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5.2 Volumes
Volumes
In trying to find the volume of a solid we face the same type of problem as in finding areas.
We have an intuitive idea of what volume means, but we must make this idea precise by using calculus to give an exact definition of volume.
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Volumes
We start with a simple type of solid called a cylinder (or, more precisely, a right cylinder).
As illustrated in Figure 1(a), a cylinder is bounded by a plane region B1, called the base, and a congruent
region B2 in a parallel plane.
The cylinder consists of all points on line segments that are perpendicular to the base and join B1 to B2. If the area of the base is A and the height of the cylinder (the distance from B1 to B2) is h, then the volume V of the cylinder is defined as V = Ah.
Figure 1(a)
Volumes
In particular, if the base is a circle with radius r, then the cylinder is a circular cylinder with volume V = πr2h [see
Figure 1(b)], and if the base is a rectangle with length l and width w, then the cylinder is a rectangular box (also called a rectangular parallelepiped) with volume V = lwh
[see Figure 1(c)].
Figure 1(b) Figure 1(c)
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Volumes
For a solid S that isn’t a cylinder we first “cut” S into pieces and approximate each piece by a cylinder. We estimate the volume of S by adding the volumes of the cylinders. We
arrive at the exact volume of S through a limiting process in which the number of pieces becomes large.
We start by intersecting S with a plane and obtaining a plane region that is called a cross-section of S.
Volumes
Let A(x) be the area of the cross-section of S in a plane Px perpendicular to the x-axis and passing through the point x, where a ≤ x ≤ b. (See Figure 2. Think of slicing S with a
knife through x and computing the area of this slice.)
The cross-sectional area A(x) will vary as x increases from a to b.
Figure 2
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Volumes
Let’s divide S into n “slabs” of equal width ∆x by using the planes Px
1, Px
2, . . . to slice the solid. (Think of slicing a loaf of bread.)
If we choose sample points xi∗ in [xi– 1, xi], we can
approximate the i th slab Si (the part of S that lies between the planes Px
i – 1 and Px
i) by a cylinder with base area A(xi∗) and “height” ∆x. (See Figure 3.)
Figure 3
Volumes
The volume of this cylinder is A(xi∗) ∆x, so an
approximation to our intuitive conception of the volume of the i th slab Si is
V(Si) ≈ A(xi∗) ∆x
Adding the volumes of these slabs, we get an
approximation to the total volume (that is, what we think of intuitively as the volume):
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Volumes
This approximation appears to become better and better as n → . (Think of the slices as becoming thinner and
thinner.)
Therefore we define the volume as the limit of these sums as n → .
But we recognize the limit of Riemann sums as a definite integral and so we have the following definition.
Volumes
When we use the volume formula it is
important to remember that A(x) is the area of a moving
cross-section obtained by slicing through x perpendicular to the x-axis.
Notice that, for a cylinder, the cross-sectional area is constant: A(x) = A for all x. So our definition of volume
gives ; this agrees with the formula
V = Ah.
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Example 1
Show that the volume of a sphere of radius r is .
Solution:
If we place the sphere so that
its center is at the origin, then the plane Px intersects the sphere in a circle whose radius (from the Pythagorean Theorem) is
(See Figure 4.) So the cross-sectional area is
A(x) = πy2
Figure 4
= π(r2 – x2)
Example 1 – Solution
Using the definition of volume with a = –r and b = r, we have
(The integrand is even.)
cont’d
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Volumes
Figure 5 illustrates the definition of volume when the solid is a sphere with radius r = 1.
From the result of Example 1, we know that the volume of the sphere is π, which is approximately 4.18879.
Figure 5
Approximating the volume of a sphere with radius 1
Volumes
Here the slabs are circular cylinders, or disks, and the three parts of Figure 5 show the geometric interpretations of the Riemann sums
when n = 5, 10, and 20 if we choose the sample points xi∗ to be the midpoints
Notice that as we increase the number of approximating
cylinders, the corresponding Riemann sums become closer to the true volume.
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Volumes
The solids are all called solids of revolution because they are obtained by revolving a region about a line. In general, we calculate the volume of a solid of revolution by using the basic defining formula
and we find the cross-sectional area A(x) or A(y) in one of the following ways:
•
If the cross-section is a disk, we find the radius of the disk (in terms of x or y) and useA = π(radius)2
Volumes
• If the cross-section is a washer, we find the inner radius rin and outer radius rout from a sketch (as in Figure 10) and compute the area of the washer by subtracting the area of the inner disk from the area of the outer disk:
A = π(outer radius)2 – π(inner radius)2
Figure 10