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x 1 A ( x )= x 1
x 1
A(x)=
x 1
(
)
2
= (x 1)
V =
2
5
A(x)dx=
2
5
(x 1)dx=
1
2 x
2
x
5
2
=
25
2
5 4
2 +2
= 15
2
y
A(y)=
y
(
)
2
V =
0
4
A(y)dy=
0
4
y
(
)
2
dy=
0
4
ydy=
1
2 y
2
4
0
=8
4. A cross section is circular with radius
, so its area is
.
5. A cross section is a disk with radius
, so its area is
.
1
Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.2 Volumes y=x
2/3
x=y
3/2
y
3/2
1
A(y)= (1)
2
y
3/2
(
)
2
=
1 y
3
(
)
V =
0
1
A(y)dy=
0
1
1 y
3
(
)
dy=
y 1
4 y
4
1
0
= 3
4
1
x
1 x
A(x)= (1 x)
2
1
x
(
)
2
=
1 2<i>x+x
2
(
)
1 2 x +x
(
)
=
3<i>x+x
2
+2 x
(
)
V =
0
1
A(x)dx=
0
1
3<i>x+x
2
+2 x
(
)
dx =
3
2 x
2
+ 1
3 x
3
+ 4
3 x
3/2
1
0
=
3
2 +
5
3
= 6
10.
, so a cross section is a washer with inner radius
and outer radius , and its
area is
.
11. A cross section is a washer with inner radius
and outer radius
, so its area is
.
2
Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.2 Volumes R
1
OC
x=0
V =
0
1
A(y)dy=
0
1
(1)
2
3
y
(
)
2
dy=
0
1
(1 y
2/3
)dy=
y 3
5 y
5/3
1
0
=
1 3
5
= 25
R
1
AB
x=1
V =
0
1
A(y)dy=
1
0
1
3
y
(
)
2
dy=
0
1
(1 2<i>y
1/3
+y
2/3
)dy
=
y 3
2 y
4/3
+ 3
5 y
5/3
1
0
=
1 3
2 +
3
5
= 10
R
1
BC
y=1
V =
0
1
A(x)dx=
0
1
(1)
2
(1 x
3
)
2
dx=
0
1
1 (1 2<i>x
3
+x
6
) dx
=
0
1
(2<i>x
3
x
6
)dx=
1
2 x
4
1
7 x
7
1
0
=
1
2
1
7
= 514
V =
0
/4
(1 tan
3
x)
2
dx
20.
about
(the line
):
21.
about
(the line
):
22.
about
(the line
):
31.
3
Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.2 Volumes y=(x 2)
4
8<i>x y=16
(x 2)
4
=8<i>x 16=8(x 2)
(x 2)
4
8(x 2)=0
(x 2) (x 2)
3
8 =0
x 2=0
x 2=2
x=2
4.y=(x 2)
4
x 2=
4
y
x=2+
4
y
x
2
8<i>x y=16
8<i>x=y+16
x= 1
8 y+2
V =
0
16
10
1
8 y+2
2
10
2+
4
y
(
)
2
{
}
dy
V =
0
10
A(x)dx M
5
= 10 0
5
[ A(1)+ A(3)+ A(5)+ A(7)+ A(9)]
=2(0.65+0.61+0.59+0.55+0.50)=2(2.90)=5.80m
3
x
2
+y
2
=r
2
x
2
=r
2
y
2
V =
r h
r
r
2
y
2
(
)
dy=
r
2
y y
3
3
r
r h
=
r
3
r
3
3
r
2
(r h)
r h
(
)
3
3
{
}
=
2
3 r
3
1
3 (r h) 3<i>r
2
r h
(
)
2
{
}
= 1
3
2<i>r
3
(r h) 3<i>r
2
r
2
2<i>rh+h
2
(
)
{
}
= 1
3
2<i>r
3
r h
(
)
2<i>r
2
+2<i>rh h
2
{
}
= 1
3
2<i>r
3
2<i>r
3
2<i>r
2
h+rh
2
+2<i>r
2
h+2<i>rh
2
h
3
(
)
32.
and
intersect when
or
or
[ since
].
.
46.
49.
4
Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.2 Volumes = 1
3
3<i>rh
2
h
3
(
)
= 1
3
h
2
(3<i>r h)
h
2
r h
3
y
(y)=3 3
2 y.
S
V =
0
2
A(y)dy=
0
2
2
y
( )
2
2
dy= 8
0
2
3 3
2 y
2
dy
= 8
3
0
u
2
2
3
du
u=3 3
2 y,du=
3
2 dy
=
12
1
3 u
3
0
3
= 12 ( 9)= 34
S
x
S
PQ
2
=r
2
x
2
A(x)=4 r
2
x
2
(
)
S
V =
r
r
A(x)dx=4
r
r
r
2
x
2
(
)
dx
= 8
0
r
r
2
x
2
(
)
dx=8
r
2
x 1
3 x
3
r
0
= 16
3 r
3
, or, equivalently,
58. A typical cross section perpendicular to the axis in the base has length
This
length is the diameter of a cross sectional semicircle in , so
64. Each cross section of the solid in a plane perpendicular to the
axis is a square (since the
edges of the cut lie on the cylinders, which are perpendicular). One quarter of this square and one
eighth of are shown. The area of this quarter square is
. Therefore,
and the volume of is
5
Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.2 Volumes y=R cx
2
y
x
y
x= 1
2 h
R c
1
2 h
2
=R d=r
y
x>0.
V = 2
0
h/2
y
2
dx=2
0
h/2
R cx
2
(
)
2
dx=2
R
2
x 2
3 Rcx
3
+ 1
5 c
2
x
5
h/2
0
= 2
1
2 R
2
h
1
12 Rch
3
+ 1
160 c
2
h
5
V = 1
3
h
2<i>R
2
+
R
2
1
2 Rch
2
+ 3
80 c
2
h
4
.
R
2
1
2 Rch
2
+ 3
80 c
2
h
4
=
R 1
4 ch
2
2
1
40 c
2
h
4
= R d
(
)
2
2
5
1
4 ch
2
2
=r
2
2
5 d
2
.
V
V = 1
3
h
2<i>R
2
+r
2
2
5 d
2
69. (a) The radius of the barrel is the same at each end by symmetry, since the function
is
even. Since the barrel is obtained by rotating the graph of the function about the
axis, this radius
is equal to the value of at
, which is
.
(b)
The barrel is symmetric about the
axis, so its volume is twice the volume of that part of the
barrel for
Also, the barrel is a volume of rotation, so
Trying to make this look more like the expression we want, we rewrite it as
But
Substituting this back into , we see that
, as required.
6
Stewart Calculus ET 5e 0534393217;6. Application of Integration; 6.2 Volumes