
Integral
calculus 

Applications
of the definite integral

The
volume
of a solid of revolution

The
volume
of a sphere

The
volume
of a spherical segment

The
volume
of a cone







The
volume
of a solid of revolution

A solid figure generated by revolving
a given curve around an axis of revolution is called the solid of
revolution.

The volume generated by the region of
the coordinate plane bounded by the segment of a curve y
= f (x)
between x
= a, y
= b and the xaxis, revolving around
the xaxis,
is shown in the left figure below.

Note that the infinitesimal volume of
the cylinder representing an element of the integration,

dV = area
´
width = py^{2}
´
dx.


The volume generated by the segment of
a curve x = g
(y) between y
= c and y
= d, revolving around
the
yaxis,
is shown in the right figure above.


The
volume
of a sphere

Example:
Find the volume of a
sphere generated by a semicircle 

revolving around
the 

xaxis.

Solution: Since the endpoints of the
diameter, lying 
on the
xaxis,
are 
r and r,
then






The
volume
of a spherical segment

Example:
Find the volume of a
spherical segment generated by the portion of the right semicircle
between y = a
and y = a
+ h, revolving around
the yaxis,
as
is shown in the below figure.

Solution: Since the right semicircle
equation 

then



or,
by substituting r_{1}^{2}
= r^{2} 
a^{2} and
r_{2}^{2} =
r^{2} 
(a + h)^{2} 





The
volume
of a cone

Example: Find the volume of a
right circular cone generated by the line (segment) passing through
the origin and the point (h,
r), where h
denotes the height of the cone and r
is the radius of its base, revolving around the xaxis,
as shows the below figure.

Solution: The equation of the
generating line



Example:
Find the volume of
a solid of revolution generated by a plane bounded by the segment of a curve
y
=
 x^{2}
+ 3x and the xaxis,
revolving around
the xaxis,
as shows the below figure.

Solution: The limits of the integration
 x^{2}
+ 3x
= 0,

x(x
+ 3)
= 0, x_{1}
= 0 and x_{2}
= 3
then,






Example:
Find the volume of
a solid of revolution generated by one cycle of the cycloid x =
r (t 
sin t),
y
= r (1 
cos t)
and the xaxis,
revolving around
the xaxis,
as shows the below figure.


Solution: Since y^{2}
= r^{2}(1 
cos t)^{2},
dx = r (1

cos t)dt
the limits of the integration
0 <
t <
2p,
then



Example:
Find the volume of
a solid of revolution generated by the arc of the sinusoid y
= sin x between
x = 0 and x
= p/2,
revolving around
the yaxis,
as shows the below figure.



Solution: Since curve
rotates around
the yaxis,
we should apply the inverse of the sine, i.e., we use
x
= g (y) form or x
= arcsin y = sin^{1}y.
Thus, the limits of the integration,
for x_{1}
= 0, y_{1}
= 0 and for x_{2}
= p/2, y_{2}
= 1 then, applying
the integration by parts twice










Functions
contents G 



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