In the previous section,
we completed a discussion on area bounded by a curve and a line. In
this section, we will see area bounded by a curve and another curve.
The basic details can be written in 2 steps:
1. In fig.24.16 below,
♦ The red curve represent y = f(x)
♦ The green curve represent y = g(x)
 |
| Fig.24.16 |
•
The curves intersect at two points:
♦ At the first point,
x-coordinate is 'a'
♦ At the second point, x-coordinate is 'b'
(We do not want their y-coordinates)
2. We want to find the area bounded by the two curves. In the fig., it is shaded in blue color.
•
Consider the thin yellow strip of width dx. This strip is at a distance of x from O.
♦ So the y-coordinate of the bottom end of this strip will be g(x).
♦ Also, the y-coordinate of the top end of this strip will be f(x).
•
So height of this strip will be $\small{[f(x)~-~g(x)]}$
•
So area of this strip will be $\small{[f(x)~-~g(x)]dx}$
•
Therefore, area A of the blue region can be obtained as:
$\small{A~=~\int_a^b{\left[f(x)~-~g(x) \right]dx}}$
Note:
In the interval [a,b], f(x) ≥ g(x)
So in the blue region, wherever we place the yellow strip, it's height will be always $\small{[f(x)~-~g(x)]}$
Alternate method:
This can be written in 3 steps:
1. In the fig.24.16 above, consider the area bounded by four items:
♦ The curve y = f(x)
♦ The vertical line x = a
♦ The vertical line x = b
♦ The horizontal line y = 0 (x-axis)
•
We will call this area as A1. We can write:
$\small{A_1~=~\int_a^b{\left[f(x) \right]dx}}$
2. In the same fig.24.16 above, consider the area bounded by four items:
♦ The curve y = g(x)
♦ The vertical line x = a
♦ The vertical line x = b
♦ The horizontal line y = 0 (x-axis)
•
We will call this area as A2. We can write:
$\small{A_2~=~\int_a^b{\left[g(x) \right]dx}}$
3. From the fig.24.16, we have:
A2 + Blue area = A1
•
So we get:
Blue area = A1 − A2
$\small{~=~\int_a^b{\left[f(x) \right]dx}~-~\int_a^b{\left[g(x) \right]dx}}$
$\small{~=~\int_a^b{\left[f(x)~-~g(x) \right]dx}}$
Let us see another case. It can be explained in 2 steps:
1. In fig.24.17 below,
♦ The red curve represent y = f(x)
♦ The green curve represent y = g(x)
 |
| Fig.24.17 |
•
The curves intersect at three points:
♦ At the first point, x-coordinate is 'a'
♦ At the second point, x-coordinate is 'c'
♦ At the third point, x-coordinate is 'b'
(We do not want their y-coordinates)
2. We want to find the area bounded by the two curves. That is., we want (A1 + A2) in the fig.
•
A1 can be calculated just as we did in the previous fig.24.16.
We get: $\small{A_1~=~\int_a^c{\left[f(x)~-~g(x) \right]dx}}$
•
Let us find A2:
Consider the thin yellow strip of width dx. This strip is at a distance of x from O.
♦ So the y-coordinate of the bottom end of this strip will be f(x).
♦ Also, the y-coordinate of the top end of this strip will be g(x).
•
So height of this strip will be $\small{[g(x)~-~f(x)]}$
•
So area of this strip will be $\small{[g(x)~-~f(x)]dx}$
•
Therefore, area A2 of the violet region can be obtained as:
$\small{A_2~=~\int_c^b{\left[f(x)~-~g(x) \right]dx}}$
Note:
In the interval [c,b], g(x) ≥ f(x)
So in the violet region, wherever we place the yellow strip, it's height will be always $\small{[g(x)~-~f(x)]}$
Solved example 24.12
Find the area of the region bounded by the two parabolas y = x2 and y2 = x.
