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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?

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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.

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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.
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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.
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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.
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Related Practice
Textbook Question

14–25. {Use of Tech} Areas of regions Determine the area of the given region.


The region bounded by y = x²,y = 2x²−4x, and y = 0

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Textbook Question

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


The region bounded by the curves y = sec x and y=2, for 0 ≤ x ≤ π/3, is revolved about the x-axis. What is the volume of the solid that is generated? 

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Textbook Question

Comparing volumes Let R be the region bounded by y=1/x^p and the x-axis on the interval [1, a], where p>0 and a>1 (see figure). Let Vₓ and Vᵧ be the volumes of the solids generated when R is revolved about the x- and y-axes, respectively.


d. Find a general expression for Vᵧ in terms of a and p. Note that p=2 is a special case. What is Vᵧ when p=2?

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Textbook Question

Two methods The region R in the first quadrant bounded by the parabola y = 4-x² and coordinate axes is revolved about the y-axis to produce a dome-shaped solid. Find the volume of the solid in the following ways:


b. Apply the shell method and integrate with respect to x.

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Textbook Question

Position, displacement, and distance A projectile is launched vertically from the ground at t=0, and its velocity in flight (in m/s) is given by v(t)=20−10t. Find the position, displacement, and distance traveled after t seconds, for 0≤t≤4.

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Textbook Question

An area function Consider the functions y = x²/a and y = √x/a, where a>0. Find A(a), the area of the region between the curves.

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