In the previous section, we completed a discussion on integrals. In this chapter, we will see application of integrals.
Area under simple curves
One important application of integrals is the calculation of area under simple curves. Some basics about this application, can be written in steps:
1. In our earlier classes, we have seen the methods for finding areas of simple geometric figures like squares, rectangles, triangles, trapeziums etc.,
• In those simple geometric figures, all sides are straight lines.
2. Now consider fig.24.1 below:
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| Fig.24.1 |
• AB, BC and CD are straight lines.
• But AD is a curve. AD is portion of the graph of the function $\small{f(x) = x^3 + 2x + 3}$
• In such a situation, we use integral calculus to find the area ABCDA.
3. Consider fig.24.2(a) below:
 |
| Fig.24.2 |
• Recall how we calculated the area bounded by the 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 assumed that, the required area is composed of infinite number of very thin strips of width dx. The sum of the areas of all those infinite strips will give the required area. One such sample strip is shown in the fig.24.2(a).
• This sample strip is situated at a distance of ‘x’ from the origin. So height of this sample strip will be y=f(x). Consequently, area of this sample strip will be ydx, which is same as f(x) dx.
•
Note that, since there are infinite strips, the width dx is infinitesimal.
•
Area of the sample strip will be so small that, we denote the area as dA. So dA = f(x) dx.
•
Symbolically, we denoted the sum of all the strips as: $\small{\int_a^b{\left[f(x) \right]dx}}$
•
So we can write: $\small{A~=~\int_a^b{\left[f(x) \right]dx}~=~\int_a^b{\left[dA \right]}}$
4. Consider fig.24.2(b) above.
•
Here also, the area is bounded by the same 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 are tempted to apply the same equation: $\small{A~=~\int_a^b{\left[f(x) \right]dx}}$
But the calculated area will be −ve. We know that, area has no sign. So in such a situation, we must take the absolute value. We can write:
$\small{A~=~\left|\int_a^b{\left[f(x) \right]dx} \right|}$
5. Consider fig.24.3(a) below:
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| Fig.24.3 |
• The area is bounded by the four items:
♦ The curve x = g(x)
♦ The horizontal line y = c
♦ The horizontal line y = d
♦ The vertical line x=0 (y-axis)
• Here we can modify the equation that we saw in step (3) above. We get: $\small{A~=~\int_c^d{\left[g(y) \right]dy}~=~\int_c^d{\left[dA \right]}}$
6. Consider fig.24.3(b) above.
• Here we are tempted to apply the same equation as in step (5) above: $\small{A~=~\int_c^d{\left[g(y) \right]dy}}$
But the calculated area will be −ve. We know that, area has no sign. So in such a situation, we must take the absolute value. We can write:
$\small{A~=~\left|\int_c^d{\left[g(y) \right]dy} \right|}$
7. Consider fig.24.4 below:
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| Fig.24.4 |
•
We want the area is bounded by the four items:
♦ The curve y = f(x)
♦ The vertical line x = a
♦ The vertical line x = b
♦ The horizontal line y=0 (x-axis)
•
Such an area is the sum of two areas:
♦ violet area (A1)
♦ yellow area (A2)
•
If we use the equation $\small{A~=~\int_a^b{\left[f(x) \right]dx}}$, we will be getting $\small{-A_1~+~A_2}$
•
Two avoid this error, we calculate A1 and A2 separately. Then we find the sum: A = |A1| + A2
Now we will see a solved example
Solved example 24.1
Find the area enclosed by the circle $\small{x^2~+~ y^2~=~a^2}$
Solution:
1. First we write the given equation in the form: y = f(x)
$\small{x^2~+~ y^2~=~a^2}$
$\small{\Rightarrow y^2~=~a^2~-~x^2}$
$\small{\Rightarrow y~=~f(x)~=~\pm \sqrt{a^2~-~x^2}}$