Solution:
1. First we write the given equations in the form: y = f(x) and y = g(x)
$\small{y~=~f(x)~=~x^2}$
$\small{y~=~g(x)~=~\pm \sqrt x}$
2. Solving the two equations, we find that, they intersect at (0,0) and (1,1).
The point (1,1) is marked as point A in fig.24.18 below:
 |
| Fig.24.18 |
The region bounded by the two curves is shaded in blue color. So we are interested in the interval [0,1]
3. From the fig., it is clear that $\small{g(x)~\ge~f(x)}$ in the interval [0,1].
4. So, if we assume a thin vertical strip of width dx in the magenta region, then the height of that strip will be $\small{[g(x)~-~f(x)]}$
So area of the strip will be $\small{[g(x)~-~f(x)]dx}$
5. Therefore, area A of the blue region can be obtained as:
$\small{A~=~\int_0^1{\left[g(x)~-~f(x) \right]dx}~=~\int_0^1{\left[\sqrt x~-~x^2 \right]dx}~=~\frac{1}{3}}$ sq.units
(The reader may write all steps related to the integration process)
Solved example 24.13
Find the area lying above the x-axis and included between the circle x2 + y2 = 8x and inside of the parabola y2 = 4x.
Solution:
1. Consider the given equation $\small{x^2~+~y^2~=~8x}$
This can be rearranged as:
$\small{x^2~-8x~+~16~-~16~+~y^2~=~0}$
$\small{\Rightarrow (x^2~-8x~+~16)~+~y^2~=~16}$
$\small{\Rightarrow (x-4)^2~+~y^2~=~4^2}$
This is the equation of a circle with center (4,0) and radius 4 units.
2. Next we write the two equations in the form y = f(x) and y = g(x).
For the circle, we can write: $\small{y~=~f(x)~=~\pm \sqrt{4^2~-~(x-4)^2}}$
• In the above equation, if we input all x values from the interval [0,8], we will get the ordered pairs, which when plotted, will give the red circle in fig.24.19 below:
 |
| Fig.24.19 |
For the parabola, we can write: $\small{y~=~g(x)~=~\pm 2 \sqrt{x}}$
• In the above equation, if we input all x values from the interval $\small{[0,\infty]}$, we will get the ordered pairs, which when plotted, will give the green parabola in fig.24.19 above.
3. Let us solve the equations of the two curves: $\small{x^2~+~y^2~=~8x~\text{and}~y^2~=~4x}$
We get: $\small{x^2~+~4x~=~8x \Rightarrow x^2~-~4x~=~0 }$
$\small{ \Rightarrow x(x-4)~=~0 \Rightarrow x = 0~\text{and}~x=4}$
♦ When x = 0, we get: y = 0
♦ When x = 4, we get: y = 4
So the two curves intersect at (0,0) and (4,4)
♦ (0,0) is the origin O
♦ (4,4) is marked as 'A' in the fig.
4. We are asked to find the area of OABCO. It is shaded in magenta color.
• We see that,
Area of (OABCO + ODAO) = Area of the semicircle.
(Point D is an arbitrary point, just to define the curved shape between O and A along the circle)
• In other words
magenta + blue = $\small{\frac{\pi r^2}{2}~=~\frac{\pi (4)^2}{2}~=~8 \pi}$ sq.units
5. So our next aim is to find the area of the blue region. This region is bounded by two items:
♦ The curve y = f(x)
♦ The curve y = g(x)
In the interval [0,4], f(x) ≥ g(x)
Therefore, area of blue region
$\small{~=~\int_0^4{\left[f(x)~-~g(x) \right]dx}~=~\int_0^4{\left[\sqrt{4^2~-~(x-4)^2} ~-~2 \sqrt{x} \right]dx}}$
$\small{~=~\int_0^4{\left[\sqrt{4^2~-~(x-4)^2} \right]dx}~-~\int_0^4{\left[2 \sqrt{x} \right]dx}}$
• Here we use $\small{\sqrt{4^2~-~(x-4)^2} ~\rm{and}~2 \sqrt{x}}$
Instead of $\small{-\sqrt{4^2~-~(x-4)^2} ~\rm{and}~-2 \sqrt{x}}$
This is because, in the first quadrant, y values are +ve.
6. The integration can be done as shown below:
The first term is: $\small{\int_0^4{\left[\sqrt{4^2~-~(x-4)^2} \right]dx}}$
• Here we can use the standard integral:
$\small{ \int{\big[\sqrt{a^2-u^2} \big]dx}~=~\frac{u}{2}\sqrt{a^2 - u^2} ~+~\frac{a^2}{2} \sin^{-1}\frac{u}{a}~+~\rm{C}}$
• In our present case, a = 4 and u = x-4
• So we get:
$\small{\int_0^4{\left[\sqrt{4^2~-~(x-4)^2} \right]dx}}$
$\small{~=~\left[\frac{x-4}{2}\sqrt{4^2 - (x-4)^2} ~+~\frac{4^2}{2} \sin^{-1}\frac{x-4}{4} \right]_0^4}$
$\small{~=~\left[0 ~+~0 \right]~-~\left[(-2)\sqrt{0} ~+~8 \sin^{-1}(-1) \right]}$
$\small{~=~(-1)(8) \left[\sin^{-1}(-1) \right]~=~(8) \left[\sin^{-1}(1) \right]~=~\frac{8 \pi}{2}~=~4 \pi}$ sq.units
7. The second term is:
$\small{\int_0^4{\left[2 \sqrt{x} \right]dx}}$
This gives: $\small{\frac{32}{3}}$ sq.units
8. So from step (5), we get:
Area of blue region = $\small{4 \pi~-~\frac{32}{3}}$ sq.units.
