Problem 22 - Calculus Early Transcendentals

Question

Let \(R\) be the region bounded by the following curves. Use the shell method to find the volume of the solid generated when \(R\) is revolved about the \(x\) -axis. $$y=x^{3}, y=8, \text { and } x=0$$

Step-by-Step Solution

Answer

Question: Find the volume of the solid generated when the region R bounded by the curves y=x^3, y=8, and x=0 is revolved about the x-axis using the shell method. Answer: The volume of the solid generated is \(\frac{256\pi}{5}\) cubic units.

Step 1: Identify the Bounds and Function for Integration

As mentioned earlier, we want to integrate over the range of our region. Since the region is bounded by \(x=0\) and the intersection point of \(y=x^3\) and \(y=8\), we need to find that intersection point. The intersection point occurs when \(x^3=8\): $$x^3 = 8$$ $$x = 2$$ Therefore, our bounds for integration will be from \(x=0\) to \(x=2\).

Step 2: Determine the Height and Radius of Each Shell

Now we need to find the formula for the height and radius of each cylindrical shell. The height of a shell at a given \(x\) position will be the difference between the \(y\) values of the two functions at that point: $$h(x) = 8 - x^3$$ The radius of a shell at a given \(x\) position is equal to the distance between the \(x\)-axis and the shell. In this case, the distance is simply the \(x\)-coordinate: $$r(x) = x$$

Step 3: Set Up the Integral for the Volume of a Single Shell

Now that we have the height and radius expressions, we can set up the integral for the volume of a single shell. The formula for the volume of a cylindrical shell is: $$V_{shell} = 2\pi rh$$ In our case, \(r(x)=x\) and \(h(x)=8-x^3\), so we have: $$V_{shell}(x) = 2\pi x (8-x^3)$$

Step 4: Use the Integral to Find the Total Volume of the Solid

Now we need to integrate the volume of a single shell from \(x=0\) to \(x=2\) to find the total volume: $$V_{total} = \int_{0}^{2} 2\pi x (8-x^3) \, dx$$ To solve this integral, we multiply the integrand and apply some basic rules for integration: $$V_{total} = \int_{0}^{2} 16\pi x - 2\pi x^4 \, dx$$ $$V_{total} = \left[8\pi x^2 - \frac{2\pi x^5}{5}\right]_0^2$$ $$V_{total} = 8\pi (2^2)^2 - \frac{2\pi (2^5)^2}{5}$$ $$V_{total} = 64\pi - \frac{64\pi}{5}$$ Finally, we have the total volume of the solid when the region \(R\) is revolved about the x-axis using the shell method: $$V_{total} = 64\pi \left(1 - \frac{1}{5}\right)$$ $$V_{total} = \frac{256\pi}{5}$$

Key Concepts

Volume of Solids

When dealing with three-dimensional objects, understanding how to calculate their volume is fundamental. In calculus, finding the volume of solids formed by rotating a region around an axis is a common problem. This is particularly useful in engineering and architecture. The problem states that the volume of the solid is formed by revolving a specific region in the plane, defined by curves, around the x-axis.

In our exercise, the region is bounded by the curves of the function \(y=x^3\), a horizontal line \(y=8\), and the vertical line \(x=0\). The volume of the solid generated by revolving this region about the x-axis can be computed using methods like disk, washer, or cylindrical shells. Here, the shell method is utilized which can be quite efficient for problems involving vertical lines or strips parallel to the axis of revolution. Understanding these techniques enables us to approach and solve many practical situations in calculus and beyond.

Cylindrical Shells

The method of cylindrical shells is an elegant technique in calculus for finding the volume of certain solids of revolution. In effect, it visualizes the solid as a collection of many thin, hollow cylinders stacked together.

Each cylindrical shell can be imagined as a small portion of the solid, defined by revolving a rectangle parallel to the axis of rotation (in this case, the x-axis). The height of this cylinder corresponds to the y-value difference, i.e., the span of the rectangle, which in our exercise is calculated as \(h(x) = 8 - x^3\).

The formula to find the volume of a cylindrical shell is given by \(V_{shell} = 2\pi rh\), where \(r\) is the radius of the cylinder and \(h\) is the height. In the problem given, \(r(x) = x\) is the distance of the shell from the axis of rotation. This setup is then integrated along the bounds of \(x = 0\) and \(x = 2\) to account for the entire volume. The cylindrical shells method is particularly useful when faced with regions shaped more vertically, or when the revolution pivot doesn’t yield straightforward application of disk/washer methods.

Definite Integral

A definite integral is a core concept in calculus and plays a pivotal role in calculating the exact areas, volumes, and other measures when limits are crucial. The essence of definite integration is summing up an infinite number of infinitesimally small quantities to find a total value over a given interval.

In the context of finding volume with the shell method, the definite integral enables aggregation of all infinitely small shell volumes from \(x=0\) to \(x=2\). The integral is set up as \(\int_{0}^{2} 2\pi x (8-x^3) \, dx\). This integral embodies the summation of all small cylindrical shell volumes, allowing us to compute the total volume when the curve is revolved.

The actual computation involves simplifying the integrand, integrating with respect to \(x\), and applying the limits from the lower bound to the upper bound. Thus, definite integrals are not only mathematical tools but also serve as bridges for converting continuous, infinite processes into finite, tangible results in geometric contexts like volume via the shell method.