微积分教学资料-chapter 5 25页PPT文档

Solution
By solving the system of equations
y 2 2 x We find that the points of intersection

y

x

4
are(2, -2) and (8, 4).So
y
yx4
A 4(y41y2)dy
2
oa x b x
2.The volume of the solid obtained by rotating
the region bounded by yf(x)y , g(x),f(x)g(x)
xa, and x b, about the xaxis
The cross-sectional area is
2
y2 2x
(8,4)
(y2 4yy3)
4 18
o
x
2
6 2
(2,2)
or
2
A20 2xdx
8
2[ 2x(x4)]dx
y
yx4
4 y2 2x
(8,4)
o
x
-2 (2,2)
18
Volumes
Definition of volume Let s be a solid that lies between
y
x2 y2 16
4
V
1 (16x2)dx
42 3
21316x13x344
x
128 33
the volume of a solid of revolution Solids of revolution:
、
1. The volume of the solid obtained by rotating
202[x22(y)x12(y)d ] y
22 [3 (4 y 2)2 (3 4 y 2)2 ]d 0
24 2 4y2dy 0
242
Example The region enclosed by the curves
yxandyx2is rotated about the x-axis. Find
x=1 ,its volume is
V 1A (x )d x1(x 4 5 x 2 4 x )d x 8
0
0
1 5
谢谢！
xin [a,b] is
b
Aa[f(x)g(x)]dx
The Riemann sum
n
[f (xi)g(xi)]x
i1
Therefore
n
A
lim
n
i1
[
f
(xi)

g(xi)]x
b
a [ f (x)g(x)]dx
Example 1 Find the area of the region bounded Above by y e x bounded below by yx
1
A ( 0
xx2)dx
[32
3
x2
x33]
1

0
1 3
or
1
A ( 0
yy2)dy
[32
3
y2
y33]
1 0

1 3
Example 3 Find the area enclosed by the curves
y 2 2 xa n d b y th elin ey x 4
Solution
1
V 0 A ( x )d x
1
0 xdx

x2 2
1 0

2
Example Find the volume of the solid obtained by rotating about the y axis the region bounded by yx3,y8andx0.
2
Solution y sin x

A 2 sinxcosxdx 0
4 (cos x sin x )dx 0

ycosx

2
(sin
x

cos
x)dx
0

4
42


[sinxcosx]0 4[cosxsinx]2
4
2 22
The area A of the region bounded by the curves
b
Aa f(x)g(x)dx
c
b
a(g(x)f(x))dxc(f(x)g(x))dx
y
0a
y f (x)
y g(x)
c
bx
Ebxyamthpelecu2rveFsindyth esin arxe,y a oc f o ts hx e,x re g0 ioa nn d bx ou nd.ed
2. the region bounded by yf(x), y 0 xa,
and x b, about the xaxis
The cross-sectional area is
A (x) [ra d iu s]2 [f(x)]2 y yf(x)
V b [ f ( x)]2 dx a b [ f ( x)]2dx a
then the volume of s is
Vlni m i n1A(xi*)xabA(x)dx A(x)
oax
bx
1) Partition:
分点为：a x 0 x 1 x 2 x i 1 x i x n b ,
xxi xi 1bn -a(i1 ,2 , n).
x(y),x(y) And the lines yc,yd, where
and are continuous and (y)(y) for all
y in [ c , d ] is
Acd[(y)(y)]dy
y
d
x(y)
x(y)
c
o
x
Example 3 Find the area enclosed by the curves
xaandxb. If the cross-sectional area of s in
the plane p x Through x and perpendicular to the
x a x i s , i s A ( x ) , w h e r e A i s a c o n t i n u o u s f u n c t i o n ,
and bounded on the sides by x0anxd1
Solution
y
y ex
yx
0
x
A 1(ex x)dx 0

(ex

1 2
x2 )
1 0
e1.5
The area A between the curves yf(x)a n dyg(x) And between xaandxbis
Chapter 5 Applications of integrals
Areas between curves
y
y f(x)
A
yg(x)
a
b
x
The area A of the region bounded by the curves yf(x)y ,g(x) And the lines xa,xb, where f and g are continuous and f(x)g(x) for all
the volume of the resulting solid.
Solution
The cross-sectional area is
y2
A(x)(x22)2(x2)2
y x (1,1)
(x45x24x)

(0,0)
y x2
The solid lies between x=0 and
rotating the region bounded by (x3)2y24
about the y-axis
Leabharlann Baidu
y 2
x2 3 4y2
Solution
o1 3
x
A (y)[x 1(y)]2[x2(y)]2
-2
x1 3 4y2
V 22[x22(y)x12(y)]dy
A(x)[outradius]2[innerradius]2
[f(x)]2[g(x)]2 y
y f(x)
A
yg(x)
Vb[(f(x))2(g(x))2]dx a a
b
3. V d[(y)]2dy c
y d
x(y)
c
o
x
Example Find the volume of the solid obtained by rotating about the x-axis the region under the curve y x from 0 to 1.

(0,0)
y x2
x=1 ,its volume is
V1A (x)d x1(x2x4)d x2
0
0
1 5
Example The region enclosed by the curves
yxandyx2is rotated about the y=2.Find
Solution
The cross-sectional area is
2
A(y)x2(3y)2y3
The solid lies between y=0 and y=8 ,its volume is
V8A(y)dy8y2 3dy96
0
0
5
Example Find the volume of the solid obtained by
the volume of the resulting solid.
Solution
The cross-sectional area is
A (x )(x )2 (x 2 )2(x 2 x 4 )
y x (1,1)
The solid lies between x=0 and
b
a A(x)dx
Example A wedge is cut out of a circular cylinder of radius 4 by two planes.0ne plane is perpendicular to the axis of the cylinder.the other intersects the first at
A(
x
* i
)
2) Approximation:
V i A(xi*)x
xi*[xi1,xi],
x
* i
(i1,2, ,n)
oa
x i 1
xi b x
3) Sum:
n
n
V Vi A(xi*)xi
i 1
i1
4) Limit:
n
V

lim
n
i1
A(xi*)x
an angle of 3 0 along a diameter of the cylinder.find the volume of the wedge.
Solution
The cross-sectional area is
A (x)116x2 16x2tan30 2
4 30
o
x 30
4
y2xandyx2
Solution
y
y2 x (1,1)
By solving the system of equations
y x2
o
1
x
y2 x
y

x2
We find that the points of intersection are(0, 0) and (1, 1).So