Volumes by Cylindrical Shell

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Lawrence Parks
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1 Volumes by Cylindrical Shell Problem: Let f be continuous and nonnegative on [a, b], and let R be the region that is bounded above by y = f(x), below by the x - axis, and on the sides by the lines x = a and x = b. Find the volume of the solid of revolution S that is generated by revolving the region R about the y - axis.

2 We are going to use a different method than before for this problem, since the disks and washers method is not convenient or not possible for this type of area rotated in this direction. Rather than go through the rectangular approximation method again, we go directly to the infinitesimal approach. We begin by slicing the area into infinitesimal vertical rectangles. These rectangles are rotated about an axis parallel to them rather than perpendicular to them.

3 Each little infinitesimal rectangle generates a cylinder. Its height is y = f(x), its radius is x and its thickness is dx. Thus the total infinitesimal volume of the cylinder is dv= πxydx= πxf() xdx

4 The cylinders, when added together, produce the rotated solid. b b Its volume is V= πxydx= πxf() xdx a a

5 Example: Use cylindrical shells to find the volume of the solid generated when the region enclosed by the curves y= xx, = 4, x= 9, y= 0 is revolved around the y-axis. x dx y Solution. dv = πxydx = πxf(x)dx = π x xdx. Thus V= πx xdx π x 4π ( 43 3) 844π = = =

6 We can also use cylindrical shells when rotating about the x-axis, if appropriate. Again, the infinitesimal rectangle being rotated is parallel to the axis of rotation. d x y dy dv = πyxdy = πyg(y)dy c Thus d d V= πyxdy= πyg( ydy ) c c

7 Example. Use cylindrical shells to find the volume generated when the area bounded by the curves y = 1 x, y = 0, and x = 0 is revolved around the x axis. y x dy dv = πyxdy = π y 1 ydy V= πy 1 ydy= π [ 1 u] u( du) = π [ 1 u] udu [ ] = π 1 u udu π u u 8π = =

8 As with disks and washers, we can also use cylindrical shells for volumes generated when the area between two curves is rotated about an axis. As in the diagram below, if such an area is to be rotated about the y-axis, then the infinitesimal rectangle [f(x) - g(x)]dx will generate an infinitesimal cylinder whose volume is dv = πx[f(x) g(x)]dx dx f(x) g(x)

9 Example. Use cylindrical shells to find the volume of the solid generated when the region R in the first quadrant enclosed between y = x and y = x is revolved about the y-axis.

10 Solution: The infinitesimal area (x x )dx is revolved around the y-axis to get an infinitesimal cylinder whose volume is dv = πx(x x )dx. Thus the volume is: x x V= π x x dx π π = =

11 Finally, we can also use cylindrical shells when we revolve the area between two curves about the x axis. Example. Use cylindrical shells to find the volume of the solid generated when the region R under y = x over the interval [0, ] is revolved around the x-axis.

12 Solution. dv = π y( ydy ) 4 4 V = π y( ydy ) π y y ) dy π = = y y = π 16 3π =

13 The choice of method to use depends on which method works, or if both work, which produces the simplest integral. Example. Consider the area between y= x and y= x Evaluate the volume of the solid generated by rotating about the y axis. Use both methods.

14 y I (1, 1) y II (1, 1) y= x y= x x= y x= y x x Solution. I. (cylindrical shells) ( ) V πx x x dx π x x dx π x x = 3π = = = II.(washers) ( ) ( ) 1 4 y y V= π y y dy π y y dy π 3π = = =

15 Example. Find the volume of the solid generated by revolving around the y axis the area between the curves y = x, x = 0, and y= 4 x. This is at. Thus Solution: Here we choose shells, because the choice of washers would lead to a much more difficult problem. The curve and the line y = x meet where 4 x = x ; x= 4 x ; x=. 4 V= x 4 x x dx udu x π = π π dx 0 0

16 3 4 3 x V= π udu x dx u 8π π = π π 3 =

17 Example. Find the volume of the solid generated by revolving around the x axis the area between the curves y = x, y = 0, and y= x+. Solution: Here we choose shells, because the choice of washers would lead to a much more difficult problem. The curve and the line y = x meet where x+ = x; x= x+ ; x x = 0 This is at and 1. Thus 3 4 y y V= π y y ( y ) dy π y 16π = + =

18 Example. Find the volume of the solid generated by revolving around the y axis the area enclosed by y = x - x and y = 0. Solution. Here dv = πxydx. Thus x V= πx( x x ) dx π ( x x ) dx π x 8π = = =

19 Example. Find the volume of the solid generated by revolving around the x axis the area enclosed by xy = 4 and x + y = 5. Solution (Shells). x(5 x) = 4, so 0 = x 5x + 4 = (x 1)(x 4) 4 dv = π y (5 y) dy y (5 y y V = π y y ) 4 dy= π 4y = 9π 3 1 1

20 xy = 4 x + y = 5 Solution (Washers). Here dv 4 π (5 x) = dx x V (5 ) = π x π x 10x 5 π dx x = + x x 3 16π = π 5x + 5x + = 9π 3 x 1 1

PROBLEM SET. Practice Problems for Exam #1. Math 1352, Fall 2004. Oct. 1, 2004 ANSWERS

PROBLEM SET. Practice Problems for Exam #1. Math 1352, Fall 2004. Oct. 1, 2004 ANSWERS PROBLEM SET Practice Problems for Exam # Math 352, Fall 24 Oct., 24 ANSWERS i Problem. vlet R be the region bounded by the curves x = y 2 and y = x. A. Find the volume of the solid generated by revolving