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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
All textbooksBriggs 3rd EditionCh. 6 - Applications of IntegrationProblem 6.4.49b
Chapter 6, Problem 6.4.49b

Volume of a sphere Let R be the region bounded by the upper half of the circle x²+y² = r² and the x-axis. A sphere of radius r is obtained by revolving R about the x-axis.


b. Repeat part (a) using the disk method.

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Identify the region R bounded by the upper half of the circle \(x^{2} + y^{2} = r^{2}\) and the x-axis. Since it is the upper half, the function describing the curve is \(y = \sqrt{r^{2} - x^{2}}\) for \(x\) in \([-r, r]\).
Set up the volume integral using the disk method. When revolving around the x-axis, the volume is given by \(V = \pi \int_{a}^{b} [f(x)]^{2} \, dx\), where \(f(x)\) is the radius of the disk at position \(x\).
Substitute \(f(x) = \sqrt{r^{2} - x^{2}}\) into the formula, so the volume integral becomes \(V = \pi \int_{-r}^{r} (\sqrt{r^{2} - x^{2}})^{2} \, dx\).
Simplify the integrand: \((\sqrt{r^{2} - x^{2}})^{2} = r^{2} - x^{2}\), so the integral is \(V = \pi \int_{-r}^{r} (r^{2} - x^{2}) \, dx\).
Evaluate the definite integral \(\int_{-r}^{r} (r^{2} - x^{2}) \, dx\) by integrating term-by-term and then multiply the result by \(\pi\) to find the volume of the sphere.

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Key Concepts

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

Disk Method

The disk method is a technique for finding the volume of a solid of revolution by slicing the solid perpendicular to the axis of rotation. Each slice forms a disk whose volume is approximated by π(radius)²(thickness). Integrating these volumes over the interval gives the total volume.
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Disk Method Using y-Axis

Equation of a Circle and Region Definition

The region R is bounded by the upper half of the circle x² + y² = r² and the x-axis, meaning y = √(r² - x²) for x in [-r, r]. Understanding this curve is essential to set up the integral for the volume calculation.
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Parameterizing Equations of Circles & Ellipses

Volume of Revolution about the x-axis

Revolving a region around the x-axis generates a 3D solid. The volume is found by integrating the cross-sectional areas (disks) perpendicular to the x-axis, where each radius corresponds to the y-value of the function defining the region.
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Finding Volume Using Disks
Related Practice
Textbook Question

Equal integrals Without evaluating integrals, explain the following equalities. (Hint: Draw pictures.)


b. ∫²₀(25−(x²+1)²) dx = 2∫₁⁵ y√y−1 dy

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Textbook Question

Different axes of revolution Suppose R is the region bounded by y=f(x) and y=g(x) on the interval [a, b], where f(x)≥g(x).


b. How is this formula changed if x0>b?

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Textbook Question

Use the region R that is bounded by the graphs of y=1+√x,x=4, and y=1 complete the exercises.


Region R is revolved about the y-axis to form a solid of revolution whose cross sections are washers.


b. What is the inner radius of a cross section of the solid at a point y in [1, 3]?

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Textbook Question

A right circular cylinder with height R and radius R has a volume of VC=πR^3 (height = radius).


b. Find the volume of the hemisphere that is inscribed in the cylinder with the same base as the cylinder. Express the volume in terms of VC.

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Textbook Question

Determine whether the following statements are true and give an explanation or counterexample. 


b. If f is not one-to-one on the interval [a, b], then the area of the surface generated when the graph of f on [a, b] is revolved about the x-axis is not defined.

Textbook Question

Let R be the region bounded by the curve y=cos^−1x and the x-axis on [0, 1]. A solid of revolution is obtained by revolving R about the y-axis (see figures). 


b. Find an expression for the area A(y) of a cross section of the solid at a point y in [0,π/2].