
Integral
calculus 

Applications
of the definite integral

The
area of a region in the plane

The
area between the graph of a curve and the coordinate axis

The
area between the graph of a curve and the coordinate axis examples

The
area bounded by a parametric curve

The
area of the sector of a curve in polar coordinates

The
area of the sector of a curve given in Cartesian (or rectangular) coordinates

The
area of the sector of a parametric curve







The
area of a region in the plane

The
area between the graph of f (x) and the xaxis

If
given a continuous nonnegative function f
defined over an interval [a,
b] then, the area A
enclosed by the
curve y
= f (x),
the vertical lines, x
= a and x
= b and the xaxis,
is defined as



The
area between the graph of a curve and the coordinate axis examples

Example:
Find the area between
the graph of f
(x) = 
(1/3)x^{3} + 3x
and the xaxis
over the interval defined by two nonnegative successive
roots of the given cubic.

Solution: To
find roots we should solve f
(x) = 0,



Example:
Find the area between
the curve x = 
y^{2} + y
+ 2 and the yaxis.

Solution: Since
given curve is the parabola whose axis of symmetry is
parallel to the xaxis
we first calculate its
yintercepts
by setting x =
0 to determine the limits of
integration,



Thus,
the area 





Example:
Find the area of the
ellipse b^{2}x^{2} +
a^{2}y^{2} = a^{2}b^{2 }
that is symmetrical about
the coordinate axes
and that intersects the coordinate axes at the
points (+ a,
0) and (0,
+
b).

Solution: Let
write the ellipse in the explicit form y
= f (x)


Thus,
we calculate the area A
using the solution of 
the
indefinite integral example 48. that is 




Hence, the area of the ellipse is abp,
and by setting b
= a we get the area of
the circle
a^{2}p.


Example:
Find the area A
enclosed
by the parabola, the line and the xaxis
shown in the below figure.

Solution: First we should find
the equation of the parabola and the line using given points.

The parabola is translated in the
direction of the xaxis
by x_{0}
= 1, so
that


and the line 

Thus,
the area 





The
area bounded by a parametric curve

If
a curve is given by the parametric equations x
= f (t) and
y = g (t), then
the area enclosed by the curve,
the vertical lines, x
= a and x
= b and the xaxis,
we
obtain from


where,
f (t_{1}) = a
and f (t_{2})
=
b,
g (t)
>
0
inside
[t_{1},
t_{2}].


Example:
Find the area bounded
by the
ellipse x = a
cos t,
y = b
sin t,
( 0 <
t <
2p
).

Solution:
As the ellipse is symmetrical about
the coordinate axes we will calculate quarter of the area.

By substituting x
= 0 and x
= a into the
equation x = a
cos t and
solving for
t,
we get the limits of integration
t_{1} =
p/2
and t_{2} =
0
respectively.

Therefore,


so that, the area of the ellipse
A = abp.


The
area in polar coordinates

Polar coordinates (
r, q )
locate a point in a plane by means of the length r,
of the line joining the point to the origin or pole
O,
and angle q
swept out by that line from the polar axis.

Recall the area A
of a circular sector bounded by two radii and an arc, is defined
by the proportion

A : r^{2}
p
= q
: 2p
thus, A
= (1/2) r^{2}q

where r
is the length of the radius and q
is the central angle subtended by the arc.

Suppose given a curve in polar
coordinates by r
= f (q
)
or
r
= r
(q
)
then, the area of the region bounded by the curve and the radii that correspond to
q
_{1} = a
and q
_{2} = b,
is given by



Example:
Find the area of the
region enclosed by the lemniscate of Bernoulli whose polar equation
is
r^{2} = a^{2}
cos2q,
shown in the below figure.

Solution: As the
lemniscate consists of two symmetrical loops meeting at a node, we will calculate quarter
of the
area lying inside 0
<
q
<
p/4.


Thus,
the area of the lemniscate A
= a^{2}. 




The
area of the sector of a curve given in Cartesian or rectangular coordinates

We
use the formulas for conversion between Cartesian and polar
coordinates,


to find the area of the sector bounded by
two radii and the
arc P_{0}P,
of a curve y
= f (x) given in the

Cartesian coordinates, where x
is the value inside the interval [a,
b]. Then


therefore,


is the area of the sector bounded by
two radii and the arc P_{0}P
of a curve y
= f (x).


The
area of the sector of a parametric curve

The area of the sector bounded by two
radii and the arc P_{0}P
of a parametric curve is given by


where the parametric values, t_{0}
and t
relate to the endpoints, P_{0}
and P
of the arc, respectively.


Example:
Find the area of the
sector of the rectangular hyperbola x^{2}

y^{2 }= 1
enclosed by the
xaxis, the arc of the right branch of the
hyperbola between the vertex and the point
P(cosh
t,
sinh
t), and the
line joining this point to the origin, as shown in
the below figure.

Solution: Since the hyperbolic functions satisfy the identity
cosh^{2}
t

sinh^{2 }t^{
}= 1 then, the point
x = cosh
t and
y = sinh
t
that lies on the
hyperbola x^{2}

y^{2 }= 1,
describes the right branch of the
hyperbola as parameter
t increases from

oo
to +
oo.

The point traces the hyperbola passing from negative infinity, through
P_{0}
(t = 0) to positive
infinity.

As,
x'
(t) = sinh
t
and
y' (t) = cosh
t, 
so
that x(t)
y' (t) = cosh^{2}
t
and
y(t) x' (t) = sinh^{2 }t, 
and
by using the above formula for the area of the
sector 
of
a parametric curve 

we
get the area of the sector P_{0}OP
of the rectangular hyperbola. 



Therefore, the parameter (or argument)
t,
of hyperbolic functions, is twice the area of the sector P_{0}OP,
i.e., t = 2A.
That is why the inverse hyperbolic functions are also called the
area functions.












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