第6.2节 旋转体的体积

6.2 VOLUMES OF SOLIDS OF REVOLUTION

Integrals are used in this section to find the volume of a solid of revolution. A solid of revolution is generated by taking a region in the first quadrant of the plane and rotating it in space about the x - or y-axis (Figure 6.2.1).

Figure 6.2.1 Solids of Revolution

We shall work with the region under a curve and the region between two curves. We use one method for rotating about the axis of the independent variable and another for rotating about the axis of the dependent variable.

For areas our starting point was the formula

area = base × height

for the area of a rectangle. For volumes of a solid of revolution our starting point is the usual formula for the volume of a right circular cylinder (Figure 6.2.2)

Figure 6.2.2

DEFINITION

The volume of a right circular cylinder with height h and base of radius r is

V = π r² h.

DISC METHOD : For rotations about the axis of the independent variable.

Let us first consider the region under a curve. Let R be the region under a curve y= f(x) from x=a to x = b, shown in figure 6.2.3(a). x is the independent

Figure 6.2.3

Variable in this case. To keep R in the first quadrant we assume 0 ≤ a ≤b and 0 ≤ f(x). Rotate R about the x-axis, generating the solid of revolution S shown in Figure 6.2.3(b).

This volume is given by the formula below.

VOLUME BY DISC METHOD V= ______ π(f(x)²dx.

To justify this formula we slice the region R into vertical strips of infinitesimal width Δx. This slices the solid S into discs of infinitesimal thickness Δx. Each disc is almost a cylinder of height Δx whose base is a circle of radius f(x) (Figure 6.2.4). Therefore

Δ V= _______ π(f(x))² Δ x.

Figure 6.2.4 Disc Method

EXAMPLE 1 Find the volume of a right circular cone with height h and base of radius r .

It is convenient to center the cone on the x-axis with its vertex at the origin as shown in Figure 6.2.5. This cone is the solid generated by rotating about the x- axis the triangular region R under the line y= (r/h) x, 0 ≤ x ≤ h.

Figure 6.2.5

Since x is the independent variable we use the Disc Method. The volume formula gives

Or

Now we consider the region R between two curves y= f(x) and y= g(x) from x=a to x=b. Rotating R about the x-axis generates a solid of revolution S shown in Figure 6.2.6(c).

Figure 6.2.6

Let R1 be the region under the curve y= f(x) shown in Figure 6.2.6(b) , and R2, the region under the curve y= g(x), shown in Figure 6.2.6(a). Then S can be found by removing the solid of revolution S1 generated by R1 from the solid of revolution S2 generated by R2. Therefore

Volume of S = volume of S2 - volume of S1.

This justifies the formula

We combine this into a single integral.

VOLUME BY DISC METHOD V = _____ π[(g(π))² - (f(x))²] dx.

Another way to see this formula is to divide the solid into annular discs (washers) with inner radius f(x) and outer radius g(x), as illustrated in Figure 6.2.7.

Figure 6.2.7

EXAMPLE 2 The region R between the curves y=2-x² and y=x² is rotated about the x-axis

generating a solid S. Find the volume of S.

The curves y=2-x² and y=x² cross at x=___ 1. The region is sketched in

Figure 6.2.8. The volume is

Figure 6.2.8

Warning : When using the disc method for a region between two curves, the correct formula is

or

A common mistake is to subtract f(x) from g(x) before squaring.

Wrong: V= ____ π(g(x) - f(x))² dx.

Wrong:(for Example 2):

V = ____ π(2-x²) - x²)-x²)²dx =_____π(2 - 2x²)² dx.

=_______ π(4-8x² +4 x4)dx = 64π / 15.

CYLINDRICAL SHELL METHOD:

For rotations about the axis of the dependent variable.

Let us again consider the region R under a curve y=f(x) from x= a to x=b, so that x is still the

independent variable. This time rotate R about the y-axis to generate a solid of revolution S

(Figure 6.2.9).

VOLUME BY CYLINDRICAL SHELL METHOD V= ____ 2πx f (x) dx.

Let us justify this formula. Divide R into vertical strips of infinitesimal width Δx as shown in Figure 6.2.10. When a vertical strip is rotated about the y-axis it generates a cylindrical shell of thickness Δx and volume ΔV. This cylindrical shell is the difference between an outer cylinder of radius x+Δx and an inner cylinder of radius Δx. Both cylinders have height infinitely close to f(x). Thus compared to Δx,

ΔV ≈ outer cylinder - inner cylinder

≈ π(x + Δx)² f(x) - πx² f(x)

= π(x² + 2xΔx +(Δx)² -x²) f(x)

= π(2xΔx +(Δx)² f(x) ≈ π 2xΔx f(x),

Whence ΔV ≈ 2π x f(x)Δx, (compared to Δx).

By the Infinite Sum Theorem,

V=______ 2π x f(x)dx.

EXAMPLE 3 The region R between the line y= 0 and the curve y= 2x - x² is

rotated about the y-axis to form a solid of revolution S. Find the volume of S.

We use the cylindrical shell method because y is the dependent variable. We

see that the curve crosses the x-axis x=0 and x=2, and sketch the region in

Figure 6.2.11. The volume is

Figure 6.2.11

Now let R be the region between the curves y=f(x) and y= g(x) for a ≤ x ≤b, and generate the solid S by rotating R about the y=axis. The volume of S can be found by subtracting the volume of the solid S1 generated by the region under y= f(x) from the volume of the solid S2 generated by the region under y=g(x) ( Figure 6.2.12). The formula for the volume is

V= S2 - S1 = ____ 2πxg(x)dx - _____ 2π x f(x) dx.

Combining into one integral, we get

VOLUME BY CYLINDRICAL SHELL METHOD V = ___ 2πx (g(x) -f(x)) dx.

