In the previous section, we saw area bounded by a curve and a line. We saw some solved examples also. In this section, we will see some solved examples related to the topics that we discussed so far in this chapter.
Solved example 24.8
Find the area of the region bounded by the ellipse $\small{\frac{x^2}{16}~+~\frac{y^2}{9}~=~1}$.
Solution:
1. First we write the given equation in the form: y = f(x)
$\small{\frac{x^2}{16}~+~ \frac{y^2}{9}~=~1}$
$\small{\Rightarrow \frac{y^2}{9}~=~1~-~\frac{x^2}{16}}$
$\small{\Rightarrow
y^2~=~9 \left(1~-~\frac{x^2}{16} \right)~=~9 \left(\frac{16~-~x^2}{16}
\right)~=~\frac{9}{16}\left(16~-~x^2 \right)}$
$\small{\Rightarrow y~=~f(x)~=~\pm \frac{3}{4} \sqrt{4^2~-~x^2}}$
•
In the above equation, if we input all x values from the interval
[−4,4], we will get the ordered pairs, which when plotted, will give the
red ellipse in fig.24.12 below:
 |
| Fig.24.12 |
2. In the fig.24.12, we want the area of the portion shaded in violet color. This portion is named as OABOA.
• The portion shaded in violet color is bounded by four items:
♦ The curve y = f(x)
♦ The vertical line x = 0 (y-axis)
♦ The horizontal line y=0 (x-axis)
•
Consider the thin vertical strip of width dx. This strip is situated at
a distance of x from O. So height of the strip will be f(x). So area of
the strip will be f(x) dx.
Therefore, area (A) of the violet portion can be obtained as:
$\small{A~=~\int_0^4{\left[f(x) \right]dx}~=~\int_0^4{\left[\frac{3}{4} \sqrt{4^2~-~x^2} \right]dx}}$
•
Here we discard $\small{f(x)~=~-\frac{3}{4} \sqrt{4^2~-~x^2}}$ and
take $\small{f(x)~=~+\frac{3}{4} \sqrt{4^2~-~x^2}}$. This is because,
OABOA is in the first quadrant. In this quadrant, all y values are +ve.
3. This integration can be done as shown below:
• First we find the indefinite integral F:
$\small{F~=~\frac{3}{4} \int{\left[\sqrt{4^2~-~x^2} \right]dx}}$
• This is a standard integral (see section 23.19). We have:
$\small{F~=~\int{\left[\sqrt{a^2~-~x^2} \right]dx}~=~\frac{x}{2} \sqrt{a^2~-~x^2}~+~\frac{a^2}{2} \sin^{-1}\frac{x}{a}}$
Therefore, $\small{A~=~\left[\frac{x}{2} \sqrt{4^2~-~x^2}~+~\frac{4^2}{2} \sin^{-1}\frac{x}{4} \right]_0^{4}}$
$\small{~=~\frac{3}{4}
\left[\frac{4}{2} \sqrt{4^2~-~4^2}~+~\frac{4^2}{2} \sin^{-1}\frac{4}{4}
\right]~-~\frac{3}{4} \left[\frac{0}{2}
\sqrt{4^2~-~0^2}~+~\frac{4^2}{2} \sin^{-1}\frac{0}{4} \right]}$
$\small{~=~\frac{3}{4} \left[0~+~8 \sin^{-1}1 \right]~-~\frac{3}{4} \left[0~+~\frac{4^2}{2} \sin^{-1}0 \right]}$
$\small{~=~\frac{3}{4} \left[8 \left(\frac{\pi}{2} \right) \right]~-~\frac{b}{a} \left[0~+~0 \right]~=~3 \pi}$
4.
In the fig.24.12 above, we see that, the ellipse is symmetrical about
both x and y axes. So the total area of the ellipse is four times the
area of OABOA.
• We can write:
Whole area enclosed by the ellipse
$\small{~=~(4)(3) \pi~=~12 \pi}$
Alternate method:
1. In a previous section of this chapter, we derived the formula for area of the ellipse. See solved example 24.2.
•
We saw that:
Area enclosed by the ellipse $\small{\frac{x^2}{a^2}~+~\frac{y^2}{b^2}~=~1}$ is $\small{\pi a b}$.