• In the above equation, if we input all x values from the interval [−a,a], we will get the ordered pairs, which when plotted, will give the red circle in fig.24.5 below:
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| Fig.24.5 |
2. Consider fig.24.5(a) above. 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 vertical line x = a
♦ 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:
$\small{A~=~\int_0^a{\left[f(x) \right]dx}~=~\int_0^a{\left[\sqrt{a^2~-~x^2} \right]dx}}$
• Here we discard $\small{f(x)~=~-\sqrt{a^2~-~x^2}}$ and take $\small{f(x)~=~+\sqrt{a^2~-~x^2}}$. 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:
$\small{F~=~\int{\left[\sqrt{a^2~-~x^2} \right]dx}}$
• This is a standard integral (see section 23.19). We have:
$\small{F~=~\int{\left[\sqrt{a^2~-~x^2} \right]dx}~=~\frac{x}{2} \sqrt{a^2~-~x^2}~+~\frac{a^2}{2} \sin^{-1}\frac{x}{a}}$
$\small{~=~\left[\frac{a}{2} \sqrt{a^2~-~a^2}~+~\frac{a^2}{2} \sin^{-1}\frac{a}{a} \right]~-~\left[\frac{0}{2} \sqrt{a^2~-~0^2}~+~\frac{a^2}{2} \sin^{-1}\frac{0}{a} \right]}$
$\small{~=~\left[0~+~\frac{a^2}{2} \sin^{-1}1 \right]~-~\left[0~+~\frac{a^2}{2} \sin^{-1}0 \right]}$
$\small{~=~\left[\frac{a^2}{2} \left(\frac{\pi}{2} \right) \right]~-~\left[0~+~0 \right]~=~\frac{\pi a^2}{4}}$
4. In the fig.24.5(a) above, we see that, the circle is symmetrical about both x and y axes. So the total area of the circle is four times the area of OABOA.
•
We can write:
Whole area enclosed by the circle
$\small{~=~(4)\frac{\pi \,a^2}{4}~=~\pi\,a^2}$
•
Note that, this result is in full agreement with the well known formula: Area of a circle = $\small{\pi r^2}$.
In our present case, the radius is 'a'.
Alternate method:
1. First we write the given equation in the form: x = f(y)
$\small{x^2~+~ y^2~=~a^2}$
$\small{\Rightarrow x^2~=~a^2~-~y^2}$
$\small{\Rightarrow x~=~f(y)~=~\pm \sqrt{a^2~-~y^2}}$
• In the above equation, if we input all y values from the interval [−a,a], we will get the ordered pairs, which when plotted, will give the red circle in fig.24.5(a) and (b) above.
2. Consider fig.24.5(b) above. We want the area of the portion shaded in blue color. This portion is named as OBCOB.
• The portion shaded in blue color is bounded by four items:
♦ The curve x = f(y)
♦ The horizontal line y = 0 (x-axis)
♦ The horizontal line y = a
♦ The vertical line x = 0 (y-axis)
• Consider the thin horizontal strip of thickness dy. This strip is situated at a distance of y from O. So length of the strip will be f(y). So area of the strip will be f(y) dy.
Therefore, area (A) of the blue portion can be obtained as:
$\small{A~=~\int_0^a{\left[f(y) \right]dy}~=~\int_0^a{\left[-\sqrt{a^2~-~y^2} \right]dy}}$
• Here we discard $\small{x~=~f(y)~=~\sqrt{a^2~-~y^2}}$ and take $\small{x~=~f(y)~=~-\sqrt{a^2~-~y^2}}$. This is because, OBCOB is in the second quadrant. In this quadrant, all x values are −ve.
3. This integration can be done as shown below:
• First we find the indefinite integral F:
$\small{F~=~\int{\left[-\sqrt{a^2~-~y^2} \right]dy}~=~(-1)\int{\left[\sqrt{a^2~-~y^2} \right]dy}}$
• This is a standard integral (see section 23.19). We have:
$\small{F~=~(-1)\int{\left[\sqrt{a^2~-~y^2} \right]dy}~=~(-1)\left[\frac{x}{2} \sqrt{a^2~-~y^2}~+~\frac{a^2}{2} \sin^{-1}\frac{y}{a} \right]}$
Therefore, $\small{A~=~(-1)\left[\frac{y}{2} \sqrt{a^2~-~y^2}~+~\frac{a^2}{2} \sin^{-1}\frac{y}{a} \right]_0^{a}}$
$\small{~=~(-1)\left[\frac{a}{2} \sqrt{a^2~-~a^2}~+~\frac{a^2}{2} \sin^{-1}\frac{a}{a} \right]~-~(-1)\left[\frac{0}{2} \sqrt{a^2~-~0^2}~+~\frac{a^2}{2} \sin^{-1}\frac{0}{a} \right]}$
$\small{~=~(-1)\left[0~+~\frac{a^2}{2} \sin^{-1}1 \right]~+~\left[0~+~\frac{a^2}{2} \sin^{-1}0 \right]}$
$\small{~=~(-1)\left[\frac{a^2}{2} \left(\frac{\pi}{2} \right) \right]~+~\left[0~+~0 \right]~=~-\frac{\pi a^2}{4}}$
• Area cannot be −ve. We must take the absolute value. So we get:
