In the present section, we apply the approach for finding areas of plane regions bounded by plane curves.

**Evaluation of a
Bounded Plane Area by Integration**

In the
beginning of this chapter, we have already introduced definite integral by a
geometrical approach. In that approach, we have noted that, whenever the
integrand of the definite integral is non-negative, the definite integral
yields the geometrical area. In the present section, we apply the approach for
finding areas of plane regions bounded by plane curves.

** **

**Case
(i)**

Let *y* = *f*
( *x* ), *a* â‰¤ *x* â‰¤
*b* be the equation of the portion of
the continuous curve that lies above the *x
*âˆ’* *axis (that is, the portion lies either in the first quadrant
or in the second quadrant) between the lines *x* = *a* and *x* = *b*
. See Fig.9.8. Then, *y* â‰¥
0 for every point of the portion of the curve. Consider the region bounded by
the curve, *x *âˆ’* *axis, the ordinates* x *=* a *and* x *=* b *. It is important to note that *y *does not change its sign in the
region. Then, the area* A *of the
region is found as follows:

Viewing
in the positive direction of the *y* âˆ’
axis, divide the region into elementary vertical strips (thin rectangles) of
height *y* and width Î”*x* **.** Then, *A* is the limit sum of the areas of the
vertical strips.

Hence,
we get

**Case
(ii)**

Let *y* = *f*
( *x* ), *a* â‰¤ *x* â‰¤
*b* be the equation of the portion of the
continuous curve that lies below the *x*
âˆ’
axis (that is, the portion lies either
in the third quadrant or in the fourth quadrant).
Then, *y* â‰¤
0 for every point of the portion of the curve. It is important to note that *y* does not change its sign in the region. Consider the region bounded by the curve,
*x
*âˆ’* *axis, the ordinates* x *=* a *and* x *=* b *. See Fig.9.9. Then, the area
*A* of the region is found as follows:

Viewing
in the negative direction of the *y* âˆ’
axis, divide the region into elementary vertical strips (thin rectangles) of
height |y| = âˆ’ *y* and width Î”*x* **.** Then, *A* is the limit of the sum of the areas
of the vertical strips. Hence, we get

A = lim âˆ‘_{aâ‰¤xâ‰¤b}
âˆ’yÎ”*x *= âˆ’^{b}âˆ«_{a} y*dx* = | ^{b}âˆ«_{a} *ydx* | .

**Case
(iii)**

Let *y* = *f*
( *x* ), *a* â‰¤ *x* â‰¤
*b* be the equation of the portion of
the continuous curve that lies above as well as below the *x* âˆ’ axis (that is, the portion may lie in all quadrants).

Draw the
graph of *y *=* f *(*
x*)* *in the *XY *âˆ’* *plane. The graph lies alternately above
and below the *x* âˆ’
axis and it is intercepted between the ordinates *x *=* a *and* x *=* b*.

Divide
the interval[*a* , *b*] into subintervals [*a* ,
*c*_{1} ] , [*c*_{1} , *c*_{2} ] , . . . , [*c _{k}*
,

Hence
the geometrical area of the region bounded by the graph of *y* = *f* ( *x*) , the *x*-axis, the lines *x *=* a *and* x *=* b *is given by

For
instance, consider the shaded region in Fig. 9.10. Here *A*_{1} , *A*_{2}
, *A*_{3} , and *A*_{4 }denote geometric_{ }areas
of the individual parts. Then, the total area is given by_{}

** **

**Case
(iv)**

Let *x* = *f*
( *y* ), *c* â‰¤ *y* â‰¤
*d* be the equation of the portion of the
continuous curve that lies to the right side of *y
*âˆ’* *axis (that is, the portion lies either in the first quadrant
or in the fourth quadrant). Then, *x *â‰¥* *0 for every point of the portion of the curve. It is
important to note that *x *does not
change its sign in the region.

