Problem 8 - Calculus Early Transcendentals: Pearson New International Edition

Question

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. \(y=x^{2}, y=3 x\); about the \(y\) -axis

Step-by-Step Solution

Answer

The volume of the solid is \( 13.5\pi \).

Step 1: Sketch the Region R

First, let's visualize the region bounded by the curves. We have two functions: \( y = x^2 \) and \( y = 3x \). These are parabolic and linear respectively. Plot the curves on the xy-plane, and identify the region where these curves intersect. Solve the equation \( x^2 = 3x \) to find the points of intersection: \( x^2 - 3x = 0 \) which factors to \( x(x - 3) = 0 \). Thus, the intersection points are \( x = 0 \) and \( x = 3 \). The region \( R \) is bounded by these curves from \( x = 0 \) to \( x = 3 \).

Step 2: Show a Typical Rectangular Slice

Next, visualize a typical vertical slice of the region \( R \). This slice is positioned at a general point \( x \) and has a thickness \( \Delta x \). Since we are revolving around the y-axis, this slice forms a cylindrical shell. The height of the slice is \( 3x - x^2 \), and the distance from the y-axis to the slice is \( x \).

Step 3: Write the Formula for Shell Volume

The cylindrical shell has a height equal to the difference in y-values (\(3x - x^2\)), a circumference of \( 2\pi x \) (because the shell's radius is \( x \) and it revolves around the y-axis), and a thickness of \( \Delta x \). Hence, the approximate volume of one shell is calculated by: \[\Delta V = 2\pi x (3x - x^2) \Delta x\]

Step 4: Set Up the Integral

We now set the integral to sum the volume of these shells from \( x = 0 \) to \( x = 3 \). The volume \( V \) is: \[V = \int_0^3 2\pi x (3x - x^2) \, dx\]Simplify the integrand: \[V = \int_0^3 (6\pi x^2 - 2\pi x^3) \, dx\]

Step 5: Evaluate the Integral

Calculate the definite integral: \[V = \int_0^3 (6\pi x^2 - 2\pi x^3) \, dx = \left[ 6\pi \frac{x^3}{3} - 2\pi \frac{x^4}{4} \right]_0^3\]Simplify further: \[= \left[ 2\pi x^3 - \frac{\pi}{2} x^4 \right]_0^3\]Evaluate at the bounds: \[= \left( 2\pi (3)^3 - \frac{\pi}{2} (3)^4 \right) - \left( 2\pi (0)^3 - \frac{\pi}{2} (0)^4 \right)\]\[= (54\pi - 40.5\pi)\]\[= 13.5\pi\]Thus, the volume of the solid is \( 13.5 \pi \).

Step 6: Answer

The volume of the solid generated when the region is revolved around the y-axis is \( 13.5\pi \).

Key Concepts

Cylindrical Shells Method

The cylindrical shells method is a powerful technique for finding the volume of a solid of revolution. This method is particularly useful when the solid is generated by rotating a region around the y-axis. The idea is to divide the region into thin vertical slices and revolve these slices to form cylindrical shells.

The key characteristics of a cylindrical shell include:

• The radius, which is the distance from the y-axis to the slice. For a slice at coordinate \( x \), the radius is \( x \).
• The height, given by the difference between the two functions within the region. In this case, the height is \( 3x - x^2 \).
• The circumference, which is calculated as \( 2\pi x \).
• The thickness represented by a small change in \( x \), \( \Delta x \).

The approximate volume \( \Delta V \) of each cylindrical shell can be defined as \( \Delta V = 2\pi x(3x - x^2)\Delta x \). By integrating across the entire region, we accumulate the volume of each shell to find the total volume.

Calculating Integrals

Integration is a mathematical technique used to calculate the total accumulation of quantities such as area under a curve, length, or volume. When using the cylindrical shells method to find the volume of a solid of revolution, setting up and evaluating integrals is essential.

In our original problem, we formulated an integral to calculate the volume of a solid formed by revolving the region bounded by \( y = x^2 \) and \( y = 3x \) around the y-axis. The function we integrated was \( 2\pi x(3x - x^2) \), simplified to \( 6\pi x^2 - 2\pi x^3 \). This process involved evaluating the definite integral: \[ \int_0^3 (6\pi x^2 - 2\pi x^3) \, dx \] This computes the total volume by summing the volumes of an infinite number of infinitesimally thin shells.

Remember, integrating a polynomial involves using the power rule: increase the exponent by one and divide the coefficient by this new exponent. Apply this to simplify the expression and proceed to substitution using the limits of integration.

Region Bounded by Curves

To use the cylindrical shells method, it is crucial to identify the region enclosed by different curves. In the given problem, the region \( R \) is bound by the parabolic curve \( y = x^2 \) and the linear line \( y = 3x \).

Finding the intersection points where these two curves meet helps in determining the limits of integration. Solving \( x^2 = 3x \) gives \( x = 0 \) and \( x = 3 \). These points of intersection define our region in the xy-plane that we wish to revolve.

Understanding the region bounded by curves makes it easier to visualize each slice in the cylindrical shells method. Thus, sketching the region and recognizing where the curves intersect is a fundamental step in solving such volume problems.

• Intersection points provide the integration limits.
• Curves dictate the function differences forming shell heights.

Visualizing this clearly helps avoid mistakes when setting up the problem and ensures accurate volume calculation.

Definite Integrals

Definite integrals calculate the net area under a curve between two bounds, and in the context of cylindrical shells, they determine the exact volume of the solid formed. To obtain a definite integral, we integrate the function over a specific interval, substituting the upper and lower bounds after integration.

For the problem at hand, we solve:
\[ \int_0^3 (6\pi x^2 - 2\pi x^3) \, dx = \left[ 2\pi x^3 - \frac{\pi}{2} x^4 \right]_0^3 \]
Substitute the limits into your antiderivative function: evaluate at \( x = 3 \) and subtract the result when evaluated at \( x = 0 \). Here's what it looks like in practice: - Substitute \( x = 3 \) into the antiderivative to get a value.
- Substitute \( x = 0 \) to find the lower bound's contribution (often zero if the function passes through the origin).
This process gives our final answer: the total volume, \( 13.5\pi \). By carefully working through this process, the definite integral provides an exact solution for the volume.