$\small{A~=~ \left|-\frac{\pi a^2}{4} \right|~=~\frac{\pi a^2}{4}}$
4. Since the circle is symmetrical about both x and y axes, total area of the circle is four times the area of OBCOB.
• We can write:
Whole area enclosed by the circle
$\small{~=~(4)\frac{\pi \,a^2}{4}~=~\pi\,a^2}$
• We get the same result in both methods. For this problem, there is a total of eight methods (vertical or horizontal strip in each of the four quadrants) available. The reader may try all the eight methods to get a good understanding about the process.
Solved example 24.2
Find the area enclosed by the ellipse $\small{\frac{x^2}{a^2}~+~ \frac{y^2}{b^2}~=~1}$
Solution:
1. First we write the given equation in the form: y = f(x)
$\small{\frac{x^2}{a^2}~+~ \frac{y^2}{b^2}~=~1}$
$\small{\Rightarrow \frac{y^2}{b^2}~=~1~-~\frac{x^2}{a^2}}$
$\small{\Rightarrow y^2~=~b^2 \left(1~-~\frac{x^2}{a^2} \right)~=~b^2 \left(\frac{a^2~-~x^2}{a^2} \right)~=~\frac{b^2}{a^2}\left(a^2~-~x^2 \right)}$
$\small{\Rightarrow y~=~f(x)~=~\pm \frac{b}{a} \sqrt{a^2~-~x^2}}$
• In the above equation, if we input all x values from the interval [−a,a], we will get the ordered pairs, which when plotted, will give the red ellipse in fig.24.6 below:
 |
| Fig.24.6 |
2. Consider fig.24.6(a) above. 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 vertical line x = a
♦ 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:
$\small{A~=~\int_0^a{\left[f(x) \right]dx}~=~\int_0^a{\left[\frac{b}{a} \sqrt{a^2~-~x^2} \right]dx}}$
• Here we discard $\small{f(x)~=~-\frac{b}{a} \sqrt{a^2~-~x^2}}$ and take $\small{f(x)~=~+\frac{b}{a} \sqrt{a^2~-~x^2}}$. 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:
$\small{F~=~\frac{b}{a} \int{\left[\sqrt{a^2~-~x^2} \right]dx}}$
• 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, $\small{A~=~\left[\frac{x}{2} \sqrt{a^2~-~x^2}~+~\frac{a^2}{2} \sin^{-1}\frac{x}{a} \right]_0^{a}}$
$\small{~=~\frac{b}{a} \left[\frac{a}{2} \sqrt{a^2~-~a^2}~+~\frac{a^2}{2} \sin^{-1}\frac{a}{a} \right]~-~\frac{b}{a} \left[\frac{0}{2} \sqrt{a^2~-~0^2}~+~\frac{a^2}{2} \sin^{-1}\frac{0}{a} \right]}$
$\small{~=~\frac{b}{a} \left[0~+~\frac{a^2}{2} \sin^{-1}1 \right]~-~\frac{b}{a} \left[0~+~\frac{a^2}{2} \sin^{-1}0 \right]}$
$\small{~=~\frac{b}{a} \left[\frac{a^2}{2} \left(\frac{\pi}{2} \right) \right]~-~\frac{b}{a} \left[0~+~0 \right]~=~\frac{\pi ab}{4}}$
4. In the fig.24.6(a) 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
$\small{~=~(4)\frac{\pi \,ab}{4}~=~\pi\,ab}$
Alternate method:
1. First we write the given equation in the form: x = f(y)
$\small{\frac{x^2}{a^2}~+~ \frac{y^2}{b^2}~=~1}$
$\small{\Rightarrow \frac{x^2}{a^2}~=~1~-~\frac{y^2}{b^2}}$
$\small{\Rightarrow x^2~=~a^2 \left(1~-~\frac{y^2}{b^2} \right)~=~a^2 \left(\frac{b^2~-~y^2}{b^2} \right)~=~\frac{a^2}{b^2}\left(b^2~-~y^2 \right)}$
$\small{\Rightarrow x~=~f(y)~=~\pm \frac{a}{b} \sqrt{b^2~-~y^2}}$
• In the above equation, if we input all y values from the interval [−b,b], we will get the ordered pairs, which when plotted, will give the red ellipse in fig.24.6 above.
