable involved, which in this instance is \( y \).

- Applying the limits (0 to 27) to our antiderivative to pinpoint precise bounds of our region.

Ultimately, definite integration transforms a seemingly complex problem of infinite disks into a singular computation.

Solid of Revolution

When a region is rotated around an axis, the resultant 3D figure is referred to as a "Solid of Revolution." This exercise involves taking the portion of a 2D plane bound by the specified curves and creating a 3D object by spinning it around the y-axis.

- To imagine this, think of a potter's wheel: as the clay spins, it morphs into a vessel or bowl depending on the original shape.
- The equation \( x = y^{2/3} \) coupled with bounds \( y = 27 \) and \( x = 0 \) delineate the shape of our clay in this analogy.
Upon rotation, each point in the region traces a circle with varying radius, thus generating the 3D surface.
This solid occupies space due to its circular cross-sectional disks, which when stacked, form the bulk of this smooth, curved solid.

Understanding solids of revolution not only helps in calculating volumes but also gives a tangible picture of abstract mathematical concepts.

Graphical Representation of Bounded Regions

Graphical representation is paramount in visualizing mathematical concepts. It clarifies understanding by providing a visual frame of the region involved.
In this exercise, sketching the region involves:- Drawing the curve \( x = y^{2/3} \), which outlines the boundary shape between \( x \) and \( y \).
- Highlighting lines \( y = 27 \) (the horizontal upper boundary) and \( x = 0 \) (the vertical boundary at the y-axis).
By plotting these, we visualize a bounded shape that can be turned into a solid via rotation.

- Illustrating horizontal slices in the sketch further aids in understanding how each horizontal disk constitutes the complete solid.

This graphical method transforms what might seem a daunting algebraic task into an approachable and solvable problem, laying the groundwork for setup and solution of the integral, and ultimately finding the volume of the solid.