7.3 Day One: Volumes by Slicing7.3 Day One: Volumes by Slicing

x2

π x2

2

π x2

8

34

⎛⎝⎜

⎞⎠⎟ x2

154

⎛⎝⎜

⎞⎠⎟x2

33

3

Find the volume of the pyramid:

Consider a horizontal slice through the pyramid.

sdh

The volume of the slice is s2dh.

If we put zero at the top of the pyramid and make down the positive direction, then s=h.

0

3

h

2sliceV h dh=

3 2

0V h dh=∫

33

0

1

3h= 9=

This correlates with the formula:1

3V Bh=

19 3

3= ⋅ ⋅ 9=

→

Method of Slicing:

1

Find a formula for V(x).(Note that I used V(x) instead of A(x).)

Sketch the solid and a typical cross section.

2

3 Find the limits of integration.

4 Integrate V(x) to find volume.

→

xy

A 45o wedge is cut from a cylinder of radius 3 as shown.

Find the volume of the wedge.

You could slice this wedge shape several ways, but the simplest cross section is a rectangle.

If we let h equal the height of the slice then the volume of the slice is: ( ) 2V x y h dx= ⋅ ⋅

Since the wedge is cut at a 45o angle:x

h45o h x=

Since2 2 9x y+ = 29y x= −

→

xy

( ) 2V x y h dx= ⋅ ⋅h x=

29y x= −

( ) 22 9V x x x dx= − ⋅ ⋅

3 2

02 9V x x dx= −∫

29u x= −2 du x dx=−

( )0 9u =( )3 0u =

10

2

9V u du=−∫

93

2

0

2

3u=

227

3= ⋅ 18=

Even though we started with a cylinder, π does not enter the calculation!

→

Cavalieri’s Theorem:

Two solids with equal altitudes and identical parallel cross sections have the same volume.

Identical Cross Sections

π

Cavalieri’s Theorem: Volume of a Sphere

π

( )

( )( )

[ ]

2

0

0

0

2

V sin2

11 cos 2x

2 2

sin

4 2

4

4

= x dx

= dx

2x = x +

=

=

π

π

π

π

π−

π ⎡ ⎤⎢ ⎥⎣ ⎦

ππ

π

∫

∫

7.3 Disk and Washer Methods

0

1

2

1 2 3 4

y x= Suppose I start with this curve.

My boss at the ACME Rocket Company has assigned me to build a nose cone in this shape.

So I put a piece of wood in a lathe and turn it to a shape to match the curve.

→

0

1

2

1 2 3 4

y x=How could we find the volume of the cone?

One way would be to cut it into a series of thin slices (flat cylinders) and add their volumes.

The volume of each flat cylinder (disk) is:

2 the thicknessrπ ⋅

In this case:

r= the y value of the function

thickness = a small change

in x = dx

π ( )2x dx

→

0

1

2

1 2 3 4

y x=The volume of each flat cylinder (disk) is:

2 the thicknessrπ ⋅

If we add the volumes, we get:

( )24

0x dxπ∫

4

0 x dxπ=∫

42

02x

π= 8π=

π ( )2x dx

→

This application of the method of slicing is called the disk method. The shape of the slice is a disk, so we use the formula for the area of a circle to find the volume of the disk.

If the shape is rotated about the x-axis, then the formula is:

2 b

aV y dxπ= ∫

2 b

aV x dyπ= ∫A shape rotated about the y-axis would be:

→

The region between the curve , and the

y-axis is revolved about the y-axis. Find the volume.

1x

y= 1 4y≤ ≤

0

1

2

3

4

1

y x

1 1

2

3

4

1.707

2=

1.577

3=

1

2

We use a horizontal disk.

dy

The thickness is dy.

The radius is the x value of the function .1

y=

24

1

1 V dy

yπ⎛ ⎞

= ⎜ ⎟⎜ ⎟⎝ ⎠

∫

volume of disk

4

1

1 dy

yπ=∫

4

1ln yπ= ( )ln 4 ln1π= −

02ln 2π= 2 ln 2π=

→

The natural draft cooling tower shown at left is about 500 feet high and its shape can be approximated by the graph of this equation revolved about the y-axis:

2.000574 .439 185x y y= − +x

y

500 ft

( )500 22

0.000574 .439 185 y y dyπ − +∫

The volume can be calculated using the disk method with a horizontal disk.

