13. The solid whose base is the triangle with vertices 10, 02, 12, 02, and 10, 22, and whose cross sections perpendicular to the base and parallel to the y-axis are semicircles
14. The pyramid with a square base 4 m on a side and a height of 2 m (Use calculus.)
15. The tetrahedron (pyramid with four triangular faces), all of whose edges have length 4
16. A circular cylinder of radius r and height h whose axis is at an angle of p>4 to the base
17–26. Disk method Let R be the region bounded by the following curves. Use the disk method to find the volume of the solid generated when R is revolved about the x-axis.
17. y = 2x, y = 0, x = 3 (Verify that your answer agrees with the volume formula for a cone.)
y ! 2x
5. Why is the disk method a special case of the general slicing method?
6. The region R bounded by the graph of y = f 1×2 Ú 0 and the x-axis on 3a, b4 is revolved about the line y = -2 to form a solid of revolution whose cross sections are washers. What are the inner and outer radii of the washer at a point x in 3a, b4?
Basic Skills 7–16. General slicing method Use the general slicing method to find the volume of the following solids.
7. The solid whose base is the region bounded by the curves y = x2 and y = 2 – x2, and whose cross sections through the solid per- pendicular to the x-axis are squares
8. The solid whose base is the region bounded by the semicircle y = 21 – x2 and the x-axis, and whose cross sections through the solid perpendicular to the x-axis are squares
y x !1 ! x2 y ”
9. The solid whose base is the region bounded by the curve y = 1cos x and the x-axis on 3-p>2, p>24, and whose cross sections through the solid perpendicular to the x-axis are isosceles right triangles with a horizontal leg in the xy-plane and a vertical leg above the x-axis
x y ! cos x
10. The solid with a circular base of radius 5 whose cross sections perpen- dicular to the base and parallel to the x-axis are equilateral triangles
y ! 2 ” x2
y ! x2
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6.3 Volume by Slicing 431
28. y = x, y = 24 x y ! x
y ! 4 x
29. y = ex>2, y = e-x>2, x = ln 2, x = ln 3 y ! ex/2
y ! e”x/2
xln 2O ln 3
30. y = x, y = x + 2, x = 0, x = 4
y ! x ” 2
y ! x
31. y = x + 3, y = x2 + 1
32. y = 1sin x, y = 1, x = 0 33. y = sin x, y = 1sin x, for 0 … x … p>2 34. y = ! x ! , y = 2 – x2
35–40. Disks ,washers about the y-axis Let R be the region bounded by the following curves. Use the disk or washer method to find the vol- ume of the solid generated when R is revolved about the y-axis.
35. y = x, y = 2x, y = 6
y ! x
y ! 2x
y ! 6
18. y = 2 – 2x, y = 0, x = 0 (Verify that your answer agrees with the volume formula for a cone.)
y ! 2 ” 2x
19. y = e-x, y = 0, x = 0, x = ln 4
y ! e”x
20. y = cos x on 30, p>24, y = 0, x = 0 1Recall that cos2 x = 1 211 + cos 2×2.2
y ! cos x
21. y = sin x on 30, p4, y = 0 (Recall that sin2 x = 1211 – cos 2×2.2 22. y = 225 – x2, y = 0 (Verify that your answer agrees with the
volume formula for a sphere.)
23. y = 124 1 – x2, y = 0, x = 0, and x = 12
24. y = sec x, y = 0, x = 0, and x = p4
25. y = 121 + x2, y = 0, x = -1, and x = 1
26. y = 124 1 – x2, y = 0, x = -12, and x = 12
27–34. Washer method Let R be the region bounded by the following curves. Use the washer method to find the volume of the solid gener- ated when R is revolved about the x-axis.
27. y = x, y = 21x R
y ! 2!x
y ! x
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434 Chapter 6 Applications of Integration
We divide 3a, b4 into n subintervals of length ∆x = 1b – a2>n and identify an arbi- trary point xk
* on the kth subinterval, for k = 1, c, n. Now observe the rectangle built on the kth subinterval with a height of f 1xk*2 and a width ∆x (Figure 6.40). As it revolves about the y-axis, this rectangle sweeps out a thin cylindrical shell.
