Problem 70 - Calculus Early Transcendentals

Question

Consider the region \(R\) in the first quadrant bounded by \(y=x^{1 / n}\) and \(y=x^{n},\) where \(n>1\) is a positive number. a. Find the volume \(V(n)\) of the solid generated when \(R\) is revolved about the \(x\) -axis. Express your answer in terms of \(n\). b. Evaluate \(\lim _{n \rightarrow \infty} V(n) .\) Interpret this limit geometrically.

Step-by-Step Solution

Answer

Answer: The limit as n approaches infinity of the volume V(n) is \(\frac{\pi}{2}\).

Step 1: Set up the integral for the volume using the disk method

Using the disk method, the volume of the solid generated by rotating the region R about the x-axis can be obtained by integrating the area of the disks. The area A of each disk is given by \(A = \pi{(\text{outer radius})^2 - (\text{inner radius})^2}\) The outer radius is \(y=x^{1/n}\), and the inner radius is \(y=x^n\). Therefore, the area of each disk is \(A = \pi(x^{2/n}-x^{2n})\). To find the volume, we have to integrate this function over the range of x, which is from \(0\) to \(1\): $$ V(n) = \int_{0}^1 \pi(x^{2/n}-x^{2n}) dx $$

Step 2: Integrate the volume function

To find the volume, we need to integrate the function: $$ \begin{aligned} V(n) &=\int_{0}^{1} \pi\left(x^{2 / n}-x^{2 n}\right) \mathrm{d} x \\ &=\pi \int_{0}^{1}\left(x^{2 / n}-x^{2 n}\right) \mathrm{d} x \\ &=\pi\left[\frac{n}{2+n} x^{2+n}|_{0} ^{1}-\frac{1}{2 n+1} x^{2 n+1}|_{0} ^{1}\right] \end{aligned} $$ After evaluating the integral at the limits, we get: $$ V(n)=\pi\left(\frac{n}{2+n}-\frac{1}{2n+1}\right) $$

Step 3: Find the limit of the volume as n approaches infinity

To find the limit as n approaches infinity, we need to evaluate the following expression: $$ \lim _{n \rightarrow \infty} V(n)=\lim _{n \rightarrow \infty} \pi\left(\frac{n}{2+n}-\frac{1}{2n+1}\right) $$ As n approaches infinity, the limit simplifies to: $$ \lim _{n \rightarrow \infty} V(n)=\pi\left(\frac{1}{2}-0\right) $$ So, the limit as n approaches infinity of V(n) is: $$ \lim _{n \rightarrow \infty} V(n)=\frac{\pi}{2} $$

Step 4: Geometric interpretation of the limit

The volume as n approaches infinity represents the volume of the solid when the inner function \(y=x^n\) gets closer and closer to the x-axis. In other words, y tends to 0 as x ranges from 0 to 1. Hence, the closer this curve gets to the x-axis, the lesser the inner region will be. As a result, the volume of the solid generated by the region R will converge to the volume of the outer solid, which is generated by the outer function \(y=x^{1/n}\). The limit \(\lim _{n \rightarrow \infty} V(n)=\frac{\pi}{2}\) tells us that the maximum volume of the solid that can be generated by the region R, when the difference between the outer and inner radius is smallest possible (i.e., when \(y=x^n\) converges to the x-axis), is equal to \(\frac{\pi}{2}\).

Key Concepts

Disk Method

Understanding the disk method is crucial for calculating the volume of solids of revolution in calculus. Imagine slicing the solid into thin, flat disks perpendicular to the x-axis, like cutting a loaf of bread. Each disk's thickness is infinitesimally small, akin to a calculus concept known as dx, which represents a tiny change in the x-direction.

The area of one side of the disk is the key to finding its volume. If you have a region bounded by two curves and revolve it around the x-axis, the