Question

The base of a solid is the region \(R\) bounded by \(y=\sqrt{x}\) and \(y=x^{2} .\) Each cross section perpendicular to the \(x\) -axis is a semicircle with diameter extending across \(R\). Find the volume of the solid.

Short answer

The volume of the solid is \( \frac{9\pi}{560} \).

Step-by-step solution

Identify the Bounds of the Region

Since the curves given are in terms of y, find the points where the curves intersect to determine the bounds of the region R. Set the equations equal to each other: \( \sqrt{x} = x^2 \). Squaring both sides gives \( x = x^4 \) or \( x^4 - x = 0 \). Factoring gives \( x(x^3 - 1) = 0 \), so \( x = 0 \) and \( x = 1 \). Thus, the region R is bounded by \( x=0 \) and \( x=1 \).

Determine the Diameter of the Semicircle

The diameter of each semicircular cross section is the vertical distance between the two curves for a given x. This distance is \( \sqrt{x} - x^2 \).

Calculate the Area of the Semicircular Cross Section

The area of a semicircle is given by \( \frac{1}{2}\pi\left(\frac{d}{2}\right)^2 \), where \( d \) is the diameter. With \( d = \sqrt{x} - x^2 \), the radius is \( \frac{\sqrt{x} - x^2}{2} \). So, the area is \( \frac{1}{2}\pi\left(\left(\frac{\sqrt{x} - x^2}{2}\right)^2\right) = \frac{\pi}{8}\left(\sqrt{x} - x^2\right)^2 \).

Set Up the Integral for the Volume

The volume V of the solid is the integral of the cross-sectional area from x=0 to x=1. Thus, set up the integral: \( V = \int_{0}^{1} \frac{\pi}{8}\left(\sqrt{x} - x^2\right)^2 \, dx \).

Simplify and Evaluate the Integral

Simplify \( \left(\sqrt{x} - x^2\right)^2 = x - 2x^{5/2} + x^4 \). Integrate each term: \( \int_{0}^{1} \frac{\pi}{8}(x - 2x^{5/2} + x^4) \, dx \). This becomes \( \frac{\pi}{8} \left[ \frac{x^2}{2} - \frac{4x^{7/2}}{7} + \frac{x^5}{5} \right]_{0}^{1} \), resulting in \( \frac{\pi}{8} \left( \frac{1}{2} - \frac{4}{7} + \frac{1}{5} \right) \).

Calculate the Final Volume

Evaluating the expression \( \frac{1}{2} - \frac{4}{7} + \frac{1}{5} = \frac{35}{70} - \frac{40}{70} + \frac{14}{70} = \frac{9}{70} \). Thus, the volume is \( \frac{\pi}{8} \times \frac{9}{70} = \frac{9\pi}{560} \).

Key concepts

Cross Sections

In geometry, a cross section is essentially a slice of a three-dimensional object. Imagine cutting through a solid with a perfectly sharp knife at a right angle to its base. The resulting shape is the cross section. For this problem, each cross section is a semicircle. These semicircles are formed perpendicular to the x-axis, meaning each slice is parallel to the y-axis.
To define these semicircular cross sections, you first need to know where exactly to slice the solid. This comes from understanding the shape of the region over which the solid extends. As given, the region is bounded by the curves \( y = \sqrt{x} \) and \( y = x^2 \). The intersection of these curves determines where the cross sections start and stop, thus giving us the full range of this unique geometric shape.

Definite Integrals

Definite integrals are a fundamental tool in mathematics, especially when calculating volumes and areas. Just like calculating the area under a curve, we use integrals to sum up all these tiny pieces to get the total volume of the object.
The integral used here takes into account the area of each semicircular cross section along the x-axis from 0 to 1. It's essentially adding up an infinite number of infinitely thin semicircular slices, carefully summing their areas to find the entire volume.
In this specific exercise, the integral is set up as: \[ V = \int_{0}^{1} \frac{\pi}{8} \left(\sqrt{x} - x^2\right)^2 \, dx \]Here, every tiny piece represented by \( dx \) is considered to compute the total volume.

Intersection Points

Intersection points are crucial in determining the bounds of the region of integration. This is where the curves \( y = \sqrt{x} \) and \( y = x^2 \) cross, setting the limits for the entire solid.
By setting the two equations equal to each other, \( \sqrt{x} = x^2 \), and solving, we determine that the intersection occurs at \( x = 0 \) and \( x = 1 \). These points are critical because they define the region where our semicircular cross sections exist. Without these limits, we wouldn't know where our solid begins or ends. They provide an essential framework for building the integral and calculating the volume efficiently.

Semicircular Areas

The area of a semicircle is computed a bit differently than a full circle. While the area of a full circle is \( \pi r^2 \), a semicircle is simply half that, \( \frac{1}{2}\pi r^2 \).
In the exercise, the diameter of each semicircle is given by the distance between the curves, \( \sqrt{x} \) and \( x^2 \). This diameter is determined by subtracting the two: \( \sqrt{x} - x^2 \).
For a semicircle, the radius \( r \) is half of this diameter, or \( \frac{\sqrt{x} - x^2}{2} \). So, the area for each semicircle is calculated as:\[ \frac{1}{2} \pi \left( \frac{\sqrt{x} - x^2}{2} \right)^2 \]This formula is crucial as it forms the expression that needs integrating to determine the solid’s volume.