Finding Volumes

Finding Volumes

Chapter 6.2February 22, 2007

In General:

Vertical Cut: Horizontal Cut:

A =topfunction

⎛⎝⎜

⎞⎠⎟−

bottomfunction

⎛⎝⎜

⎞⎠⎟dxa

b

∫ a≤b

A =rightfunction

⎛⎝⎜

⎞⎠⎟−

leftfunction

⎛⎝⎜

⎞⎠⎟dyc

d

∫ c≤d

In-class Integrate:

Set up an integral to find the area of the region bounded by:

y =5e−x, y=2e3x, and x ≥0

Find the area of the region bounded by

Bounds? In terms of y: [-2,1]

Points: (0,-2), (3,1)Right Function?

Left Function?

Area?

x =2 + yx =4 −y2

[(4 −y2−2

1

∫ )−(2 + y)]dy

x + y2 =4 and x−y=2

Volume & Definite Integrals

We used definite integrals to find areas by slicing the region and adding up the areas of the slices.

We will use definite integrals to compute volume in a similar way, by slicing the solid and adding up the volumes of the slices.

For Example………………

Blobs in SpaceVolume of a blob:


Cross sectional area at height h: A(h)


Cross sectional area at height h: A(h)

Volume =

ExampleSolid with cross sectional area A(h) = 2h at height h.Stretches from h = 2 to h = 4. Find the volume.




Volumes:

We will be given a “boundary” for the base of the shape which will be used to find a length.

We will use that length to find the area of a figure generated from the slice .

The dy or dx will be used to represent the thickness.

The volumes from the slices will be added together to get the total volume of the figure.

Find the volume of the solid whose bottom face is the circle and every cross section perpendicular to the x-axis is a square.

x2 + y2 ≤1

Bounds?

Top Function?

Bottom Function?

[-1,1]

y = 1−x2

y =− 1−x2

Length? 1−x2 − − 1−x2( )

=2 1− x2


x2 + y2 ≤1

We use this length to find the area of the square.

Length? =2 1− x2

Area? 2 1−x2( )2

4 1−x2( )

4 1−x2( )dx−1

1

∫

Volume?


x2 + y2 ≤1

What does this shape look like?

4 1−x2( )dx−1

1

∫Volume?

http://www.ies.co.jp/math/java/calc/renshi/renshi.html

Find the volume of the solid whose bottom face is the circle and every cross section perpendicular to the x-axis is a circle with diameter in the plane.

x2 + y2 ≤1

Length? =2 1− x2

Area?

p 1− x2( )

2

p 1 − x2( )

p 1− x2( )dx−1

1

∫Volume?

2 1−x2

2Radius:

Using the half circle [0,1] as the base slices perpendicular to the x-axis are isosceles right triangles.

x2 + y2 ≤1

Length? =2 1− x2

Area?12

2 1−x2( ) 2 1−x2( )

Volume? 2 1−x2( )dx0

1

∫

Bounds? [0,1]

Visual?

http://www.ies.co.jp/math/java/calc/rhansan/rhansan.html

The base of the solid is the region between the curve and the interval [0,π] on the x-axis. The cross sections perpendicular to the x-axis are equilateral triangles with bases running from the x-axis to the curve.

y =2 sinx

Bounds?

Top Function?

Bottom Function?

[0,π]

y =2 sinx

y =0

Length? 2 sin x

Area of an equilateral triangle?

Area of an Equilateral Triangle?

S

S

S

S/2

S

S

Sqrt(3)*S/2

S/2

Area = (1/2)b*h

=12

⎛⎝⎜

⎞⎠⎟ S( )

32S

⎛⎝⎜

⎞⎠⎟=

34S2

The base of the solid is the region between the curve and the interval [0,π] on the x-axis. The cross sections perpendicular to the x-axis are equilateral triangles with bases running from the x-axis to the curve.

y =2 sinx

Bounds?

Top Function?

Bottom Function?

[0,π]

y =2 sinx

y =0

Length? 2 sin x

Area of an equilateral triangle?3

4(2 sin x )2

34

(2 sin x )2dx0

p

∫Volume?

34

(S)2⎛⎝⎜

⎞⎠⎟

Find the volume of the solid whose bottom face is the circle and every cross section perpendicular to the x-axis is a square with diagonal in the plane.

x2 + y2 ≤1

We used this length to find the area of the square whose side was in the plane….

Length? =2 1− x2

Area with the length representing the diagonal?

Area of Square whose diagonal is in the plane?

D

SS

S2 + S2 =D 2

2S2 =D 2

S2 =D 2

2⇒ S =

D2

Find the volume of the solid whose bottom face is the circle and every cross section perpendicular to the x-axis is a square with diagonal in the plane.

x2 + y2 ≤1

Length of Diagonal? =2 1− x2

Length of Side?

2 1−x2

2= 2 1− x2

2(1−x2 )Area?

Volume? 2 1−x2( )dx−1

1

∫

(S =D2)

Solids of Revolution

We start with a known planar shape and rotate that shape about a line resulting in a three dimensional shape known as a solid of revolution. When this solid of revolution takes on a non-regular shape, we can use integration to compute the volume.

For example……

The Bell!

Volume:

Area:

p f (x)( )2

p f (x)( )2 dxa

b

∫

Find the volume of the solid generated by revolving the region defined by , x = 3 and the x-axis about the x-axis.

y = x

Bounds?

Length? (radius)

Area?

Volume?

[0,3]

y = x

p x( )2

p xdx0

3

∫

Find the volume of the solid generated by revolving the region defined by , x = 3 and the x-axis about the x-axis.

y = x

Solids of RevolutionRotate a region about an axis.