6.4 Volume by Shells You can solve many challenging volume problems using the disk>washer method. There are, however, some volume problems that are difficult to solve with this method. For this reason, we extend our discussion of volume problems to the shell method, which—like the disk>washer method—is used to compute the volume of solids of revolution. Cylindrical Shells Let R be a region bounded by the graph of f, the x-axis, and the lines x = a and x = b, where 0 … a 6 b and f 1×2 Ú 0 on 3a, b4. When R is revolved about the y-axis, a solid is generated (Figure 6.39) whose volume is computed with the slice-and-sum strategy.
➤ Why another method? Suppose R is the region in the first quadrant bounded by the graph of y = x2 – x3 and the x-axis (Figure 6.38). When R is revolved about the y-axis, the resulting solid has a volume that is difficult to compute using the washer method. The volume is much easier to compute using the shell method.
y ! x2 ” x3
Revolving region R about the y-axis…
… produces a solid.
Revolving the kth rectangle about the y-axis…
… produces a cylindrical shell with height f (xk*) and thickness !x.
When the kth cylindrical shell is unwrapped (Figure 6.41), it approximates a thin rect- angular slab. The approximate length of the slab is the circumference of a circle with ra- dius xk
*, which is 2pxk *. The height of the slab is the height of the original rectangle f 1xk*2
and its thickness is ∆x; therefore, the volume of the kth shell is approximately
2pxk * # f 1xk*2 # ∆x = 2pxk* f 1xk*2∆x.
length height thickness
Summing the volumes of the n cylindrical shells gives an approximation to the volume of the entire solid:
V ≈ a n
k = 1 2pxk
* f 1xk*2∆x.
(1)1* (1)1* ()*
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6.4 Volume by Shells 435
As n increases and as ∆x approaches 0 (Figure 6.42), we obtain the exact volume of the solid as a definite integral:
V = lim nS∞
k = 1 2p xk
* f 1xk*2∆x = Lba 2px f 1×2 dx. shell shell circumference thickness
Circumference ! 2! ” radius ! 2!xk*
Radius ! xk*
Height ! f (xk*)
Thickness ! #x
Length ! 2! ” radius ! 2!xk*
Increase the number of shells.
Volume ! lim k!1
” 2!xk* f (xk*)#x
! ! 2!x f (x) dx
n ! $
Before doing examples, we generalize this method as we did for the disk method. Suppose that the region R is bounded by two curves, y = f 1×2 and y = g1x2, where f 1×2 Ú g1x) on 3a, b4 (Figure 6.43). What is the volume of the solid generated when R is revolved about the y-axis?
➤ Rather than memorizing, think of the meaning of the factors in this formula: f 1×2 is the height of a single cylindrical shell, 2px is the circumference of the shell, and dx corresponds to the thickness of a shell. Therefore, 2px f 1×2 dx represents the volume of a single shell, and we sum the volumes from x = a to x = b. Notice that the integrand for the shell method is the function A1x2 that gives the surface area of the shell of radius x, for a … x … b.
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436 Chapter 6 Applications of Integration
The situation is similar to the case we just considered. A typical rectangle in R sweeps out a cylindrical shell, but now the height of the kth shell is f 1xk*2 – g1xk*2, for k = 1, c, n. As before, we take the radius of the kth shell to be xk*, which means the volume of the kth shell is approximated by 2pxk
*1 f 1xk*2 – g1xk*22∆x (Figure 6.43). Sum- ming the volumes of all the shells gives an approximation to the volume of the entire solid:
V ≈ a n
k = 1 2pxk
* 1 f 1xk*2 – g1xk*22∆x. shell shell circumference height
Taking the limit as n S ∞ (which implies that ∆x S 0), the exact volume is the definite integral
V = lim nS∞
k = 1 2pxk
*1 f 1xk*2 – g1xk*22∆x = Lba 2px1 f 1×2 – g1x22 dx. We now have the formula for the shell method.
➤ An analogous formula for the shell method when R is revolved about the x-axis is obtained by reversing the roles of x and y:
V = L d
c 2py1 p 1y2 – q1y22 dy.
We assume R is bounded by the curves x = p1y2 and x = q1y2, where p1y2 Ú q1y2 on 3c, d4.
y ” f (x)
y ” g(x)
Shell height ” f (xk*) # g(xk*)
Shell radius ” xk*
Volume of kth shell ! 2!xk*( f (xk*) # g(xk*))!x
EXAMPLE 1 A sine bowl Let R be the region bounded by the graph of f 1×2 = sin x2, the x-axis, and the vertical line x = 1p>2 (Figure 6.44). Find the volume of the solid generated when R is revolved about the y-axis.
