Problem 62 - Essential Calculus

Question

Consider the region satisfying the inequalities. (a) Find the area of the region. (b) Find the volume of the solid generated by revolving the region about the \(x\) -axis. (c) Find the volume of the solid generated by revolving the region about the \(y\) -axis. $$ y \leq \frac{1}{x^{2}}, y \geq 0, x \geq 1 $$

Step-by-Step Solution

Answer

The area of the region is 1. The volume generated by revolving the region about the x-axis is \(\pi\), while revolving about the y-axis is \(2 \pi\).

Step 1: Identify Area Under the Curve

The area under the curve is the integral of \(y\) with respect to \(x\) from 1 to infinity. We need to solve the definite integral \(\int_{1}^{\infty} \frac{1}{x^{2}} dx\).

Step 2: Solve the Integral

To solve the integral, apply the power rule: \(\int x^n = \frac{1}{n+1} x^{n+1}\). Applying this rule, the solution is \([-x^{-1}]\) evaluated from 1 to \(\infty\). The upper limit \(\infty\) becomes 0, since as \(x\) grows towards \(\infty\), \(x^{-1}\) tends to 0. Thus, the definite integral is 1.

Step 3: Revolve Around the x-axis and Determine Volume

The volume of a solid generated by rotating the region is found using disk method: \(\pi \int_{a}^{b} [f(x)]^2 dx \) where \( f(x) \) is the function being revolved. For this problem, \( f(x) = \frac{1}{x^2} \), so the volume becomes \(\pi \int_{1}^{\infty} [\frac{1}{x^{2}}]^2 dx = \pi \int_{1}^{\infty} \frac{1}{x^{4}} dx\). Solve the integral similarly to Step 2, the volume around the \(x\)-axis is \(\pi\).

Step 4: Revolve Around the y-axis and Determine Volume

To find the volume by revolving around the \(y\)-axis, we use the same disk method as in step 3, but with respect to \(y\) instead of \(x\). We need to rewrite the function in y: \(x = y^{-1/2} \). Our equation becomes \(\pi \int_{0}^{1} [y^{-1/2}]^2 dy = \pi \int_{0}^{1} y^{-1} dy\). Solving the integral gives us a volume of \(2 \pi\).