Comparing volumes Let R be the region bounded by y=1/x^p and the x-axis on the interval [1, a], where p>0 and a>1 (see figure). Let Vₓ and Vᵧ be the volumes of the solids generated when R is revolved about the x- and y-axes, respectively.


c. Find a general expression for Vₓ in terms of a and p. Note that p=1/2 is a special case. What is Vₓ when p=1/2?

Verified step by step guidance

1. Identify the region R bounded by the curve \(y = \frac{1}{x^p}\), the x-axis, and the vertical lines \(x=1\) and \(x=a\), where \(p > 0\) and \(a > 1\).
2. To find the volume \(V_x\) generated by revolving the region R about the x-axis, use the disk method. The formula for the volume is: \[V_x = \pi \int_1^a \left( y \right)^2 \, dx\] Since \(y = \frac{1}{x^p}\), substitute this into the integral.
3. Rewrite the integral explicitly: \[V_x = \pi \int_1^a \left( \frac{1}{x^p} \right)^2 \, dx = \pi \int_1^a x^{-2p} \, dx\]
4. Evaluate the integral \(\int_1^a x^{-2p} \, dx\). For \(p \neq \frac{1}{2}\), use the power rule for integration: \[\int x^m \, dx = \frac{x^{m+1}}{m+1} + C\] where \(m = -2p\). So, \[\int_1^a x^{-2p} \, dx = \left[ \frac{x^{-2p+1}}{-2p+1} \right]_1^a\]
5. For the special case \(p = \frac{1}{2}\), the exponent becomes \(-2p = -1\), so the integral becomes: \[\int_1^a x^{-1} \, dx = \left[ \ln|x| \right]_1^a = \ln a\] Thus, the volume \(V_x\) for \(p = \frac{1}{2}\) is: \[V_x = \pi \ln a\]

Key Concepts

Volume of Solids of Revolution (Disk/Washer Method)

This method calculates the volume of a solid formed by revolving a region around an axis. When revolving around the x-axis, the volume is found by integrating π times the square of the function (radius) with respect to x over the given interval. It is essential to set up the integral correctly using the function y = 1/x^p.

Handling Parameters in Integrals

The problem involves parameters p and a, which affect the shape and bounds of the region. Understanding how to integrate functions with variable exponents and parameters is crucial. This includes recognizing special cases, such as p = 1/2, where the integral may require different techniques or simplifications.

Improper Integrals and Convergence

Since the function y = 1/x^p involves powers and the interval starts at 1, it is important to consider the behavior of the integral for different values of p. Some values of p may lead to convergent or divergent integrals, affecting the existence and finiteness of the volume. Recognizing these conditions ensures correct evaluation.