Mathematics
Determining Volumes by Slicing
Determining volumes by slicing is a method used in calculus to find the volume of a three-dimensional object by dividing it into infinitely thin slices. The slices are then integrated to find the total volume of the object. This method is commonly used in real-world applications such as engineering and physics.
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4 Key excerpts on "Determining Volumes by Slicing"
eBook - PDF- Howard Anton, Irl C. Bivens, Stephen Davis(Authors)
- 2022(Publication Date)
- Wiley(Publisher)
In this section we will see that the same basic principle can be used to find volumes of certain three-dimensional solids. Volumes by Slicing Recall that the underlying principle for finding the area of a plane region is to divide the region into thin strips, approximate the area of each strip by the area of a rectangle, add the approximations to form a Riemann sum, and take the limit of the Riemann sums to produce an integral for the area. Under appropriate conditions, the same strategy can be used to find the volume of a solid. The idea is to divide the solid into thin slabs, approximate the volume of each slab, add the approximations to form a Riemann sum, and take the limit of the Riemann sums to produce an integral for the volume (Figure 6.2.1). Sphere cut into horizontal slabs Right pyramid cut into horizontal slabs Right circular cone cut into horizontal slabs Right circular cone cut into vertical slabs ▴ Figure 6.2.1 What makes this method work is the fact that a thin slab has a cross section that does not vary much in size or shape, which, as we will see, makes its volume easy to approximate (Figure 6.2.2). Moreover, the thinner the slab, the less variation in its cross sections and the better the approximation. Thus, once we approximate the volumes of the slabs, we can set up a Riemann sum whose limit is the volume of the entire solid. We will give the details shortly, but first we need to discuss how to find the volume of a solid whose cross sections do not vary in size and shape (i.e., are congruent). In a thin slab, the cross sections do not vary much in size and shape. Cross section ▴ Figure 6.2.2 One of the simplest examples of a solid with congruent cross sections is a right circular cylinder of radius r, since all cross sections taken perpendicular to the central axis are circular regions of radius r.
eBook - PDFCalculus
Early Transcendental Single Variable
- Howard Anton, Irl C. Bivens, Stephen Davis(Authors)
- 2016(Publication Date)
- Wiley(Publisher)
VOLUMES BY SLICING Recall that the underlying principle for finding the area of a plane region is to divide the region into thin strips, approximate the area of each strip by the area of a rectangle, add the approximations to form a Riemann sum, and take the limit of the Riemann sums to produce an integral for the area. Under appropriate conditions, the same strategy can be used to find the volume of a solid. The idea is to divide the solid into thin slabs, approximate the volume of each slab, add the approximations to form a Riemann sum, and take the limit of the Riemann sums to produce an integral for the volume (Figure 6.2.1). Figure 6.2.1 What makes this method work is the fact that a thin slab has a cross section that does not Figure 6.2.2 vary much in size or shape, which, as we will see, makes its volume easy to approximate (Figure 6.2.2). Moreover, the thinner the slab, the less variation in its cross sections and the better the approximation. Thus, once we approximate the volumes of the slabs, we can set up a Riemann sum whose limit is the volume of the entire solid. We will give the details shortly, but first we need to discuss how to find the volume of a solid whose cross sections do not vary in size and shape (i.e., are congruent). One of the simplest examples of a solid with congruent cross sections is a right circular cylinder of radius r, since all cross sections taken perpendicular to the central axis are circular regions of radius r. The volume V of a right circular cylinder of radius r and height h can be expressed in terms of the height and the area of a cross section as V = πr 2 h = [area of a cross section] × [height] (1) This is a special case of a more general volume formula that applies to solids called right cylinders. A right cylinder is a solid that is generated when a plane region is translated along a line or axis that is perpendicular to the region (Figure 6.2.3).
eBook - PDFCalculus
Single Variable
- Howard Anton, Irl C. Bivens, Stephen Davis(Authors)
- 2022(Publication Date)
- Wiley(Publisher)
5.2.2 Volume Formula Let S be a solid bounded by two parallel planes perpendicular to the x-axis at x = a and x = b. If, for each x in [a, b], the cross-sectional area of S perpendicular to the x-axis is A(x), then the volume of the solid is V = b a A(x) dx (3) provided A(x) is integrable. It is understood in our cal- culations of volume that the units of volume are the cubed units of length [e.g., cubic inches (in 3 ) or cubic meters (m 3 )]. There is a similar result for cross sections perpendicular to the y-axis. 5.2.3 Volume Formula Let S be a solid bounded by two parallel planes perpendicular to the y-axis at y = c and y = d. If, for each y in [c, d], the cross-sectional area of S perpendicular to the y-axis is A(y), then the volume of the solid is V = d c A(y) dy (4) provided A(y) is integrable. 5.2 Volumes by Slicing; Disks and Washers 283 In words, these formulas state: The volume of a solid can be obtained by integrating the cross-sectional area from one end of the solid to the other. O C B y h − y h s (b) (a) x-axis y-axis y B(0, h) C ( a, 0 ) O 1 2 1 2 a 1 2 FIGURE 5.2.7 Example 1 Derive the formula for the volume of a right pyramid whose altitude is h and whose base is a square with sides of length a. Solution As illustrated in Figure 5.2.7a, we introduce a rectangular coordinate system in which the y-axis passes through the apex and is perpendicular to the base, and the x-axis passes through the base and is parallel to a side of the base. At any y in the interval [0, h] on the y-axis, the cross section perpendicular to the y-axis is a square.
eBook - PDFCalculus
Late Transcendental
- Howard Anton, Irl C. Bivens, Stephen Davis(Authors)
- 2016(Publication Date)
- Wiley(Publisher)
Under appropriate conditions, the same strategy can be used to find the volume of a solid. The idea is to divide the solid into thin slabs, approximate the volume of each slab, add the approximations to form a Riemann sum, and take the limit of the Riemann sums to produce an integral for the volume (Figure 5.2.1). Figure 5.2.1 What makes this method work is the fact that a thin slab has a cross section that does not Figure 5.2.2 vary much in size or shape, which, as we will see, makes its volume easy to approximate (Figure 5.2.2). Moreover, the thinner the slab, the less variation in its cross sections and the better the approximation. Thus, once we approximate the volumes of the slabs, we can set up a Riemann sum whose limit is the volume of the entire solid. We will give the details shortly, but first we need to discuss how to find the volume of a solid whose cross sections do not vary in size and shape (i.e., are congruent). One of the simplest examples of a solid with congruent cross sections is a right circular cylinder of radius r, since all cross sections taken perpendicular to the central axis are circular regions of radius r. The volume V of a right circular cylinder of radius r and height h can be expressed in terms of the height and the area of a cross section as V = πr 2 h = [area of a cross section] × [height] (1) This is a special case of a more general volume formula that applies to solids called right cylinders. A right cylinder is a solid that is generated when a plane region is translated along a line or axis that is perpendicular to the region (Figure 5.2.3). Figure 5.2.3 If a right cylinder is generated by translating a region of area A through a distance h, then h is called the height (or sometimes the width) of the cylinder, and the volume V of the cylinder is defined to be V = A · h = [area of a cross section] × [height] (2) 286 Chapter 5 / Applications of the Definite Integral in Geometry, Science, and Engineering (Figure 5.2.4).
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