Consider
the region bounded by the curve, *y *âˆ’* *axis, the lines *y *=* c *and* y *=* d *. The region is sketched as in Fig.
9.11. Then, the area *A *of the region
is found as follows:

Viewing
in the positive direction of the *x* âˆ’
axis, divide the region into thin horizontal strips (thin rectangles) of length
*x* and width Î”*y* **.** Then, *A* is the limit of the sum of the areas
of the horizontal strips. Hence, we get *A*
=
lim âˆ‘_{câ‰¤yâ‰¤d} *xÎ” y* =
* ^{d}*âˆ«

**Case
(v)**

Let *x* = *f*
( *y* ), *c* â‰¤ *y* â‰¤
*d* be the equation of the portion of the
continuous curve that lies to the left side of *y* âˆ’ axis (that is, the portion lies either in the second
quadrant or in the third quadrant). Then, *x*
â‰¤
0 for every point of the portion of the curve. It is important to note that *x* does not change its sign in the region. Consider the region bounded by
the curve, *y *âˆ’* *axis, the lines* y *=* c *and *y *=* d *. The region is sketched as in Fig.
9.12. Then, the area *A* of the region
is found as follows:

Viewing
in the positive direction of the *x* âˆ’
axis, divide the region into thin horizontal strips (thin rectangles) of length
|*x*| = âˆ’*x* and width Î”*y* **.** Then, *A* is the limit of the sum of the areas
of the horizontal strips.

**Case
(vi)**

Let *x* = *f*
( *y* ), *c* â‰¤ *y* â‰¤
*d* be the equation of the portion of
the continuous curve that lies to the right as well as to the left of the *y* âˆ’ axis (that is, the portion may lie
in all quadrants). Draw the graph of *x*
=
*f* ( *y*) in the *XY* âˆ’
plane. The graph lies alternately to the right and to the left of the *y* âˆ’ axis and it is intercepted between
the lines *y* =
*c* and *y* = *d* . Divide the
interval [*c* , *d* ] into subintervals [*c*
, *a*_{1} ] , [*a*_{1} , *a*_{2} ] , . . . , [*a _{k}*
,

Hence the
geometrical area A of the region bounded by the graph of *x* =* f *( y) , the y-axis,
the lines* y = c* and* y *= *d*
is given by

For
instance, consider the shaded region in Fig. 9.13. Here, *B*_{1} , *B*_{2}
, *B*_{3}
and *B*_{4} denote geometric
areas of the individual parts. Then the
total area *B*
of the region bounded by the curve
*x* = *f*(*y*), *y* âˆ’ axis and the lines* y *= *c* and* y *= *d*
is given by

*B* = *B*_{1}+*B*_{2}+*B*_{3}+*B*_{4}

** **

**Example 9.47**

Find the
area of the region bounded by the line 6*x*
+
5*y* = 30 , *x* âˆ’ axis and the lines *x*
= âˆ’1
and *x* = 3 **.**

**Solution**

The
region is sketched in Fig. 9.14. It lies above the *x* âˆ’ axis. Hence, the required area is given by

** **

**Example 9.48**

Find the
area of the region bounded by the line 7*x*
âˆ’
5*y* = 35 , *x* âˆ’ axis and the lines *x*
= âˆ’2
and *x* = 3**.**

**Solution**

The
region is sketched in Fig. 9.15. It lies below the *x *âˆ’* *axis. Hence, the required area is given by

** **

**Example 9.49**

Find the
area of the region bounded by the ellipse

**Solution**

The
ellipse is symmetric about both major and minor axes. It is sketched as in
Fig.9.16. So, viewing in the positive direction of *y *-axis, the required area* A *is
four times the area of the region bounded by the portion of the ellipse in the
first quadrant

**Note**

Viewing
in the positive direction of* x *-axis,
the required* y * area A is four times the area of the region
bounded by the portion of the ellipse in the first quadrant

* y*-axis,
*y* = 0 and *y = b*. Hence, by taking horizontal strips (see Fig.9.17), we get

**Note**

Putting *b* = *a*
in the above result, we get that the area of the region enclosed by the circle *x*^{2}+ *y*^{2} = *a*^{2} is *Ï€**a*^{2} .

** **

**Example 9.50**

Find the
area of the region bounded between the parabola *y*^{2} = 4*ax* and its
latus rectum.

**Solution**

The
equation of the latus-rectum is* x *= a
. It intersects the parabola at the
points L ( *a*, 2*a*) and L_{1} ( *a*,
âˆ’2*a*) . The required area is sketched
in Fig. 9.18. By symmetry, the required area A is twice the area bounded by the* *portion of the parabola

*y* = 2 âˆš*a*âˆš*x *,*
x *-axis,* x *= 0 and* x *= a.

