Skip to main content
Ch. 6 - Applications of Integration
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Not the one you use?Change textbook
Chapter 6, Problem 6.R.44

43–55. Volumes of solids Choose the general slicing method, the disk/washer method, or the shell method to answer the following questions.


What is the volume of the solid whose base is the region in the first quadrant bounded by y = √x,y = 2-x, and the x-axis, and whose cross sections perpendicular to the base and parallel to the y-axis are semicircles?

Verified step by step guidance
1
First, identify the region that forms the base of the solid. The base is bounded by the curves \(y = \sqrt{x}\), \(y = 2 - x\), and the \(x\)-axis in the first quadrant. To find the limits of integration, determine the points of intersection between \(y = \sqrt{x}\) and \(y = 2 - x\) by solving \(\sqrt{x} = 2 - x\).
Rewrite the equation \(\sqrt{x} = 2 - x\) by squaring both sides to eliminate the square root: \(x = (2 - x)^2\). Then solve the resulting quadratic equation to find the \(x\)-values where the curves intersect. These \(x\)-values will serve as the bounds for the integral.
Since the cross sections perpendicular to the base and parallel to the \(y\)-axis are semicircles, the diameter of each semicircle corresponds to the vertical distance between the two curves at a given \(x\). Express the diameter \(d(x)\) as the difference between the upper curve and the lower curve: \(d(x) = (2 - x) - \sqrt{x}\).
The area of a semicircle with diameter \(d\) is given by \(A = \frac{\pi}{8} d^2\) because the radius \(r = \frac{d}{2}\) and the area of a full circle is \(\pi r^2\), so the semicircle area is half of that: \(\frac{1}{2} \pi r^2 = \frac{\pi}{8} d^2\). Write the area function \(A(x) = \frac{\pi}{8} \left[(2 - x) - \sqrt{x}\right]^2\).
Set up the integral for the volume by integrating the cross-sectional area \(A(x)\) with respect to \(x\) over the interval determined by the intersection points. The volume \(V\) is given by \(V = \int_{a}^{b} A(x) \, dx = \int_{a}^{b} \frac{\pi}{8} \left[(2 - x) - \sqrt{x}\right]^2 \, dx\), where \(a\) and \(b\) are the \(x\)-values found in step 2.

Verified video answer for a similar problem:

This video solution was recommended by our tutors as helpful for the problem above.
Video duration:
8m
Was this helpful?

Key Concepts

Here are the essential concepts you must grasp in order to answer the question correctly.

Region Bounded by Curves

Understanding the region bounded by the given curves y = √x, y = 2 - x, and the x-axis is essential. This involves finding the points of intersection and sketching the area in the first quadrant to determine the limits of integration and the shape of the base of the solid.
Recommended video:
05:06
Finding Area When Bounds Are Not Given

Cross Sections and Volume by Slicing

The volume of the solid is found by integrating the area of cross sections perpendicular to the base. Since the cross sections are semicircles, the area formula involves the radius derived from the base width at each slice, and integrating these areas over the interval gives the total volume.
Recommended video:
05:38
Introduction to Cross Sections

Setting up the Integral with Respect to y or x

Choosing the correct variable of integration is crucial. Since cross sections are perpendicular to the base and parallel to the y-axis, slices are taken along the x-axis, requiring expressing the base width as a function of y or x and setting up the integral accordingly to compute the volume.
Recommended video:
03:39
Integrals of Natural Exponential Functions (e^x)