Chapter 7

7.1 Area between Two Curves

7.2 Volumes by Slicing; Disks and Washers

7.3 Volumes by Cylindrical Shells

7.4 Length of a Plane Curve

7.5 Area of a Surface of Revolution

7.6 Average Value of a Function and its Appliations.

7.1 Area Between Two Curves

Area Formula

Here is a systematic procedure that you can follow to set up the area formula.

Example: Find the area of the region bounded above by y=x+6, bounded belowBy , and bounded on the sides by the lines x=0 and x=2. 2y x

Solution: from the graphs of y=x+6 and , we have 2y x

2 32 2 200

34 34[( 6) ] [ 6 ] 02 3 3 3x xA x x dx x

Example: Find the area of the region that is enclosed between and y=x+6.2y x

Sometimes the sides of the region will be points, when this occurs you will have toDetermine the points of intersection to obtain the limits of integration.

Solution: A sketch of the region shows that the lower boundary is and upperBoundary is y=x+6.

2y x

2

6y xy x

Which yields . Thus x=-2 and x=3. 2 6x x

2 33 2 322

125[( 6) ] [ 6 ]2 3 6x xA x x dx x

To find the endpoints, we equate

Second Area Problem

Area Formula

Section 7.2 Volumes By Slicing; Disks and Washers

Xa b

SCross Section

Cross section area = A (x)

Right Cylinder

A right cylinder is a solid that is generated when a plane region is translated along a line or axis that is perpendicular to the region.

If a right cylinder is generated y translating a region of area A through a distant h, then h is called the height of the cylinder, and the volume V of the cylinder is defined to be

V=Ah =[area of a cross section] X [height]

• To solve this problem, we begin by dividing the interval [a, b] into n subintervals, thereby dividing the solid into n slabs.

• If we assume that the width of the kth subinterval is xk, then the volume of the kth slab can be approximated by the volume A(xk

*)xk of a right cylinder of width (height) xk and cross-sectional area A(xk

*), where xk* is a point in

the kth subinterval.

• Adding these approximations yields the following Riemann sum that approximates the volume V:

• Taking the limit as n increases and the width of the subintervals approach zero yields the definite integral

*

1

( )n

k kk

V A x x

*

max 0 1

lim ( ) ( )k

n b

k k ax k

V A x x A x dx

Volumes By Slicing

Volume Formula

The Volume of a solid can be obtained by integrating the cross-sectional area from one end of the solid to the other.

Solids of Revolution

A solid of revolution is a solid that is generated by revolving a plane region abouta line that lies in the same plane as the region; the line is called the axis of revolution.

Volumes by Disks Perpendicular to the x-Axis

We can solve this problem by slicing. Observe that the cross section of the solid taken perpendicular to the x-axis at the point x is a circular disk of radius f(x). The area of the region is

A(x)= [f (x)]2

Thus the volume of the solid is

Because the cross sections are disk shaped, the application of this formula is called the method of disks.

Examples

Example: Find the volume of the solid that is obtained when the region under the curve over the interval [1, 4] is revolved about the x-axis.

Solution. The volume is

y x

4242

11

15[ ( )] 82 2 2

b

a

xV f x dx xdx

Volumes by Washers Perpendicular to the x-Axis

Not all solids of revolution have solid interiors; some have holes or channels that Create interior surfaces. Thus, we will be interested in the following problem.

Volumes by Washers Perpendicular to the x-Axis

We can solve this problem by slicing. Observe that the cross section of the solid taken perpendicular to the x-axis at the point x is a “washer-shaped” region with inner radius g(x) and outer radius f(x); hence its area is A(x)=[f(x)]2 -[g(x)]2= ([f(x)]2 -[g(x)]2)

Thus the volume of the solid is

Examples

Find the volume of the solid generated when the region between the graphs of the equations f(x)=1/2+x2 and g(x)=x over the interval [0, 2] is revolved about the x-axis.

Solution: The volume is

22 2 2 2 2

0

252 4

00

1([ ( )] [ ( )] ) ([ ] )2

1 69( ) ( )4 4 5 10

b

aV f x g x dx x x dx

x xx dx

Volumes by Disks and Washers Perpendicular to the y-Axis

The methods of disks and washers have analogs for regions that are revolved About the axis.

Section 7.3 Volumes by Cylindrical Shells

Volume of Cylindrical Shells

A cylindrical shell is a solid enclosed by two concentric right circular cylinders.The volume V of a cylindrical shell with inner radius r1, outer radius r2, and height h can be written as

V=2 [1/2(r1+r2) ] h (r2-r1)So V=2 [ average radius ] [height] [thickness]

• The idea is to divide the interval [a, b] into n subintervals, thereby subdividing the region R into n strips, R1, R2,, …, Rn.

• When the region R is revolved about the y-axis, these strips generate “tube-like” solids S1, S2, …, Sn that are nested one inside the other and together comprise the entire solid S.

• Thus the volume V of the solid can be obtained by adding together the volumes of the tubes; that is

V=V(S1)+V(S2)+…+V(Sn).

Method of Cylindrical Shells

Suppose that the kth strip extends from xk-1 to xk and that the width of the strip is xk. If we let xk* be the midpoint of the interval [xk-1, xk], and if we construct a rectangle of height f(xk*) over the interval, then revolving this rectangle about the y-axis produces a cylindrical shell of average radius xk*, height f(xk*), and thickness xk.

Then the volume Vk of this cylindrical shell is

Vk=2xk*f(xk*) xk

* *

1

2 ( )n

k k kk

V x f x x

Hence, we have

* *

max 0 1

lim 2 ( ) 2 ( )k

n b

k k k ax k

V x f x x xf x dx

Method of Cylindrical Shells

Volume by Cylindrical Shells about the y-axis

Let f be continuous and nonnegative on [a, b] (0a<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. Then the volume V of the solid of revolution that is generated by revolving the region R about the y-axis is given by

2 ( )b

aV xf x dx

Examples

Example. Use cylindrical shells to find the volume of the solid generated when the region enclosed between , and the x-axis is revolved about the y-axis.

Solution: Since , a=1, and b=4,

, 1, 4y x x x

( )f x x

4

1

4 3/2

1

45/2

1

2 ( ) 2

2

2 12425 5

b

aV xf x dx x xdx

x dx

x