324,700,000 ft≈→

0

1

2

3

4

1 2

The region bounded by and is revolved about the y-axis.Find the volume.

2y x= 2y x=

The “disk” now has a hole in it, making it a “washer”.

If we use a horizontal slice:

The volume of the washer is: ( )2 2 thicknessR rπ π− ⋅

( )2 2R r dyπ −

outerradius

innerradius

2y x=

2

yx=

2y x=

y x=

2y x=

2y x=

( )2

24

0 2

yV y dyπ

⎛ ⎞⎛ ⎞= −⎜ ⎟⎜ ⎟⎜ ⎟⎝ ⎠⎝ ⎠∫

4 2

0

1

4V y y dyπ ⎛ ⎞= −⎜ ⎟

⎝ ⎠∫4 2

0

1

4V y y dyπ= −∫

42 3

0

1 1

2 12y yπ ⎡ ⎤= −⎢ ⎥⎣ ⎦

168

3π ⎡ ⎤= −⎢ ⎥⎣ ⎦

8

3

π=

→

This application of the method of slicing is called the washer method. The shape of the slice is a circle with a hole in it, so we subtract the area of the inner circle from the area of the outer circle.

The washer method formula is: 2 2 b

aV R r dxπ= −∫

→

0

1

2

3

4

1 2

2y x=If the same region is rotated about the line x=2:

2y x=

The outer radius is:

22

yR = −

R

The inner radius is:

2r y= −

r

2y x=

2

yx=

2y x=

y x=

4 2 2

0V R r dyπ= −∫

( )2

24

02 2

2

yy dyπ ⎛ ⎞= − − −⎜ ⎟

⎝ ⎠∫

( )24

04 2 4 4

4

yy y y dyπ

⎛ ⎞= − + − − +⎜ ⎟

⎝ ⎠∫

24

04 2 4 4

4

yy y y dyπ= − + − + −∫

14 2 2

0

13 4

4y y y dyπ= − + +∫

432 3 2

0

3 1 8

2 12 3y y yπ

⎡ ⎤= ⋅ − + +⎢ ⎥

⎣ ⎦16 64

243 3

π ⎡ ⎤= ⋅ − + +⎢ ⎥⎣ ⎦

8

3

π=

π

Washer Cross Section

The region in the first quadrant enclosed by the y-axis and the graphs of y = cos x and y = sin x is revolved about the x-axis to form a solid. Find its volume.

Washer Cross Section

The region in the first quadrant enclosed by the y-axis and the graphs of y = cos x and y = sin x is revolved about the x-axis to form a solid. Find its volume.

( )4 4 2 2

0 0

4

0

43

0

V = A(x) dx = cos x - sin x dx

= cos 2x dx

sin 2x = = units

2 2

π π

π

π

π

π

π⎡ ⎤π ⎢ ⎥⎣ ⎦

∫ ∫

∫

g

7.3 The Shell Method

Find the volume of the region bounded by , , and revolved about the y-axis.

2 1y x= + 2x =0y =

0

1

2

3

4

5

1 2

2 1y x= +

We can use the washer method if we split it into two parts:

( )25 2 2

12 1 2 1y dyπ π− − + ⋅ ⋅∫

21y x− = 1x y= −

outerradius

innerradius

thicknessof slice

cylinder

( )5

14 1 4y dyπ π− − +∫

5

15 4y dyπ π− +∫

52

1

15 4

2y yπ π⎡ ⎤− +⎢ ⎥⎣ ⎦

25 125 5 4

2 2π π⎡ ⎤⎛ ⎞ ⎛ ⎞− − − +⎜ ⎟ ⎜ ⎟⎢ ⎥⎝ ⎠ ⎝ ⎠⎣ ⎦

25 94

2 2π π⎡ ⎤− +⎢ ⎥⎣ ⎦

164

2π π⋅ +

8 4π π+ 12π=→

Japanese Spider CrabGeorgia Aquarium, Atlanta

0

1

2

3

4

5

1 2

If we take a vertical slice and revolve it about the y-axis

we get a cylinder.

cross section

If we add all of the cylinders together, we can reconstruct the original object.

