90. Consider the infinite region in the first quadrant bounded by the graphs of
y = 1 / √x, y = 0, x = 0, and x = 1.
b. Find the volume of the solid formed by revolving the region (ii) about the y-axis.

Verified step by step guidance

1. First, identify the region bounded by the curves: \( y = \frac{1}{\sqrt{x}} \), \( y = 0 \), \( x = 0 \), and \( x = 1 \). This region lies in the first quadrant between \( x = 0 \) and \( x = 1 \), above the x-axis and below the curve \( y = \frac{1}{\sqrt{x}} \).
2. Since the solid is formed by revolving this region about the y-axis, we will use the method of cylindrical shells. The formula for the volume using cylindrical shells when revolving around the y-axis is: \(\n\[\n\)\[ V = 2\pi \int_{a}^{b} (\text{radius})(\text{height}) \, dx \]\(\n\]\nwhere\) the radius is the distance from the y-axis (which is \( x \)) and the height is the function value \( y = \frac{1}{\sqrt{x}} \).
3. Set up the integral with the limits of integration from \( x = 0 \) to \( x = 1 \): \(\n\[\n\)\[ V = 2\pi \int_{0}^{1} x \cdot \frac{1}{\sqrt{x}} \, dx \]\(\n\]\nSimplify\) the integrand before integrating.
4. Simplify the integrand \( x \cdot \frac{1}{\sqrt{x}} = x \cdot x^{-1/2} = x^{1 - \frac{1}{2}} = x^{\frac{1}{2}} \). So the integral becomes \(\n\)\(\n\)\[ V = 2\pi \int_{0}^{1} x^{\frac{1}{2}} \, dx \]
5. Evaluate the integral \( \int_{0}^{1} x^{\frac{1}{2}} \, dx \) using the power rule for integration: \(\n\[\n\)\[ \int x^{n} \, dx = \frac{x^{n+1}}{n+1} + C \]\(\n\]\nApply\) the limits from 0 to 1 and multiply by \( 2\pi \) to express the volume.

Key Concepts

Setting up the region bounded by curves

Understanding the region involves identifying the area enclosed by the given curves y = 1/√x, y = 0, x = 0, and x = 1 in the first quadrant. This means recognizing the limits of integration and the shape formed, which is essential before applying volume formulas.

Method of cylindrical shells for volume

When revolving a region around the y-axis, the cylindrical shells method is often used. It involves integrating the volume of thin cylindrical shells with radius equal to the x-value, height given by the function, and thickness dx, allowing calculation of the solid's volume.

Integration with respect to x

Since the region is bounded between x = 0 and x = 1, and the function is given as y in terms of x, the volume integral is set up with respect to x. Proper integration techniques must be applied to evaluate the integral and find the exact volume.