9. So from (4), we get:
Area of magenta region = $\small{8 \pi~-~4 \pi~+~\frac{32}{3}~=~4 \pi~+~\frac{32}{3}}$ sq. units
Solved example 24.14
In the fig.24.20 below, AOBA is the part of the ellipse in the first quadrant such that OA = 2 and OB = 6. Find the area between the arc AB and the chord AB.
 |
| Fig.24.20 |
Solution:
1. First we will rearrange the given equation into the form y =
f(x)
We have: $\small{\frac{x^2}{4}~+~\frac{y^2}{36}~=~1}$
$\small{\Rightarrow \frac{9 x^2~+~y^2}{36}~=~1 \Rightarrow 9x^2 ~+~y^2~=~36 \Rightarrow y~=~f(x)~=~\pm \sqrt{36~-~9x^2}}$
• In the above equation, if we input all x values from the interval [−2,2], we will get the ordered pairs, which when plotted, will give the red ellipse in fig.24.20 above.
2. We are given the coordinates of end points A and B. So we can easily write the equation of the line AB. We get:
$\small{y-0~=~\left(\frac{6-0}{0-2} \right)(x-2)}$
$\small{\Rightarrow y~=~g(x)~=~-3x + 6}$
3. We are asked to find the area of the blue region in the fig.24.20. This region is bounded by
f(x) and
g(x). Also in the interval [0,2],
f(x) ≥
g(x)
• So the area of the blue region
$\small{~=~\int_0^2{\left[f(x)~-~g(x) \right]dx}~=~\int_0^2{\left[\sqrt{36~-~9x^2} ~-~(-3x + 6) \right]dx}}$
•
This integration gives: $\small{(3\pi~-~6)}$ sq.units
(The reader may write all steps related to the integration process)
Solved example 24.15
Using integration find the area of the region bounded by the triangle whose vertices are (1,0), (2,2) and (3,1).
Solution:
1. Let the vertices of the triangle be A(1,0), B(2,2) and C(3,1). They are shown in fig.24.21 below:
 |
| Fig.24.21 |
•
We are asked to find the area bounded by the triangle ABC. In the fig above, this required area is separated into two:
♦ ABD shaded in magenta color
♦ DCB shaded in blue color
•
Point D is the intersection of two lines:
♦ Perpendicular dropped from B
♦ Side AC
•
So we want the sum of magenta and blue areas.
2. First we will write the equations of the three lines:
(i) Line AB:
$\small{y-0~=~\left(\frac{2-0}{2-1} \right)(x-1)}$
$\small{\Rightarrow y~=~f(x)~=~2x - 2}$
(ii) Line BC:
$\small{y-2~=~\left(\frac{1-2}{3-2} \right)(x-2)}$
$\small{\Rightarrow y~=~-x + 2 + 2}$
$\small{\Rightarrow y~=~g(x)~=~-x + 4}$
(iii) Line AC:
$\small{y-0~=~\left(\frac{1-0}{3-1} \right)(x-1)}$
$\small{\Rightarrow y~=~\frac{1}{2} (x-1)}$
$\small{\Rightarrow y~=~h(x)~=~\frac{x}{2} - \frac{1}{2}}$
3. Next, we will find the area of magenta region. This area is bounded by three items:
♦ The line y = f(x)
♦ The line y = h(x)
♦ The vertical line x = 2
Therefore, area of the magenta region
$\small{~=~\int_1^2{\left[f(x)~-~h(x) \right]dx}~=~\int_1^2{\left[2x-2~-~\left(\frac{x}{2} - \frac{1}{2} \right) \right]dx}}$
$\small{~=~\int_1^2{\left[2x-2~-~\frac{x}{2} + \frac{1}{2} \right]dx}~=~\int_1^2{\left[\frac{3x}{2} - \frac{3}{2} \right]dx}~=~\frac{3}{4}}$ sq.units
4. Next, we will find the area of blue region. This area is bounded by three items:
♦ The line y = g(x)
♦ The line y = h(x)
♦ The vertical line x = 2
Therefore, area of the magenta region
$\small{~=~\int_2^3{\left[g(x)~-~h(x) \right]dx}~=~\int_2^3{\left[-x+4~-~\left(\frac{x}{2} - \frac{1}{2} \right) \right]dx}}$
$\small{~=~\int_2^3{\left[-x+4~-~\frac{x}{2} + \frac{1}{2} \right]dx}~=~\int_2^3{\left[-\frac{3x}{2} + \frac{9}{2} \right]dx}~=~\frac{3}{4}}$ sq.units
5. Finally, we can calculate the sum:
Magenta area + Blue area = $\small{\frac{3}{4}~+~\frac{3}{4}~=~\frac{3}{2}}$ sq.units
Solved example 24.16
Find the area of the region enclosed between the two circles x2 + y2 = 4 and (x−2)2 + y2 = 4.