2 1y x= +

Here is another way we could approach this problem:

→

0

1

2

3

4

5

1 2

cross section

The volume of a thin, hollow cylinder is given by:

Lateral surface area of cylinder thickness⋅

=2 thicknessr hπ ⋅ ⋅ r is the x value of the function.

circumference height thickness= ⋅ ⋅

h is the y value of the function.

thickness is dx.( )2=2 1 x x dxπ +

r hthicknesscircumference

2 1y x= +

→

0

1

2

3

4

5

1 2

cross section

=2 thicknessr hπ ⋅ ⋅

( )2=2 1 x x dxπ +

r hthicknesscircumference

If we add all the cylinders from the smallest to the largest:

( )2 2

02 1 x x dxπ +∫

2 3

02 x x dxπ +∫

24 2

0

1 12

4 2x xπ ⎡ ⎤+⎢ ⎥⎣ ⎦

[ ]2 4 2π +

12π

This is called the shell method because we use cylindrical shells.

2 1y x= +

→

0

1

2

3

4

1 2 3 4 5 6 7 8

( )2410 16

9y x x= − − +

Find the volume generated when this shape is revolved about the y axis.

We can’t solve for x, so we can’t use a horizontal slice directly.

→

0

1

2

3

4

1 2 3 4 5 6 7 8

( )2410 16

9y x x= − − +

Shell method:

Lateral surface area of cylinder

=circumference height⋅

=2 r hπ ⋅Volume of thin cylinder 2 r h dxπ= ⋅ ⋅

If we take a vertical sliceand revolve it about the y-axiswe get a cylinder.

→

0

1

2

3

4

1 2 3 4 5 6 7 8

( )2410 16

9y x x= − − +Volume of thin cylinder 2 r h dxπ= ⋅ ⋅

( )8 2

2

42 10 16

9x x x dxπ⎡ ⎤− − +⎢ ⎥⎣ ⎦∫

rh thickness

160π=

3502.655 cm≈

Note: When entering this into the calculator, be sure to enter the multiplication symbol before the parenthesis.

circumference

→

When the strip is parallel to the axis of rotation, use the shell method.

When the strip is perpendicular to the axis of rotation, use the washer method.

π

Find the volume of the solid when the region bounded by the curve y = , the x-axis, and the line x = 4 is revolved about the x-axis. Find the volume of the solid using cylindrical shells.

x

( ) ( )2 2

0V = 2 y 4 - y dy

= 8

π

π

∫

2

2

Radius = y

x = y

Shell height = 4 - y

Find the volume of the solid when the region bounded by the curve y = , the x-axis, and the line x = 4 is revolved about the x-axis. Find the volume of the solid using cylindrical shells.

x

Find the volume of the solid of revolution formed by revolving the region bounded by the graph of and the y axis, 0 ≤ y ≤ 1, about the x-axis. Use the Shell Method.

x =e−y2

2

1

-1

2

Find the volume of the solid of revolution formed by revolving the region bounded by the graph of and the y axis, 0 ≤ y ≤ 1, about the x-axis. Use the Shell Method.

x =e−y2

2

1

-1

2

V =2π ye−y2

0

1

∫ dy

=−π e−y2⎡

⎣⎤⎦0

1

=π 1−1e

⎛

⎝⎜⎞

⎠⎟

=1.986

Find the volume of the solid formed by revolving the region bounded by the graphs y = x3 + x + 1, y = 1, and x = 1 about the line x = 2.

3

2

1

-1

2 4

axis ofrevolution

Find the volume of the solid formed by revolving the region bounded by the graphs y = x3 + x + 1, y = 1, and x = 1 about the line x = 2.

3

2

1

-1

2 4

axis ofrevolution

V =2π 2−x( )0

1

∫ x3 + x+1−1( ) dx

=2π −x4 + 2x3 −x2 + 2x( ) 0

2

∫ dx

=2π −x5

5+x4

4−x3

3+ x2⎡

⎣⎢

⎤

⎦⎥0

1

=2π −15+12−13+1

⎡

⎣⎢

⎤

⎦⎥

=29π15