X=y^3 /3 , find the area of surface of revolution???
Answers
A surface of revolution is obtained when a curve is rotated about an axis.
We consider two cases – revolving about the
x
−
axis and revolving about the
y
−
axis.
Revolving about the
x
−
axis
Suppose that
y
(
x
)
,
y
(
t
)
,
and
y
(
θ
)
are smooth non-negative functions on the given interval.
If the curve
y
=
f
(
x
)
,
a
≤
x
≤
b
is rotated about the
x
−
axis, then the surface area is given by
A
=
2
π
b
∫
a
f
(
x
)
√
1
+
[
f
′
(
x
)
]
2
d
x
.
Surface of revolution obtained by rotating the curve y=f(x) on the interval [a,b] around the x-axis.
Figure 1.
If the curve is described by the function
x
=
g
(
y
)
,
c
≤
y
≤
d
,
and rotated about the
x
−
axis, then the area of the surface of revolution is given by
A
=
2
π
d
∫
c
y
√
1
+
[
g
′
(
y
)
]
2
d
y
.
Surface of revolution obtained by rotating the curve x=g(y) on the interval [c,d] around the x-axis.
Figure 2.
If the curve defined by the parametric equations
x
=
x
(
t
)
,
y
=
y
(
t
)
,
with
t
ranging over some interval
[
α
,
β
]
,
is rotated about the
x
−
axis, then the surface area is given by the following integral (provided that
y
(
t
)
is never negative)
A
=
2
π
β
∫
α
y
(
t
)
√
[
x
′
(
t
)
]
2
+
[
y
′
(
t
)
]
2
d
t
.
Surface obtained by rotating the parametric curve x=x(t), y=y(t) around the x-axis.
Figure 3.
If the curve defined by polar equation
r
=
r
(
θ
)
,
with
θ
ranging over some interval
[
α
,
β
]
,
is rotated about the polar axis, then the area of the resulting surface is given by the following formula (provided that
y
=
r
sin
θ
is never negative)
A
=
2
π
β
∫
α
r
(
θ
)
sin
θ
√
[
r
(
θ
)
]
2
+
[
r
′
(
θ
)
]
2
d
θ
.
Surface obtained by rotating the polar curve r=r(theta) about the x-axis.
Figure 4.
Revolving about the
y
−
axis
The functions
g
(
y
)
,
x
(
t
)
,
and
x
(
θ
)
are supposed to be smooth and non-negative on the given interval.
If the curve
y
=
f
(
x
)
,
a
≤
x
≤
b
is rotated about the
y
−
axis, then the surface area is given by
A
=
2
π
b
∫
a
x
√
1
+
[
f
′
(
x
)
]
2
d
x
.
Surface obtained by rotating the curve y=f(x) around the y-axis.
Figure 5.
If the curve is described by the function
x
=
g
(
y
)
,
c
≤
y
≤
d
,
and rotated about the
y
−
axis, then the area of the surface of revolution is given by
A
=
2
π
d
∫
c
g
(
y
)
√
1
+
[
g
′
(
y
)
]
2
d
y
.
Surface obtained by rotating the curve x=g(y) around the y-axis.
Figure 6.
If the curve defined by the parametric equations
x
=
x
(
t
)
,
y
=
y
(
t
)
,
with
t
ranging over some interval
[
α
,
β
]
,
is rotated about the
y
−
axis, then the surface area is given by the integral (provided that
x
(
t
)
is never negative)
A
=
2
π
β
∫
α
x
(
t
)
√
[
x
′
(
t
)
]
2
+
[
y
′
(
t
)
]
2
d
t
.
Surface obtained by rotating the parametric curve x=x(t), y=y(t) about the y-axis.
Figure 7.
If the curve defined by polar equation
r
=
r
(
θ
)
,
with
θ
ranging over some interval
[
α
,
β
]
,
is rotated about the
y
−
axis, then the area of the resulting surface is given by the formula (provided that
x
=
r
cos
θ
is never negative)
A
=
2
π
β
∫
α
r
(
θ
)
cos
θ
√
[
r
(
θ
)
]
2
+
[
r
′
(
θ
)
]
2
d
θ
Surface obtained by rotating the polar curve r=r(theta) about the y-axis.
Figure 8.
I hope you are understand my solution