2. In our present case, a = 4 and b = 3.
So we get:
Area = $\small{\pi (4) (3)~=~12 \pi}$
Solved example 24.9
Find the area of the region bounded by the ellipse $\small{\frac{x^2}{4}~+~\frac{y^2}{9}~=~1}$.
Solution:
1. First we write the given equation in the form: y = f(x)
$\small{\frac{x^2}{4}~+~ \frac{y^2}{9}~=~1}$
$\small{\Rightarrow \frac{y^2}{9}~=~1~-~\frac{x^2}{4}}$
$\small{\Rightarrow
y^2~=~9 \left(1~-~\frac{x^2}{4} \right)~=~9 \left(\frac{4~-~x^2}{4}
\right)~=~\frac{9}{4}\left(4~-~x^2 \right)}$
$\small{\Rightarrow y~=~f(x)~=~\pm \frac{9}{4} \sqrt{2^2~-~x^2}}$
•
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 ellipse in fig.24.13 below:
 |
| Fig.24.13 |
2. In the fig.24.13, we want the area of the portion shaded in violet color. This portion is named as OABOA.
• The portion shaded in violet color is bounded by four items:
♦ The curve y = f(x)
♦ The vertical line x = 0 (y-axis)
♦ The horizontal line y=0 (x-axis)
Therefore, area (A) of the violet portion can be obtained as:
$\small{A~=~\int_0^2{\left[f(x) \right]dx}~=~\int_0^2{\left[\frac{3}{2} \sqrt{2^2~-~x^2} \right]dx}}$
• Here we discard $\small{f(x)~=~-\frac{3}{2} \sqrt{2^2~-~x^2}}$ and take $\small{f(x)~=~+\frac{3}{2} \sqrt{2^2~-~x^2}}$. This is because, OABOA is in the first quadrant. In this quadrant, all y values are +ve.
3. This integration can be done as shown below:
• First we find the indefinite integral F:
$\small{F~=~\frac{3}{2} \int{\left[\sqrt{2^2~-~x^2} \right]dx}}$
• This is a standard integral (see section 23.19). We have:
$\small{F~=~\int{\left[\sqrt{a^2~-~x^2} \right]dx}~=~\frac{x}{2} \sqrt{a^2~-~x^2}~+~\frac{a^2}{2} \sin^{-1}\frac{x}{a}}$
Therefore, $\small{A~=~\left[\frac{x}{2} \sqrt{2^2~-~x^2}~+~\frac{2^2}{2} \sin^{-1}\frac{x}{2} \right]_0^{2}}$
$\small{~=~\frac{3}{2} \left[\frac{2}{2} \sqrt{2^2~-~2^2}~+~\frac{2^2}{2} \sin^{-1}\frac{2}{2} \right]~-~\frac{3}{2} \left[\frac{0}{2} \sqrt{2^2~-~0^2}~+~\frac{2^2}{2} \sin^{-1}\frac{0}{2} \right]}$
$\small{~=~\frac{3}{2} \left[0~+~2 \sin^{-1}1 \right]~-~\frac{3}{2} \left[0~+~2 \sin^{-1}0 \right]}$
$\small{~=~\frac{3}{2} \left[2 \left(\frac{\pi}{2} \right) \right]~-~\frac{b}{a} \left[0~+~0 \right]~=~\frac{3 \pi}{2}}$
4. In the fig.24.13 above, we see that, the ellipse is symmetrical about both x and y axes. So the total area of the ellipse is four times the area of OABOA.
• We can write:
Whole area enclosed by the ellipse
$\small{~=~(4)\frac{3 \pi}{2}~=~6 \pi}$
Alternate method:
1. In a previous section of this chapter, we derived the formula for area of the ellipse. See solved example 24.2.
•
We saw that:
Area enclosed by the ellipse $\small{\frac{x^2}{a^2}~+~\frac{y^2}{b^2}~=~1}$ is $\small{\pi a b}$.