Solution:
1. Consider the second equation given to us: $\small{(x-2)^2~+~y^2~=~4}$
This is the equation of a circle with center (2,0) and radius 2 units.
2. Next we write the two equations in the form y = f(x) and y = g(x).
For the first circle, we can write:
$\small{y~=~f(x)~=~\pm \sqrt{4~-~x^2}}$
• In the above equation, if we input all x values from the interval [−2,2], we will get the ordered pairs, which when plotted, will give the red circle in fig.24.22 below:
 |
| Fig.24.22 |
•
For the second circle, we can write:
$\small{y~=~g(x)~=~\pm \sqrt{2^2~-~(x-2)^2}}$
• In the above equation, if we input all x values from the interval [0,4], we will get the ordered pairs, which when plotted, will give the green circle in fig.24.22 above.
3. Let us solve the equations of the two curves: $\small{x^2~+~y^2~=~4~~~\text{and}~~~(x-2)^2~+~y^2~=~4}$
We get: $\small{x = 1 ~~\rm{and}~~y = \pm \sqrt 3 }$
♦ (1,√3) is marked as 'A' in the fig.
4. We are asked to find the area enclosed between the two circles.
Note that, the two circles are symmetric about the x-axis.
So the magenta region will be half of the required area. Therefore, we need to find the area of the magenta region.
5. Consider the sum: Magenta + Blue
• This sum will be one fourth of the area of the red circle. So we can write:
Magenta + Blue = $\small{\frac{\pi (2)^2}{4}~=~\pi}$ sq.units
6. So our next task is to find the area of the blue region.
This region is bounded by three items:
♦ The curve y = f(x)
♦ The curve y = g(x)
♦ The vertical line x = 1
♦ The vertical line x = 0 (y-axis)
• So area of blue region
$\small{~=~\int_0^1{\left[f(x)~-~g(x) \right]dx}~=~\int_0^1{\left[\sqrt{4~-~x^2}~-~\sqrt{2^2~-~(x-2)^2} \right]dx}}$
$\small{~=~\int_0^1{\left[\sqrt{4~-~x^2} \right]dx}~-~\int_0^1{\left[\sqrt{2^2~-~(x-2)^2} \right]dx}}$
• Here we use $\small{\sqrt{4~-~x^2} ~\rm{and}~\sqrt{2^2~-~(x-2)^2}}$
Instead of $\small{-\sqrt{4~-~x^2} ~\rm{and}~-\sqrt{2^2~-~(x-2)^2}}$
This is because, in the first quadrant, y values are +ve.
• This integration gives:
Area of blue region = $\small{(\sqrt 3~-~\frac{\pi}{3})}$ sq.units
(The reader may write all steps related to the integration process)
7. So based on step (5), we get:
Magenta = $\small{\pi~-~\left(\sqrt 3~-~\frac{\pi}{3}\right)~=~\frac{4 \pi}{3}~-~\sqrt 3}$
8. So the required area
$\small{~=~2\left(\frac{4 \pi}{3}~-~\sqrt 3 \right)~=~\left(\frac{8 \pi}{3}~-~2 \sqrt 3 \right)}$ sq.units
The link below gives a few more solved examples:
Exercise 24.2
In the next section, we will see some miscellaneous examples.
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