What is the volume of the solid produced by revolving f(x)=x2+3x−√x,x∈[0,3]around the x-axis?
Answers
anation:
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We have drawn the given expression
f
(
x
)
=
sin
x
, for
x
=
0
to
x
=
π
. When this expression is revolved around
x
axis through
360
∘
we have solids of revolution. At each point located on the graph, the cross-section of the solid, parallel to the
y
-axis, will be a circle of radius
y
.
Let us consider a thin disc of thickness
δ
x
, at a distance
x
from the origin, with its face nearest to the
y
-axis. The radius of the other circular face of the disc will be
y
+
δ
y
.
The disc is not a perfect cylinder. It will become one when
δ
x
, and hence
δ
y
→
0
.
Thus we approximate the disc with a cylinder of thickness,
δ
x
and radius
y
.
The volume
δ
V
of the disc is then given by the volume of a cylinder,
π
r
2
h
, so that
δ
V
=
π
y
2
δ
x
.
We have ignored the volume of part of the disc which is
∝
(
δ
x
×
δ
y
)
, both being infinitesimally small.
So the volume
V
of the solid of revolution is given by the integral of volume expression over the limits of interest as both
δ
x
→
0
and
δ
y
→
0
.
V
=
∫
b
a
π
y
2
d
x
Inserting values given in the problem we obtain
V
=
∫
π
0
π
(
sin
x
)
2
.
d
x
or
V
=
∫
π
0
π
sin
2
x
.
d
x
.....(1)
-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.
To find the
∫
sin
2
x
.
d
x
From trigonometric identity
sin
2
x
=
1
2
(
1
−
cos
2
x
)
∴
we need to find
∫
1
2
(
1
−
cos
2
x
)
.
d
x
⇒
∫
1
2
d
x
−
1
2
∫
cos
2
x
.
d
x
Integrating first term and for the second term
let
u
=
2
x
. Differentiating both sides and inserting in the equation
d
u
=
2
d
x
⇒
x
2
−
1
4
∫
cos
u
.
d
u
⇒
x
2
−
1
4
∫
cos
u
.
d
u
⇒
x
2
−
1
4
sin
u
, substituting back to
x
⇒
x
2
−
1
4
sin
2
x
We have ignored the constant of integration as we are eventually dealing with definite integral.
-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-.-
Substituting the value of integral in (1)
V
=
π
[
x
2
−
1
4
sin
2
x
]
π
0
V
=
π
[
(
π
2
−
1
4
sin
2
π
)
−
(
0
2
−
1
4
sin
(
2
×
0
)
)
]
, inserting the value of
sin
2
π
and
sin
0
,
both
=
0
V
=
π
2
2