SOLUTION Revolving R about the y-axis produces a bowl-shaped region (Figure 6.45). The radius of a typical cylindrical shell is x and its height is f 1×2 = sin x2. Therefore, the volume by the shell method is
V = L b
a 2px f 1×2 dx = L1p>20 2px sin x2 dx.rr
Volume by the Shell Method Let f and g be continuous functions with f 1×2 Ú g1x2 on 3a, b4. If R is the region bounded by the curves y = f 1×2 and y = g1x2 between the lines x = a and x = b, the volume of the solid generated when R is revolved about the y-axis is
V = L b
a 2px1 f 1×2 – g1x22 dx.
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6.4 Volume by Shells 437
Now we make the change of variables u = x2, which means that du = 2x dx. The lower limit x = 0 becomes u = 0 and the upper limit x = 1p>2 becomes u = p>2. The volume of the solid is
V = L 1p>2
0 2px sin x2 dx = p L
sin u du u = x2, du = 2x dx
= p1-cos u2 ` p>2 0
= p10 – 1-122 = p. Simplify. Related Exercises 5–14
QUICK CHECK 1 The triangle bounded by the x-axis, the line y = 2x, and the line x = 1 is revolved about the y-axis. Give an integral that equals the volume of the resulting solid using the shell method.
EXAMPLE 2 Shells about the x-axis Let R be the region in the first quadrant bounded by the graph of y = 1x – 2 and the line y = 2. Find the volume of the solid generated when R is revolved about the x-axis.
SOLUTION The revolution is about the x-axis, so the integration in the shell method is with respect to y. A typical shell runs parallel to the x-axis and has radius y, where 0 … y … 2; the shells extend from the y-axis to the curve y = 1x – 2 (Figure 6.46). Solving y = 1x – 2 for x, we have x = y2 + 2, which is the height of the shell at the point y (Figure 6.47). Integrating with respect to y, the volume of the solid is
V = L 2
0 2py 1y2 + 22 dy = 2pL20 1y3 + 2y2 dy = 16p.
shell shell circumference height
➤ When computing volumes using the shell method, it is best to sketch the region R in the xy-plane and draw a slice through the region that generates a typical shell.
Interval of integration
f (x) ! sin x2
Height ! sin x 2
Shell radius ! x 1
f (x) ! sin x2
y Shell circumference ! 2!x
Shell radius ! x
Shell height ! sin x2
x2 4 6
y ” 2 (6, 2)
y ” x ! 2 x ” y2 # 2
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438 Chapter 6 Applications of Integration
Related Exercises 15–26
EXAMPLE 3 Volume of a drilled sphere A cylindrical hole with radius r is drilled symmetrically through the center of a sphere with radius a, where 0 … r … a. What is the volume of the remaining material?
SOLUTION The y-axis is chosen to coincide with the axis of the cylindrical hole. We let R be the region in the xy-plane bounded above by f 1×2 = 2a2 – x2, the upper half of a circle of radius a, and bounded below by g1x2 = – 2a2 – x2, the lower half of a circle of radius a, for r … x … a (Figure 6.48a). Slices are taken perpendicular to the x-axis from x = r to x = a. When a slice is revolved about the y-axis, it sweeps out a cylindri- cal shell that is concentric with the hole through the sphere (Figure 6.48b). The radius of a typical shell is x and its height is f 1×2 – g1x2 = 22a2 – x2. Therefore, the volume of the material that remains in the sphere is
V = L a
r 2px122a2 – x22 dx
= -2pL 0
a2 – r2 1u du u = a2 – x2, du = -2x dx
= 2pa 2 3
u3>2b ` a2 – r2 0
= 4p 3
1a2 – r223>2. Simplify.
Shell height ! y2 ” 2
Shell radius ! y
x ! y2 ” 2
Shell circumference ! 2!y
f (x) ! a2 ” x2
g(x) ! ” a2 ” x2
Shell height ! f (x) ” g(x)
Shell radius ! x
➤ In Example 2, we could use the disk> washer method to compute the volume, but notice that this approach requires splitting the region into two subregions. A better approach is to use the shell method and integrate along the y-axis.
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