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Chapter 15: Problem 6

Use integration to compute the volume of a sphere of radius \(r\).

Short Answer

Expert verified
The volume of the sphere is \(V = \pi \frac{4}{3}r^3\).

Step by step solution

01

Formula of sphere volume

Knowing that they formula for volume of a sphere is \(V = \frac{4}{3}\pi r^3\), let us think of sphere as a solid of revolution. We can acquire the same volume by revolving half of the circle \(y=\sqrt{r^2 - x^2}\), from \(-r\) to \(r\), about the x-axis.
02

Set up the integral

We utilize the formula for the volume of a solid of revolution: \(V = \pi\int_{a}^{b}[f(x)]^2 dx\). Hence, in our case, \(f(x) = \sqrt{r^2 - x^2}\), \(a = -r\), and \(b = r\). Consequently, \(V = \pi\int_{-r}^{r}(\sqrt{r^2 - x^2})^2 dx\). Simplifying, we achieve \(V = \pi\int_{-r}^{r}(r^2 - x^2) dx\).
03

Solve the integral

The integral of \(r^2 - x^2\) from \(-r\) to \(r\) is obtained as \(V = \pi [r^2x - \frac{1}{3}x^3]_{-r}^{r}\). After substituting the limits, we get \(V = \pi(2r^3 -\frac{2}{3}r^3)-(-\pi(2r^3-\frac{2}{3}r^3)) = \pi \frac{4}{3}r^3\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Integration
Understanding the basics of integration is crucial when dealing with a variety of problems in calculus, including finding the volume of complex shapes like spheres. Integration, in its essence, is the process of finding the whole when you have the parts. Imagine you want to calculate the area under a curve on a graph or, as in our exercise, the volume of a solid formed by rotating a shape around an axis.

In our sphere volume problem, integration was used to stitch together infinitesimally thin slices to form the whole sphere. Each slice is a disk, and the area of the disk varies depending on where it is sliced along the x-axis. By integrating, we are effectively summing up the areas of these disks across the whole sphere, thereby finding the total volume.
Solid of Revolution
A solid of revolution is a three-dimensional object obtained by rotating a two-dimensional shape about an axis. The shape sweeps out a volume as it spins, akin to a potter shaping clay on a wheel. In our exercise, we analyze a half-circle (a semicircle) that, when spun around the x-axis, forms a sphere.

Understanding the concept of a solid of revolution is pivotal when utilizing integration to calculate the volume of such solids. In practical terms, the method of solving for the volume of a sphere provided in our step-by-step solution is an application of this concept. It simplifies the volume calculation to a manageable one-dimensional integral by considering the two-dimensional shape that generates the solid.
Definite Integrals
Definite integrals are a powerful tool in calculus for calculating areas, volumes, and other quantities that require summation of infinitesimal elements. They differ from indefinite integrals by having specific limits; they provide the exact value over the interval specified.

When dealing with the volume of a sphere by integration, as shown in the provided step-by-step solution, we use definite integrals with limits from \( -r \) to \( r \), which correspond to the leftmost and rightmost points on the x-axis of our semicircle. By calculating the integral within these limits, we are effectively 'adding up' the volume contributions of each infinitesimal disk from one side of the sphere to the other, thus obtaining the total volume of the sphere.

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Most popular questions from this chapter

Find the volume of the solid obtained by revolving the region bounded by \(y=x-x^{2}\) and the \(x\) -axis around the \(x\) -axis.

Find the arc length of \(f(x)=x^{2} / 8-\ln x\) on [1,2] .

Let \(S\) be the region of the \(x y\) -plane bounded above by the curve \(x^{3} y=64,\) below by the line \(y=1,\) on the left by the line \(x=2,\) and on the right by the line \(x=4 .\) Find the volume of the solid obtained by rotating \(S\) around (a) the \(x\) -axis, (b) the line \(y=1,\) (c) the \(y\) -axis, (d) the line \(x=2\).

A hemispheric bowl of radius \(r\) contains water to a depth \(h\). Find the volume of water in the bowl.

Use integration to find the volume of the solid obtained by revolving the region bounded by \(x+y=2\) and the \(x\) and \(y\) axes around the \(x\) -axis.
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