Areas Between Curves

Areas Between Curves

To find the area:

• divide the area into n strips of equal width

• approximate the ith strip by a rectangle with base Δx and height f(x_i) – g(x_i).

• the sum of the rectangle areas is a good approximation

• the approximation is getting better as n→∞.

y = f(x)

y = g(x)

The area A of the region bounded by the curves y=f(x), y=g(x), and the lines x=a, x=b, where f and g are continuous and f(x) ≥ g(x) for all x in [a,b], is

 ⁻

=

^b

a

f x g x dx

A [ ( ) ( )]

2

1 2

y = − x

y2 = −x

2 2

1

2 x x dx

−

− +



2

3 2

1

1 1

2 x 3 x 2 x

−

− +

8 1 1

4 2 2

3 3 2

 − +   − − + + 

   

   

8 1 1

6 2

3 3 2

− + − −

36 16 12 2 3 6

− + − − 27

= 6 9

= 2

Example

y = x

2 y = −x y = x

2 y = −x

If we try vertical strips, we

have to integrate in two parts:

dx

dx



₀^{2 x dx +}



₂^{4 x −}

(

^{x − 2}

)

^dx

We can find the same area using a horizontal strip.

dy

Since the width of the strip is

dy

, we find the length of the strip by solving for

x

ⁱⁿ

terms of

y

^.

y = x

y

2

= x

2 y = − x

2

y + = x

( )

2 2

0

y + 2 − y dy



^{2 3 2}

0

1 1

2 y + 2 y − 3 y 8

2 4 + − 3 10

= 3

General Strategy for Area Between Curves:

1

Decide on vertical or horizontal strips. (Pick

whichever is easier to write formulas for the length of the strip, and/or whichever will let you integrate fewer times.)

Sketch the curves.

2

3 Write an expression for the area of the strip.

(If the width is

dx

, the length must be in terms of

x

^.

If the width is

dy

, the length must be in terms of

y

^.

4 Find the limits of integration. (If using

dx

, the limits are x values; if using

dy

, the limits are y values.)

5 Integrate to find area.

Volumes

To find the volume of a solid S:

• Divide S into n “slabs” of equal width Δx (think of slicing a loaf of bread)

• Approximate the ith slab by a cylinder with base area A(x_i) and “height” Δx. The volume of the cylinder is A(x_i)Δx

• the sum of the cylinder areas is a good approximation for the volume of the solid

• the approximation is getting better as n→∞.

Let S be a solid that lies between x=a and x=b. If the cross-sectional area of S in the plane P_x, perpendicular to the x-axis, is A(x), where A is an integrable

function, then the volume of S is

 ^{ =} 

=

→ =



b i a

n

i

i x

dx x

A x

x A V

i

) ( )

(

1

* 0

max

lim

x

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 disks (flat circular cylinders) and add their volumes.

The volume of each disk is:

2

the thickness

 r 

In this case:

r=

^the

y

value of the function thickness

=

a small change in

x = dx

 ( ) ^x

²

^dx

→

Example of a disk

0 1 2

1 2 3 4

y = x

The volume of each flat cylinder (disk) is:

2

the thickness

 r 

If we add the volumes, we get:

( )

²

4

0



x dx



4

0



x dx

=



4 2

2 x 0

=



= 8



 ( ) ^x

²

^dx

The region between the curve , and the

y

-axis is revolved about the

y

-axis. Find the volume.

x 1

= y 1 y 4

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

2 4

1

V 1 dy

 ^ y ^

=  

 



volume of disk

4 1

1 dy

 y

=



4

ln y 1



= ^=

(

^{ln 4 ln1−}

) 0

₂

 ln 2

= = 2 ln 2

Example of rotating the region about y-axis

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:

.000574

2

.439 185

x = y − y +

x y

500 ft

( )

500 2 2

0

.000574 y .439 y 185 dy

  − +

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

24, 700, 000 ft

3



0 1 2 3 4

1 2

The region bounded by and is

revolved about the y-axis.

Find the volume.

y = x

2

y = 2 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:

(

^{ R2 −r2}

)

^thickness

(

^{R2 r2}

)

^dy

 −

outer radius

inner radius

2 y = x 2

y = x y = x2

y = x

y = x2

2 y = x

( )

^{2 2}

4

0 2

V ⁼



 ^ y ^{− }  y ^dy

4 2

0

1

V =  ^ y − 4 y ^dy

 



4 2

0

1

V = 



y − 4 y dy ^{2 3 4}

0

1 1

2 y 12 y

 ^{ }

=  − 

8 16

 ^ 3 ^

=  − 

8 3

= 

→

Example of a washer

0 1 2 3 4

1 2

2 y = x

If the same region is

rotated about the line

x

=

2

^:

y = x2

The outer radius is:

2 2 R = − y

R

The inner radius is:

2

r = − y

r

2 y = x 2

y = x y = x2

y = x

4 2 2

V =



0 R − r dy

( )

2 2

4

0 2 2

2

y y dy

 ^{ }

=  −  − −

 



( )

4 2

0 4 2 4 4

4

y y y y dy

 ^{ }

=  − + − − +

 



4 2

0 4 2 4 4

4

y y y y dy



=



− + − + −

4 1

2 2

0

3 1 4

y 4 y y dy



=



− + +

3 4

2 3 2

0

3 1 8

2 y 12 y 3 y

 ^{ }

=  − + + 

 

16 64 24 3 3

 ^{ }

=  − + + 

8 3

= 

Volumes of Solids of Revolution

The solids we considered are examples of solids of revolution because they are obtained by revolving a region about a line. In general, we

calculate the volume of a solid of revolution by using the basic defining formula

and we find the cross-sectional area A(x) or A(y) in one of the following ways:

• If the cross-section is a disk, we find the radius of the disk (in terms of x or y) and use

A = π(radius)²

• If the cross-section is a washer, we find the inner radius r_in and outer radius r_out and compute the area of the washer by subtracting the area of the inner disk from the area of the outer disk:

A = π(outer radius)² - π(inner radius)²



 ⁼

=

^d

c b

a

A x dx V A y dy

V ( ) or ( )