Hence,
by taking vertical strips, we get

**Note**

Viewing
in the positive direction of *x*-axis,
and making horizontal strips (see Fig. 9.19), we get

**Note**

It is
quite interesting to note that the above area is equal to two-thirds the base
(latus-rectum) times the height (the distance between the focus and the
vertex). This verifies Archimedesâ€™ formula for areas of parabolic arches which
states that the area under a parabolic arch is two-thirds the area of the
rectangle having base of the arch as length and height of the arch as the breadth.
It is also equal to four-thirds the area of the triangle with base
(latus-rectum) and height (the distance between the focus and the vertex).

**Example 9.51**

Find the
area of the region bounded by the *y*
-axis and the parabola *x* =
5 âˆ’
4 *y* âˆ’ *y*^{2}.

**Solution**

The
equation of the parabola is ( *y* +
2)^{2} = âˆ’ ( *x* âˆ’ 9) . The parabola crosses the *y* -axis at (0, âˆ’5) and (0,1) .The vertex is at (9, âˆ’2)
and the axis of the parabola is *y* = âˆ’2
. The required area is sketched as in Fig. 9.20.

Viewing
in the positive direction of *x* âˆ’
axis, and making horizontal strips, the required area *A* is given by

As in
the previous problem, we again verify Archimedesâ€™ formula that the area of the
parabolic arch is equal to two-thirds the base times the height.

** **

Find the
area of the region bounded by *x *âˆ’* *axis, the sine curve *y
*=* *sin* x *, the lines *x *=* *0*
*and *x *=* *2*Ï€** *.

The
required area is sketched in Fig. 9.21. One portion of the region lies above
the *x* âˆ’ axis between *x* = 0 and *x* =
*Ï€* , and the other portion lies below *x* âˆ’ axis between *x *=* **Ï€** *and* x *=* *2*Ï€** *.

So, the
required area is given by

**Note**

If we
compute the definite integral ^{2Ï€}âˆ«_{0}sin *xdx* , we get

So ^{2Ï€}âˆ«_{0}
*f *(*x*)*dx *does not represent
the area of the region bounded by the curve*
y *=* *sin* x *,* x *âˆ’* *axis, the lines *x *=* *0* *and* x *=* *2*Ï€** *.

** **

Find the
area of the region bounded by *x* âˆ’
axis, the curve *y* =
**|**cos *x***|**, the lines *x* = 0 and *x* = *Ï€*
.

**Solution**

It lies
above the x âˆ’ axis. The required area is sketched in Fig. 9.22. So, the
required area is given by

**Case
(i)**

Let *y* = *f*
( *x*) and *y* = *g*( *x*) be the equations of two curves in the
XOY âˆ’ plane such that *f (x)* â‰¥* g *( x) for all* x *âˆˆ [*a, b*] . We want to find the area A of
the region bounded between the two
curves, the ordinates* x *= a and *x* =* b *.

The
required area is sketched in Fig. 9.23. To compute A , we divide the region
into thin vertical strips of width Î”*x* and
height *f *(* x*)* *âˆ’* g *(*
x*)* *. It is important note that *f *(*x*)* *âˆ’* g *(*x*)* *â‰¥* *0 for all *x *âˆˆ[*a*,*
b*]* *. As before, the required area
is the limit of the sum of the areas of the vertical strips. Hence, we get

* A*
= ^{b}âˆ«_{a} [ *f *(*x*) - *g*(*x*)]*dx*

**Note**

Viewing
in the positive direction of *y* âˆ’
axis, the curve *y* =
*f* ( *x*) can be termed as the upper curve (U) and the curve *y* = *g*(
*x*) as the lower curve (L). Thus, we
get *A* = * ^{b}*âˆ«

**Case
(ii)**

Let *x* = *f*
( *y*) and *x* = *g*( *y*) be the equations of two curves in the
*XOY* âˆ’ plane such that *f* ( *y*)
â‰¥
*g* ( *y*) for all *y* âˆˆ[*c*,
*d* ] . We want to find the area *A* of the region bounded between the two
curves, the lines* y *= c and* y *= d . The required area is sketched in
Fig. 9.24.