EXAMPLE 4 The region between the curves y=x and y=___ is rotated about the y-axis.

Find the volume of the solid of revolution.

We make a sketch in Figure 6.2.13 and find that the curves cross at x=0 and

x=1. We take x for the independent variable and use the Cylindrical Shell

Method.

Some regions R are more easily described by taking y as the independent variable, so that R is the region between x=f(y) and x=g(y) for c ≤ y ≤ d. The volumes of the solids of revolution are then computed by integrating with respect to y. Often we have a choice of either x or y as the independent variable.

How can one decide whether to use the Disc or Cylindrical Shell Method? The answer depends on both the axis of rotation and the choice of independent variable. Use the Disc Method when rotating about the axis of the independent variable. Use the Cylindrical Shell Method when rotating about the axis of the dependent variable.

EXAMPLE 5 Derive the formula V= ___ πr3 for the volume of a sphere by both the Disc

Method and the Cylindrical Shell Method.

The circle of radius r and center at the origin has the equation.

x² + y² = r²

The region R inside this circle in the first quadrant will generate a hemisphere of radius r when it is rotated about the x-axis( Figure 6.2.14).

First take x as the independent variable and use the Disc Method. R is the region under the curve

y=________________, 0 ≤ x ≤ r.

The hemisphere has volume

Therefore the sphere has volume

V= ____ πr3.

Now take y as the independent variable and use the Cylindrical Shell Method.

R is the region under the curve.

x =_______, 0≤ y ≤ r.

The hemisphere has volume

Putting u = r² - y² , du = -2 y dy, we get

Thus again V=____

PROBLEMS FOR SECTION 6.2

In Problems 1-10 the region under the given curve is rotated about (a) the x-axis, (b) the y-axis.

Sketch the region and find the volumes of the two solids of revolution.

1 y= x², 0≤ x ≤ 1 2 y= x3, 0≤ x ≤ 1

3 y= ___ , 0≤ x ≤ 4 4 y=_______, 2 ≤ x ≤ 4

5 y= 1-x , 0≤ x ≤ 1 6 y=x, 1≤ x ≤ 2

7 y= ____, 0≤ x ≤ 1 8 y= ____, 2≤ x ≤ 4

9 y= x -3, 1≤ x ≤ 2 10 y= 1/x, 1≤ x ≤ 2

In Problems 11-22 the region bounded by the given curves is rotated about (a) the x-axis, (b) the y-axis. Sketch the region and find the volumes of the two solids of revolution.

11 x, y ≥ 0, y =_________ 12 y = 0, y = x - x²

13 y = x, y = 2x, 0 ≤ x ≤ 3 14 y = x², y = x,

15 y = x3, y = x² 16 y = 3/x, y = 4 - x

17 x = 0, x= y- y4 18 x = y, x= 2y- y²

19 x = 0, x= y+1/ y, 1 ≤ y ≤ 2

20 x ≥ 0, y ≥ 0, 2x² + y² = 4

21 y=0 , y = x-2, y= _______

22 y=___ x , y = 1 - x, y= x - 1/x ( first quadrant )

In Problems 23-34 the region under the given curve is rotated about the x-axis. Find the volume of the solid of revolution.

23 y=____ , 0 ≤ x ≤ π

24 y= cos x ______, 0 ≤ x ≤ π/2

25 y= cos x - sin x, , 0 ≤ x ≤ π/4

26 y= sin (x/2) + cos (x/2), 0 ≤ x ≤ π

27 y = ex, 0 ≤ x ≤ 1 28 y = e1-2x, 0 ≤ x ≤ 2

29 y = xex3, 0 ≤ x ≤ 1 30 y=_____ , 0 ≤ x ≤ 3

31 y = 1/ ___, 1 ≤ x ≤ 2 32 y = ___, 0 ≤ x ≤1

33 y = ___, 1 ≤ x ≤ 4 34 y = ___, 0 ≤ x ≤ 1

In Problems 35-46 the region is rotated about the x-axis. Find the volume of the solid of revolution.

35 y = ________, π /2 ≤ x ≤ π 36 y = ________, π /6 ≤ x ≤ π/2

37 y = sin (x²), 0 ≤ x ≤ _____ 38 y = cos (x²' ), 0 ≤ x ≤ _____

39 y = ex², 0 ≤ x ≤ 1 40 y = ex/x, 1 ≤ x ≤ 2

41 y = 1/x ex, 1 ≤ x ≤ 4 42 y = xex3, 1 ≤ x ≤ 2

43

45

47 A hole of radius a is bored through the center of a sphere of radius r(a < r). Find the

volume of the remaining part of the sphere.

48 A sphere of radius r is cut by a horizontal plane at a distance c above the center of the

sphere. Find the volume of the part of the sphere above the plane (c < r ).

49 A hole of radius a is bored along the axis of a cone of height h and base of radius r.

Find the remaining volume (a < r ).

50 Find the volume of the solid generated by rotating an ellipse a²x ²+ b²y² = 1 about the x-axis.

Hint: the portion of the ellipse in the first quadrant will generate half the volume.

51 the sector of a circle shown in the figure is rotated about(a) the x-axis, (b) the y-axis.

Find the volumes of the solids of revolution.

52 The region bounded by the curves y = x², y=x is rotated about (a) the line y= -1, (b) the line

x= -2. Find the volumes of the solids of revolution.

53 Find the volume of the torus ( donut ) generated by rotating the circle of radius r with center

at(c, 0) around the y-axis ( r <0 ).

□54 (a) Find a general formula for the volume of the solid of revolution generated by rotating the

region bounded by the curves y=f(x), y=g(x), a ≤x≤b, about the line y= -k.

(b) Do the same for a rotation about the line x = -h.