Such as: Region between y = x2 and the y-axis, for 0 ≤ x ≤ 2, about the y-axis.


Such as: Region between y = x2 and the y-axis, for 0 ≤ x ≤ 2, about the y-axis.

Solids of RevolutionFor solids of revolution, cross sections are circles, so we can use the formula

Volume = p r(h)( )2 dha

b

∫

Solids of RevolutionFor solids of revolution, cross sections are circles, so we can use the formula

Usually, the only difficult part is determining r(h).A good sketch is a big help.

Volume = p r(h)( )2 dha

b

∫

ExampleRotate region between x = sin(y), 0 ≤ y ≤ π, and the y axis about the y axis.



We need to figure out whatthe resulting solid looks like.

Aside: Sketching Revolutions1. Sketch the curve; determine the region.


2. Sketch the reflection over the axis.


2. Sketch the reflection over the axis.

3. Sketch in a few “revolution” lines.




Now find r(y), the radius at height y.


Now find r(y), the radius at height y.

(Each slice is a circle.)y


y


x = sin(y)


So r(y) = sin(y).


So r(y) = sin(y).

Therefore, volume is

p r y( )( )2dy

a

b

∫


So r(y) = sin(y).


p r y( )( )2dy

a

b

∫ = p sin y( )( )2dy

a

b

∫


So r(y) = sin(y).


What are the limits?

p r y( )( )2dy

a

b

∫ = p sin y( )( )2dy

a

b

∫


So r(y) = sin(y).


What are the limits?

The variable is y, so the limits are in terms of y…

= p sin y( )( )2dy

a

b

∫p r y( )( )2dy

a

b

∫


Upper limit: y = π

Lower limit: y = 0


Volume is

= p sin y( )( )2dy

0

π

∫= p sin y( )( )2dy

a

b

∫p r y( )( )2dy

a

b

∫

=p sin y( )( )2dy

0

π

∫ =p12

−12

cos 2y( )⎛⎝⎜

⎞⎠⎟dy

0

π

∫

=py2

−14

sin 2y( )⎛⎝⎜

⎞⎠⎟

0

π

=p 2

2

ExampleRevolve the region under the curve y = 3e–x, for 0 ≤ x ≤ 1, about the x axis.


(Since we rotate about x, the slices are perpendicular to x.So x is our variable.)


First, get the region:


First, get the region:


First, get the region:When we rotate, this will become a radius.


Next, revolve the region:


Next, revolve the region:


Radius at point x is 3e–x.



Limits are from x = to x =



Limits are from x = 0 to x = 1




Volume:




Volume:




Volume:




Volume:

Remember for this method: Slices are perpendicular to the

axis of rotation. Radius is then a function of

position on that axis. Therefore rotating about x axis

gives an integral in x; rotating about y gives an integral in y.

ExampleRotate y = x2, from x = 0 to x = 4, about the x-axis.


First, sketch:


Next, rotate about the x axis:


Next, rotate about the x axis:


Next, find r(x): r(x) =


Next, find r(x): r(x) =


Next, find r(x): r(x) = x2

Limits:



Limits: x = 0 to x = 4




Volume:




Volume:




Volume:




Volume:

Find the volume of the solid generated by revolving the region defined by , on the interval [1,2] about the x-axis.

y =x2

Bounds?

Length? (radius)

Area?

Volume?

[1,2]

y =x2

p x2( )2

p x4dx1

2

∫

Find the volume of the solid generated by revolving the region defined by , on the interval [1,2] about the x-axis.

y =x2

Find the volume of the solid generated by revolving the region defined by , y = 8, and x = 0 about the y-axis.

y =x3

Bounds?

Length?

Area?

Volume?

[0,8]

x = y3

p y3( )2

p y3`( )2dy

0

8

∫

Find the volume of the solid generated by revolving the region defined by , and y = 1, about the line y = 1

y =2−x2

Bounds?

Length?

Area?

Volume?

[-1,1]

(2−x2 )−1

p 1− x2( )2

p 1− x2( )2dx

−1

1

∫

What if there is a “gap” between the axis of rotation and the function?

(Temporarily end powerpointLook at Washer2.gifWasher5.gifWasher8.gifWasher10.gif)

*Find the volume of the solid generated by revolving the region defined by , and y = 1, about the x-axis.

y =2−x2

Bounds?

Outside Radius?

Inside Radius?

Area?

[-1,1]

2 −x2

1

p 2 − x2( )2

− π 1( )2

Volume? p 2 − x2( )2

− π 1( )2( )

−1

1

∫ dx

Find the volume of the solid generated by revolving the region defined by , and y = 1, about the line y=-1.

y =2−x2

Bounds?

Outside Radius?

Inside Radius?

Area?

[-1,1]

2 −x2( )− −1( )

1− −1( )

p 3 − x2( )2

− π 2( )2

Volume? p 3 − x2( )2

− π 2( )2( )

−1

1

∫ dx

Try: Set up an integral integrating with respect

to y to find the volume of the solid of revolution obtained when the region bounded by the graphs of y = x2 and y = 0 and x = 2 is rotated around

a) the y-axis

b) the line x = 4

Summary Described how a plane region rotated

about an axis describes a solid.


about an axis describes a solid. Found volumes of solids of revolution

using the area of a circle


about an axis describes a solid. Found volumes of solids of revolution

using the area of a circle Determined the variable as given by the

axis of rotation.

Solids of RevolutionIf there is a gap between the function and the axis of rotation, we have a washer and use:

If there is NO gap, we have a disk and use:

Volume = p R(h)( )2 − (r(h))2( )dha

b

∫

p r(h)( )2 dha

b

∫Volume =

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Finding Volumes

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