2. In our present case, a = 2 and b = 3.
So we get:
Area = $\small{\pi (2) (3)~=~6 \pi}$
Solved example 24.10
Find the area of the region bounded by the line y = 3x + 2, the x-axis and the ordinates x = −1 and x = 1
Solution:
1. First we write the given equations in the form: y = f(x)
But it is already in this form. $\small{y~=~f(x)~=~3x+2}$
2. We are asked to find the area bounded by four items:
♦ The line y = f(x)
♦ The vertical line x = −1
♦ The vertical line x = 1
♦ The horizontal line y=0 (x-axis)
• So we want the area $\small{\left(A_1 + A_2 \right)}$ in fig.24.14 below. Recall that, area below the x-axis will be calculated with a −ve sign. This is the reason why, we split the region into A1 and A2.
♦ A1 is shaded in magenta color. It is named as EAD
♦ A2 is shaded in yellow color. It is named as ABC
 |
| Fig.24.14 |
• Solving the equations of f(x) and the x-axis, we get:
x = −(2/3) and y = 0
So the red line intersect the x-axis at [−(2/3),0]. This is the point A in fig.24.14
3. First we will find the area of the magenta region.
• The magenta region is bounded by three items:
♦ The line y = f(x)
♦ The vertical line x = −1
♦ The horizontal line y=0 (x-axis)
• Assume a thin vertical strip of width dx, situated at a distance of x from O. So height of the strip will be f(x). So area of the strip will be f(x) dx.
Therefore, area (A1) of the magenta portion can be obtained as:
$\small{A_1~=~\int_{-1}^{-2/3}{\left[f(x) \right]dx}~=~\int_{-1}^{-2/3}{\left[3x+2 \right]dx}}$
• This integration can be done as shown below:
(i) First we find the indefinite integral F1:
$\small{F_1~=~\int{\left[3x+2 \right]dx}~=~\frac{3x^2}{2}~+~2x}$
(ii) Therefore, $\small{A_1~=~\left[\frac{3x^2}{2}~+~2x \right]_{-1}^{-2/3}}$
$\small{~=~\left[\frac{3}{2}\left(\frac{4}{9} \right)~+~2\left(\frac{-2}{3} \right) \right]~-~\left[\frac{3(-1)^2}{2}~+~2(-1) \right]}$
$\small{~=~\left[\frac{2}{3}~-~\frac{4}{3} \right]~-~\left[\frac{3}{2}~-~2 \right]~=~\left[\frac{-2}{3} \right]~-~\left[\frac{-1}{2} \right]}$
$\small{~=~\left[\frac{-2}{3} \right]~+~\left[\frac{1}{2} \right]~=~\frac{-1}{6}}$
• Area can not be −ve. We must take the absolute value. We can write:
$\small{A_1~=~\left| \frac{-1}{6} \right|~=~\frac{1}{6}}$
4. Next we will find the area of the yellow region.
• The yellow region is bounded by three items:
♦ The line y = f(x)
♦ The vertical line x = 1
♦ The horizontal line y=0 (x-axis)
• Assume a thin vertical strip of width dx, situated at a distance of x from O. So height of the strip will be f(x). So area of the strip will be f(x) dx.
•
Therefore, area (A2) of the yellow portion can be obtained as:
$\small{A_2~=~\int_{-2/3}^1{\left[f(x) \right]dx}~=~\int_{-2/3}^1{\left[3x+2 \right]dx}}$
• This integration can be done as shown below:
(i) First we find the indefinite integral F1:
$\small{F_1~=~\int{\left[3x+2 \right]dx}~=~\frac{3x^2}{2}~+~2x}$
(ii) Therefore, $\small{A_2~=~\left[\frac{3x^2}{2}~+~2x \right]_{-2/3}^1}$
$\small{~=~\left[\frac{3(1)^2}{2}~+~2(1) \right]~-~\left[\frac{3}{2}\left(\frac{4}{9} \right)~+~2\left(\frac{-2}{3} \right) \right]}$
$\small{~=~\left[\frac{7}{2} \right]~-~\left[\frac{2}{3}~-~\frac{4}{3} \right]~=~\left[\frac{7}{2} \right]~-~\left[\frac{-2}{3} \right]~=~\frac{25}{6}}$
5. So from (3) and (4), we get:
Required area = $\small{A_1~+~A_2}$
$\small{~=~\frac{1}{6} + \frac{25}{6}~=~\frac{26}{6}~=~\frac{13}{3}}$ sq.units
Solved example 24.11
Find the area of the region bounded by the curve y = cos x between x = 0 and x = 2π.