To compute *A* , we view in
the positive direction of the *x* â€“
axis and divide the region into thin horizontal strips of width Î”*y* and height *f* ( *y*) âˆ’ *g*
( *y*) . It is important note that *f* ( *y*)
âˆ’
*g* ( *y*) â‰¥ 0 for all *y *âˆˆ[*c*,*
d *]* *. As before, the required
area is the limit of the sum of the areas of the horizontal strips. Hence, we get

*A* = ^{d}âˆ«_{c} [ *f *(*x*) - *g*(*x*)]*dx*

**Note**

Viewing
in the positive direction of *x* âˆ’
axis, the curve *x* =
*f* ( *y*) can be termed as the right curve (R) and the curve *x* = *f*
( *y*) as the left curve (L). Thus, we
get *A* = * ^{b}*âˆ«

**Example 9.54**

Find the
area of the region bounded between the parabolas* y*^{2} = 4*x *and* x*^{2} = 4*y*.

**Solution **

First,
we get the points of intersection of the parabolas. For this, we solve* y*^{2}
= 4x and* x*^{2} = 4*y *simultaneously: Eliminating* y *between them,* *we get* x*^{4} =
64x and so* x *= 0 and* x *= 4 . Then the points of intersection
are (0, 0) and (4, 4) . The required region is sketched in Fig.9.25.* *

Viewing
in the direction of* y *-axis, the
equation of the upper boundary is* y *=
2*âˆšx *for 0 â‰¤* x *â‰¤ 4 and the equation of the lower boundary is* y *= x^{2}/4 for 0 â‰¤* x *â‰¤ 4 . So, the required area Î” is

**Note **

Viewing
in the positive direction of* x *-axis,
the right bounding curve is* x*^{2}
= 4*y *and the left bounding curve is* y*^{2} = 4x . See Fig. 9.26. The
equation of the right boundary is* x *=
2*âˆšy *for 0 â‰¤* y *â‰¤ 4 and the equation of the left boundary is* x *= y^{2}/4 for 0 â‰¤*
y *â‰¤ 4 . So, the required area A is

** **

**Example 9.55**

Find the
area of the region bounded between the parabola *x*^{2} = *y* and the
curve *y* = |*x*|.

**Solution**** **

Both the
curves are symmetrical about *y* -axis.

It
intersects the parabola *x*^{2}
= *y* at ( 1,1) and ( âˆ’1, 1).

The area of the region bounded by the curves is sketched in Fig. 9.27. It lies in the first quadrant as well as in the second quadrant. By symmetry, the required area is twice the area in the first quadrant.

In the
first quadrant, the upper curve is *y = x*, 0 â‰¤* x* â‰¤ 1 and the lower curve is *y
= x*^{2}, 0 â‰¤ *x* â‰¤ 1. Hence,
the required area is given by

** **

**Example 9.56**

Find the
area of the region bounded by *y* = cos
*x, y* = sin *x*, the lines x = Ï€/4 and x = 5Ï€/4.

**Solution**

The
region is sketched in Fig. 9.28. The upper boundary of the region is *y* = sin *x* for Ï€/4 â‰¤ x â‰¤ 5Ï€/4 and the lower boundary of the region is *y* = cos *x* for Ï€/4 â‰¤ x â‰¤ 5Ï€/4. So the required area *A* is given by

** **

**Example 9.57**

The
region enclosed by the circle *x*^{2}
+
*y*^{2} =
*a*^{2} is divided into two
segments by the line *x* =
*h*. Find the area of the smaller
segment.

**Solution**

The
smaller segment is sketched in Fig. 9.29. Here 0 < *h* < *a*
. By symmetry about the *x* -axis, the
area of the smaller segment is given by

** **

**Example 9.58**

Find the
area of the region in the first quadrant bounded by the parabola *y*^{2} =
4*x* , the line *x *+* y *=* *3* *and* y *-axis.

**Solution**

First,
we find the points of intersection of *x *+* y *=* *3* *and* y*^{2}* *âˆ’* *4*x *:

* x *+* y *=* *3* *â‡’* y *=* *3* *âˆ’* x *.

