Volumes
Find the volumes of the solids in Exercises 1–18.
The solid lies between planes perpendicular to the x-axis at x = 0 and x = 1. The cross-sections perpendicular to the x-axis between these planes are circular disks whose diameters run from the parabola y = x² to the parabola y = √x.

Verified step by step guidance

1. Identify the region bounded by the two curves: the parabola \(y = x^{2}\) and the curve \(y = \sqrt{x}\), between \(x = 0\) and \(x = 1\).
2. Since the cross-sections perpendicular to the x-axis are circular disks with diameters running from \(y = x^{2}\) to \(y = \sqrt{x}\), find the length of the diameter at a general point \(x\) by subtracting the lower curve from the upper curve: \(D(x) = \sqrt{x} - x^{2}\).
3. Calculate the radius of each circular cross-section as half the diameter: \(r(x) = \frac{D(x)}{2} = \frac{\sqrt{x} - x^{2}}{2}\).
4. Write the area of the circular cross-section as a function of \(x\): \(A(x) = \pi [r(x)]^{2} = \pi \left( \frac{\sqrt{x} - x^{2}}{2} \right)^{2}\).
5. Set up the volume integral by integrating the cross-sectional area from \(x = 0\) to \(x = 1\): \(V = \int_{0}^{1} A(x) \, dx = \int_{0}^{1} \pi \left( \frac{\sqrt{x} - x^{2}}{2} \right)^{2} \, dx\). This integral will give the volume of the solid.

Key Concepts

Volume of Solids with Known Cross-Sections

This concept involves finding the volume of a solid by integrating the area of cross-sectional slices perpendicular to an axis. Each cross-section's area is expressed as a function of the variable of integration, and integrating these areas over the given interval yields the total volume.

Area of a Circular Cross-Section

When cross-sections are circular disks, their area is calculated using the formula A = πr², where r is the radius. In this problem, the diameter is given by the vertical distance between two curves, so the radius is half that distance, which must be expressed as a function of x.

Using Functions to Determine Cross-Section Dimensions

The diameters of the circular cross-sections are determined by the vertical distance between two curves, y = x² and y = √x. Understanding how to find this distance at each x-value is essential to express the diameter (and thus radius) as a function for integration.