Solution:
1. First we write the given equations in the form: y = f(x)
But it is already in this form. $\small{y~=~f(x)~=~\cos x}$
2. We are asked to find the area bounded by four items:
♦ The curve y = f(x)
♦ The vertical line x = 0
♦ The vertical line x = 2π
• So we want the area $\small{\left(A_1 + A_2 + A_3 \right)}$ in fig.24.15 below. Recall that, area below the x-axis will be calculated with a −ve sign. This is the reason why, we split the region into A1, A2and A3.
♦ A1 is shaded in yellow color. It is named as OBA
♦ A2 is shaded in magenta color. It is named as BCD
♦ A3 is shaded in green color. It is named as DEF
 |
| Fig.24.15 |
• Solving the equations of f(x) and the x-axis, we get:
cos x = 0 ⇒ x = π/2 and x = 3π/2
So the red curve intersect the x-axis at (π/2,0) and (3π/2,0). These are the point B and D in fig.24.14
3. First we will find the area of the yellow region.
• The yellow region is bounded by three items:
♦ The curve y = f(x)
♦ The vertical line x = 0 (y-axis)
♦ The horizontal line y=0 (x-axis)
• Assume a thin vertical strip of width dx, situated at a distance of x from O. So height of the strip will be f(x). So area of the strip will be f(x) dx.
• Therefore, area (A1) of the yellow portion can be obtained as:
$\small{A_1~=~\int_{0}^{\pi/2}{\left[f(x) \right]dx}~=~\int_{0}^{\pi/2}{\left[\cos x \right]dx}}$
• This integration can be done as shown below:
(i) First we find the indefinite integral F1:
$\small{F_1~=~\int{\left[\cos x \right]dx}~=~\sin x}$
(ii) Therefore, $\small{A_1~=~\left[\sin x \right]_{0}^{\pi/2}}$
$\small{~=~\left[1 \right]~-~\left[0 \right]~=~1}$ sq.unit
4. Next we will find the area of the magenta region.
• The magenta region is bounded by two items:
♦ The curve y = f(x)
♦ The horizontal line y=0 (x-axis)
• Assume a thin vertical strip of width dx, situated at a distance of x from O. So height of the strip will be f(x). So area of the strip will be f(x) dx.
Therefore, area (A2) of the magenta portion can be obtained as:
$\small{A_2~=~\int_{\pi/2}^{3\pi/2}{\left[f(x) \right]dx}~=~\int_{\pi/2}^{3\pi/2}{\left[\cos x \right]dx}}$
• This integration can be done as shown below:
(i) First we find the indefinite integral F1:
$\small{F_2~=~\int{\left[\cos x \right]dx}~=~\sin x}$
(ii) Therefore, $\small{A_1~=~\left[\sin x \right]_{\pi/2}^{3\pi/2}}$
$\small{~=~\left[-1 \right]~-~\left[1 \right]~=~-2}$
• Area can not be −ve. We must take the absolute value. We can write:
$\small{A_2~=~\left| -2 \right|~=~2}$ sq.units
5. Finally, we will find the area of the green region.
• The green region is bounded by three items:
♦ The curve y = f(x)
♦ The vertical line x = 2π
♦ The horizontal line y = 0 (x-axis)
• Assume a thin vertical strip of width dx, situated at a distance of x from O. So height of the strip will be f(x). So area of the strip will be f(x) dx.
• Therefore, area (A3) of the green portion can be obtained as:
$\small{A_3~=~\int_{3\pi/2}^{2\pi}{\left[f(x) \right]dx}~=~\int_{3\pi/2}^{2\pi}{\left[\cos x \right]dx}}$
• This integration can be done as shown below:
(i) First we find the indefinite integral F3:
$\small{F_3~=~\int{\left[\cos x \right]dx}~=~\sin x}$
(ii) Therefore, $\small{A_3~=~\left[\sin x \right]_{3\pi/2}^{2\pi}}$
$\small{~=~\left[0 \right]~-~\left[-1 \right]~=~1}$ sq.unit
6. So from (3), (4) and (5), we get:
Required area = $\small{A_1~+~A_2~+~A_3}$
= 1 + 2 + 1 = 4 sq.units
The link below gives a few more solved examples:
Exercise 24.1
In the next section, we will see area between two curves.
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