chapter 7 7.1 area between two curves 7.2 volumes by slicing; disks and washers 7.3 volumes by...
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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
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