a k = area of k th rectangle, f(c k ) – g(c k ) = height, x k = width. 6.1 area between two...

Ak = area of k th rectangle,f(ck) – g(ck ) = height, xk = width.

6.1 Area between two curves

Figure 4.23: When the formula for a bounding curve changes, the area integral changes to match. (Example 5)Find the area of the region between the curves

( ) ( ) 1f x x and g x x

A

Section 6.2 Figure 5Approximating the volume of a sphere with radius 1

(a) Using 5 disks, V 4.2726

(b) Using 10 disks, V 4.2097

(c) Using 20 disks, V 4.1940

Figure 5.6: The region (a) and solid (b) in Example 4.6. 2 Volumes – Solid of revolution

y = f(x) is rotated about x-axis on [a,b]. Find the volume of the solid generated. A cross-sectional slice is a circle and a slice is a disk. 2( )diskV R thickness

Figure 5.6: The region (a) and solid (b) in Example 4.Volumes – Solid of revolution

2( )diskV R thickness

y x is rotated about the x-axis on [0, 4]Find the volume of the solid generated.

24 4 4

2 2 2 3

00 0

1(4 0 ) 8

2 2diskV x dx xdx x units

Find the volume of the solid generated by revolving a region between the y-axis and the curve x = 2/y from y = 1 to y = 4.

Volumes by disk-y axis rotation

Find the volume of the solid generated by revolving a region between the y-axis and the curve x = 2/y from y = 1 to 4.

2( )diskV R thickness2 44 4 1

2 3

1 1 1

2 14 4 4 1 3

1 4disky

V dy y dy unitsy

Figure 5.10: The cross sections of the solid of revolution generated here are washers, not disks, so the integral A(x) dx leads to a slightly different formula. b

a

WashersIf the region revolved does not border on or cross the axis of revolution, the solid has a hole in it. The cross sections perpendicular to the axis are washers.

V = Outside Volume – Inside Volume

. The region bounded by the curve y = x2 +1 and the line y = -x + 3 is revolved about the x-axis to generate a solid. Find the volume of the solid of revolution.

The inner and outer radii of the washer swept out by one slice. Outer radius R = - x + 3 and the inner radius r = x2 +1

The inner and outer radii of the washer swept out by one slice. Outer radius R = - x + 3 and the inner radius r = x2 +1

Find the limits of integration by finding the x-coordinates of the points of intersection.

x2 + 1= - x + 3

x2 + x –2=0

( x+ 2 )(x – 1) = 0

x = -2 x = 1

Outer radius R = - x + 3 and the inner radius r = x2 +1

2 2b b

a a

Vwasher R dx r dx

2 2

23 1b b

a a

Vwasher x dx x dx

1 1

2 4 2

2 2

( 6 9) 2 1x x dx x x dx

11 5 3

4 2 2 3

2 2

117( 6 8) 3 8

5 3 5

x xx x x dx x x units

Calculation of volume

The region bounded by the parabola y = x2 and the line y = 2x in the first quadrant is revolved about the y-axis to generate a solid. Find the volume of the solid.

2

yx y and x

Drawing indicates a dy integration so solve each

equation for x as a function of y

Set = to find y limits of integration2

2 24 4 02 4

y yy y y y y y

y = 0 and y = 4 are limits

y-axis rotation

The washer swept out by one slice perpendicular to the y-axis.

2 2d d

c c

Vwasher R dy r dy

The region bounded by the parabola y = x2 and the line y = 2x in the first quadrant is revolved about the y-axis to generate a solid. Find the volume of the solid.

2 2d d

c c

Vwasher R dy r dy

2 24 4

0 02

yy dy dy

44 2 2 3

3

0 0

8

4 2 12 3

y y yy dy units

calculation

Figure 5.17: Cutting the solid into thin cylindrical slices, working from the inside out. Each slice occurs at some xk between 0 and 3 and has thickness x. (Example 1)

6. 3 Cylindrical ShellsUsed to find volume of a solid of revolution by summing volumes of thin cylindrical shells or

sleeves or tree rings.

)

Imagine cutting and unrolling a cylindrical shell to get a (nearly) flat rectangular solid. Its volume is approximately V = length height thickness.

volume of a shell

Vshell =2(radius)(height)(thickness)

The region enclosed by the x-axis and the parabola y = f(x) = 3x – x2 is revolved about the y – axis. Find the volume of the solid of revolution.

2b

shella

V rh dx

Vshell =2(radius)(height)(thickness)

3 3

2 2 3

0 0

2 (3 ) 2 3shellV x x x dx x x dx

23 0 0, 3x x x x

343 3

0

81 272 2 (27 ) 2 (0 0)

4 4 2

xx units

problem

The shell swept out by the kth rectangle.Notice this axis or revolution is parallel to the red

rectangle drawn.

The region bounded by the curve y = /x, , the x –axis and the line x = 4 is revolved about the y-axis to generate a solid. Find the volume of the solid.

y x

problem

The region, shell dimensions, and interval of integration in

2b

shella

V rh dx

34 4 42

0 0 0

2 2 2rh dx x x dx x dx

45 5 5

32 2 2

0

2 4 1282 (4 0 )

5 5 5x units

The shell swept out by the rectangle in.

Summary-Volumes-which method is best

Axis of rotationx-axis

shellparallel

dx

dy

diskperpendicular dx

dy

y-axis

2d

c

V rh dy 2b

a

V rh dx

2b

a

V r dx 2d

c

V r dy

Lengths of Plane curves

2

1b

a

dyL dx

dx

Find the length of the arc formed by12( ) 4 5 [1,8]f x x on

122

dyx

dx

2

4dy

xdx

8

1

1 4L xdx u = 1 + 4xdu = 4dxdu/4 = dx 331 3 3 333

2 2 2 2

5 5

1 1 2 1(33 5 ) 29.73

4 4 3 6L u du u units

Follow the link to the slide.

Then click on the figure to play the animation.A

Figure 6.2.5

Figure 6.2.12

Figure 6.3.7

Section 6.3 Figures 3, 4Volumes by Cylindrical Shells

Section 1 / Figure 1

Computer-generated picture of the solid in Example 9

A