2. Consider fig.24.6(a) above. We want the area of the portion shaded in blue color. This portion is named as OA'B'OA'.
• The portion shaded in blue color is bounded by four items:
♦ The curve y = f(x)
♦ The horizontal line y = 0 (x-axis)
♦ The horizontal line y = −b
♦ The vertical line x = 0 (y-axis)
• Consider the thin horizontal strip of thickness dy. This strip is situated at a distance of y from O. So length of the strip will be f(y). So area of the strip will be f(y) dy.
Therefore, area (A) of the blue portion can be obtained as:
$\small{A~=~\int_{-b}^0{\left[f(y) \right]dy}~=~\int_{-b}^0{\left[-\frac{a}{b} \sqrt{b^2~-~y^2} \right]dy}}$
• Here we discard $\small{f(y)~=~+\frac{a}{b} \sqrt{b^2~-~y^2}}$ and take $\small{f(y)~=~-\frac{a}{b} \sqrt{b^2~-~y^2}}$. This is because, OA'B'OA' is in the third quadrant. In this quadrant, all x values are −ve.
3. This integration can be done as shown below:
• First we find the indefinite integral F:
$\small{F~=~-\frac{a}{b} \int{\left[\sqrt{b^2~-~y^2} \right]dy}}$
• This is a standard integral (see section 23.19). We have:
$\small{F~=~\int{\left[\sqrt{b^2~-~y^2} \right]dy}~=~\frac{y}{2} \sqrt{b^2~-~y^2}~+~\frac{b^2}{2} \sin^{-1}\frac{y}{b}}$
Therefore, $\small{A~=~\left[\frac{y}{2} \sqrt{b^2~-~y^2}~+~\frac{b^2}{2} \sin^{-1}\frac{y}{b} \right]_{-b}^{0}}$
$\small{~=~\left(-\frac{a}{b} \right) \left[\frac{0}{2} \sqrt{b^2~-~0^2}~+~\frac{b^2}{2} \sin^{-1}\frac{0}{b} \right]~-~\left(-\frac{a}{b} \right) \left[\frac{-b}{2} \sqrt{b^2~-~(-b)^2}~+~\frac{b^2}{2} \sin^{-1}\frac{-b}{b} \right]}$
$\small{~=~\left(-\frac{a}{b} \right) \left[0~+0 \right]~+~\left(\frac{a}{b} \right) \left[0~+~\frac{b^2}{2} \sin^{-1}(-1) \right]~~\because \sin^{-1}(-1)~=~-\sin^{-1}(1) }$
$\small{~=~0~+~\left(\frac{a}{b} \right) \left[-\frac{b^2}{2} \sin^{-1}(1) \right]}$
$\small{~=~\frac{a}{b} \left[-\frac{b^2}{2} \left(\frac{\pi}{2} \right) \right]~=~\frac{-\pi ab}{4}}$
• Area cannot be −ve. We must take the absolute value. So we get:
$\small{A~=~ \left|-\frac{\pi ab}{4} \right|~=~\frac{\pi ab}{4}}$
4. In the fig.24.6(b) 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 OA'B'OA'.
• We can write:
Whole area enclosed by the ellipse
$\small{~=~(4)\frac{\pi \,ab}{4}~=~\pi\,ab}$
•
We get the same result in both methods. For this problem, there is a
total of eight methods (vertical or horizontal strip in each of the
four quadrants) available. The reader may try all the eight methods to
get a good understanding about the process.
In the next section, we will see area bounded by a curve and a line.
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