Problem 48 - Calculus Early Transcendentals

Question

Let \(R\) be the region bounded by the following curves. Let \(S\) be the solid generated when \(R\) is revolved about the given axis. If possible, find the volume of \(S\) by both the disk/washer and shell methods. Check that your results agree and state which method is easiest to apply. \(y=x-x^{4}, y=0 ;\) revolved around the \(y\) -axis

Step-by-Step Solution

Answer

Answer: The shell method is the easiest method to apply in this case.

Step 1: Sketch the region R and the curves

Firstly, sketch the region R and the curves. Here, we have the curve \(y = x - x^4\) and the line \(y = 0\) which is the x-axis. When the region R is revolved around the y-axis, we get the solid S.

Step 2: Determine the integration bounds for each method

The disk/washer method requires integrating with respect to y, so we need to find the inverse of the curve \(y = x - x^4\). However, finding the inverse of this function is not possible using elementary functions. For the shell method, integrating with respect to x, we need to determine the integration bounds where the curve intersects the x-axis. This occurs when \(y = x - x^4 = 0\): $$x = x^4 \Rightarrow x(x^3 - 1) = 0$$ This implies the curve intersects the x-axis at x = 0 and x = 1. So, the bounds of integration for the shell method will be from x = 0 to x = 1.

Step 3: Apply the disk/washer method (when applicable)

As mentioned earlier, the disk/washer method is not applicable in this case since it is not possible to find the inverse function of \(y = x - x^4\) using elementary functions.

Step 4: Apply the shell method

To apply the shell method, remember that the volume V of the solid generated by revolving a region R around the y-axis is given by: $$V = 2\pi \int_{a}^{b} x y(x) dx$$ In this case, \(a = 0\), \(b = 1\), and \(y(x) = x - x^4\). So, we have: $$V = 2\pi \int_{0}^{1} x (x - x^4) dx$$ To find the volume, we must integrate this expression: $$V = 2\pi \left[ \int_{0}^{1} x^2 dx - \int_{0}^{1} x^5 dx \right]$$ $$V = 2\pi \left[ \left. \frac{x^3}{3} \right |_0^1 - \left. \frac{x^6}{6} \right |_0^1 \right]$$ Evaluating the integral, we get: $$V = 2\pi \left[ \left( \frac{1}{3} - \frac{1}{6} \right) - \left( 0 - 0 \right) \right]$$ $$V = 2\pi \left[ \frac{1}{2} \right]$$ $$V = \pi$$ So, the volume of the solid S is \(\pi\) cubic units.

Step 5: Compare the results and determine which method is easier to apply

As we found, the disk/washer method is not applicable to this problem due to difficulties finding the inverse function. On the other hand, we successfully applied the shell method and found the volume of the solid S to be \(\pi\) cubic units. Therefore, in this case, it is clear that the shell method is the easiest method to apply for determining the volume of the solid S.

Key Concepts

Disk Method

The disk method is a technique used to find the volume of a solid of revolution. When a region is revolved about an axis, this method stacks infinitesimally thin disks along the axis.

To use it, you integrate with respect to the axis of revolution. Typically, you revolve around the x-axis, and the formula becomes:

• \( V = \pi \int_{a}^{b} [f(x)]^2 \, dx \)

However, if the axis is the y-axis, you need the inverse function, making it harder when such an inverse isn’t easily found.

In our exercise, finding the inverse of \( y = x - x^4 \) is not possible with elementary functions. Thus, applying the disk method here proved unfeasible, which often happens with more complex curves. This shows one of the limitations of the disk method when the inverse isn’t straightforward to obtain.

Shell Method

The shell method is another powerful tool for finding volumes of solids of revolution, especially effective when revolving around the y-axis.

Unlike the disk method, the shell method doesn’t require finding the inverse of the function. Instead, it integrates parallel to the axis of revolution. The formula used is:

• \( V = 2\pi \int_{a}^{b} x \, [f(x)] \, dx \)

In this exercise, it was perfectly suited for finding the volume of the solid \( S \). By integrating from \( x = 0 \) to \( x = 1 \), we calculated the volume using the function \( y(x) = x - x^4 \).

This approach was straightforward and overcame the limitations faced with the disk method, highlighting the importance of choosing the right method for specific functions.

Multiple Integrations

Integrating multiple times may sound complex, but it's a routine process when using the shell or disk methods. Here, you often find yourself breaking down integrals into manageable parts.

For the shell method, we decomposed:

• \( V = 2\pi \left[ \int_{0}^{1} x^2 \, dx - \int_{0}^{1} x^5 \, dx \right] \)

This requires separately integrating \( x^2 \) and \( x^5 \) and then evaluating at the bounds. Each represents a part of the region’s contribution to the total volume.

This breakdown simplifies complex integrals, ensuring calculations remain precise while being more digestible, a crucial aspect in solving these problems effectively.

Calculus

Calculus is the foundation for solving volume of revolution problems. Its power lies in its ability to handle continuous change and infinitely small quantities.

By using integration, calculus allows us to sum infinitesimal pieces to get the whole, such as volumes in revolved solids. This requires understanding different integration techniques suitable for various scenarios.

In this problem, calculus illuminates how different methods like the shell and disk can provide insight, even when some approaches are impeded by complex functions.

The beauty of calculus is its versatility and precision in tackling these mathematical challenges, making it an indispensable tool in the field of mathematics and beyond.