Problem 4 - 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. \(y=9-x^{2}(x \geq 0), x=0, y=0 ;\) about the \(y\)-axis

Step-by-Step Solution

Answer

The volume is \( \frac{81 \pi}{2} \).

Step 1: Sketch the Region R

First, let's sketch the region bounded by the curves. The curve given is \( y = 9 - x^2 \) for \( x \geq 0 \), along with the lines \( x = 0 \) and \( y = 0 \). This region is a part of the parabola that opens downwards and is bound in the first quadrant of the coordinate plane. Draw this region on a graph.

Step 2: Identify a Typical Slice

Considering the disk method, let's identify a typical vertical slice of the region. This slice is a thin vertical strip with thickness \( \Delta x \) located at some point \( x \) between 0 and the \( x \)-intercept of the parabola. At the top, it meets the curve \( y = 9 - x^2 \) and at the bottom, it's along \( y = 0 \).

Step 3: Approximate Volume of Shell Generated

When this slice is revolved around the \( y \)-axis, it forms a cylindrical shell. The radius of the shell is \( x \), and its height is \( y = 9 - x^2 \). Therefore, the approximate volume of the shell is given by \( \Delta V = 2\pi x (9 - x^2) \Delta x \).

Step 4: Set Up the Integral for Volume

To find the total volume formed by revolving the entire region, integrate the volume of the cylindrical shell from \( x = 0 \) to \( x = 3 \) (found by setting \( y = 9 - x^2 \) to zero, which gives \( x = 3 \)). Thus, the integral is: \[ V = \int_{0}^{3} 2 \pi x (9 - x^2) \, dx \]

Step 5: Evaluate the Integral

Evaluate the integral: \[ V = 2 \pi \int_{0}^{3} (9x - x^3) \, dx \] Compute this by integrating: \[ V = 2 \pi \left[ \frac{9x^2}{2} - \frac{x^4}{4} \right]_{0}^{3} \] Calculate the definite integral: \[ = 2 \pi \left[ \left( \frac{9(3)^2}{2} - \frac{(3)^4}{4} \right) - \left( \frac{9(0)^2}{2} - \frac{(0)^4}{4} \right) \right] \] Evaluate the expression: \[ = 2 \pi \left[ \left( \frac{81}{2} - \frac{81}{4} \right) \right] = 2 \pi \left[ \frac{162}{4} - \frac{81}{4} \right] = 2 \pi \left[ \frac{81}{4} \right] = \frac{162 \pi}{4} = \frac{81 \pi}{2} \] Thus, the volume is \( \frac{81 \pi}{2} \).

Key Concepts

Volume of Solid

When we talk about the volume of a solid in calculus, we often deal with shapes created by revolving a region around an axis. Think about sculptures or vases you've seen—these are real-world examples of solids of revolution. In calculus, we mathematically determine the volume of such solids. This involves taking a limited region on a graph, revolving it around an axis, and calculating the space it occupies. The key goal is to compute the exact size of the 3D shape formed through this revolution.

For example, imagine the area under a curve rotating around the y-axis. The volume of solid calculates the space inside this revolution. It's an extension of the concept of area, moving into three-dimensional space. Each step in the process adds precise layers to how we get the final solution: starting from sketching the region, slicing it into simpler segments, and integrating over its entire span. These steps ensure the outcome remains consistent and accurate.

Revolving Region

To find the volume of a solid, we first revolve a region around an axis. The idea here is to create a 3D object from a 2D area. By rotating a shape, say around the y-axis, it mimics the movement of a Ferris wheel, turning a flat region into full circle forms.

In our problem, the region defined by the curve is revolved about the y-axis, producing a shape akin to a bell or vase.

• This region is bounded by a parabola, which is defined by the equation \( y = 9 - x^2 \).
• It also experiences limits set by the lines \( x = 0 \) and \( y = 0 \), constraining and enclosing the region within finite boundaries.

Seeing the visualization of this process aids in understanding what it's like to "wrap" the area around an axis, resulting in a symmetric and smooth solid structure that we aim to measure.

Cylindrical Shells

The cylindrical shell method is a powerful technique to compute the volume of revolution. Imagine a bunch of paper towel rolls stacked inside a revolving figure. Each roll represents one shell. Using these imaginary shells helps calculate how much space they would cover inside the solid.

For our exercise, each typical slice of the region becomes a cylindrical shell when revolved around the y-axis.

• The **radius** of each shell is simply the distance from the y-axis, or \( x \).
• The **height** of each shell corresponds to the function value \( y = 9 - x^2 \).
• We consider a **thin slice** with thickness \( \Delta x \).

Each slice creates a shell, having a curved surface area measured by \( 2\pi x (9 - x^2)\Delta x \). By integrating these slices from start to finish along the x-region, we get the full volume of cylindrical shells.

Integral Evaluation

Integrals allow us to add up an infinite number of small quantities to compute an area or volume. In this problem, we need to evaluate the integral that sums up the volumes of all cylindrical shells. Think of this as an adding machine that combines all shell slices into a single volume over a continuous range.

The integral we set up for this shape is:

• \[ V = \int_{0}^{3} 2 \pi x (9 - x^2) \, dx \]
• This integral captures the idea of summing shell volumes from \( x = 0 \) to \( x = 3 \).
• To solve it, we perform integration on the function inside, \( 9x - x^3 \).
• After integration, we apply the limits from 0 to 3.

The integral simplifies and combines the small volumes into a comprehensive volume, which turns out to be \( \frac{81 \pi}{2} \). This final result represents the complete capacity of the solid shape formed by revolving the region.