âˆ´* y *^{2}* *=* *4*x *â‡’* *(3* *âˆ’* x *)^{2}* *=* *4*x*

â‡’* x *^{2}* *âˆ’* *10*x *+* *9*
*=* *0

â‡’* x *=* *1,* x *=* *9* *.

âˆ´* x *=1* *in*
x *+* y *=* *3* *â‡’* y *=* *2* *, and *x *=* *9*
*in* x *+* y *=* *3* *â‡’* y *= âˆ’6* *.

âˆ´(1,
2) and (9, âˆ’6) are the points of intersection.

The line
*x* + *y*
=
3 meets the *y *-axis at* *(0, 3)* *.

The
required area is sketched in Fig. 9.30.

Viewing
in the direction of *y* -axis, on the
right bounding curve is given by

** **

**Example 9.59**

Find, by
integration, the area of the region bounded by the lines 5*x* âˆ’ 2 *y* =
15, *x* + *y*
+
4 =
0 and the *x*-axis.

**Solution**

The
lines 5*x* âˆ’
2 *y* = 15, *x* + *y* +
4 =
0 intersect at (1, âˆ’5) . The line 5*x* âˆ’ 2 *y* =
15 meets the *x*-axis at (
3, 0)
. The line *x* +
*y* + 4 = 0 meets the *x*-axis at ( âˆ’4, 0) . The required area is shaded in
Fig.9.31. It lies below the *x*-axis.
It can be computed either by considering vertical strips or horizontal strips.

When we
do by vertical strips, the region has to be divided into two sub-regions by the
line *x *=* *1* *. Then, we get

When we
do by horizontal strips, there is no need to subdivide the region. In this
case, the area is bounded on the right by the line 5*x* âˆ’ 2 *y* =
15 and on the left by *x* +
*y* + 4 = 0 . So, we get

**Note**

The
region is triangular with base 7 units and height 5 units. Hence its area is
35/2 without using integration.

** **

**Example 9.60**

Using
integration find the area of the region bounded by triangle *ABC*, whose vertices *A*, *B,* and C are (
âˆ’1,1),
(
3, 2)
, and (
0, 5)
respectively.

**Solution**

See Fig.
9.32.

** **

Using
integration, find the area of the region which is bounded by *x-*axis, the tangent and normal to the
circle *x*^{2}* *+* y*^{2} = 4 drawn at (1, âˆš3) .

**Solution**

We recall
that the equation of the tangent to the circle *x*^{2}* *+* y*^{2}= *a*^{2} at ( *x*_{1} , *y*_{1} ) is *xx*_{1} + *yy*_{1} = *a*^{2} .
So, the equation of the tangent to the circle *x*^{2}* *+* y*^{2}* *=* *4* *at*
*(1,
âˆš3)
is *x* + *y*
âˆš3 =
4 ; that is, *y *= âˆ’ 1/âˆš3 (*x*
âˆ’
4) . The tangent meets the *x*-axis at
the point (4,0). The slope of the tangent is â€“1/âˆš3. So the slope of
the normal is âˆš3 and hence equation of the normal is *y* âˆ’ âˆš3 = âˆš3( *x* âˆ’1) ; that is *y* = âˆš3*x* and it passes through the origin. The
area to be found is shaded in the adjoining figure. It can be found by two
methods.

**Method 1**

Viewing
in the postive direction of *y*-axis,
the required area is the area of the region bounded by *x*-axis,* y *=
âˆš3*x* and *x* + *y* âˆš3 =
4 . So it can be obtained by applying the formula * ^{b}*âˆ«

Viewing
in the direction of *x*-axis, the
required area is the area of the region bounded between *y *=* *âˆš3*x *and* x *+* y*âˆš3* *=* *4* *,* y *=* *0*
*and* y *=* *âˆš3. So, it can be obtained by applying the formula

Here, *c* = 0 , *d* = âˆš3 , *x _{R}*
is the

Draw an
arbitrary line parallel to *y*-axis
cutting the plane region. First, find the *y*-coordinate
of the point where the line enters the region. Call it *y _{ENTRY}* . Next, find the

Draw an
arbitrary line parallel to *x*-axis
cutting the plane region.

First,
find the *x*-coordinate of the point where
the line enters the region. Call it *x _{ENTRY}*
.

Next,
find the *x*-coordinate of the point
where the line exits the region. Call it *x _{EXIT}*
. Both

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