In the previous section, we completed a discussion on area bounded by a curve and another curve. In this section, we will see some miscellaneous examples.
Solved example 24.17
Find the area of the parabola y2 = 4ax bounded by it's latus rectum.
Solution:
1. We know that:
♦ y2 = 4ax is a parabola with the vertex at the origin.
♦ The latus rectum has the equation x = a.
2. First we write the equation of the parabola in the form y = f(x). We get:
$\small{y~=~f(x)~=~\pm 2 \sqrt{ax}}$
• This is the red parabola in fig.24.23 below:
 |
| Fig.24.23 |
• Since the parabola is symmetrical about the x-axis, twice the blue area will give us the required result.
3. So our next task is to find the blue area. This area is bounded by three items:
♦ The curve y = f(x)
♦ The vertical line x = a
♦ The horizontal line y = 0 (x-axis)
• So the interval is: [0,a]
• Therefore we can write:
Blue area = $\small{\int_0^a{\left[f(x) \right]dx}~=~\int_0^a{\left[2 \sqrt{ax} \right]dx}~=~2 \sqrt{a} \int_0^a{\left[\sqrt{x} \right]dx}}$
[Here we use $\small{2 \sqrt{ax}}$
Instead of $\small{-2 \sqrt{ax}}$
This is because, in the first quadrant, y values are +ve]
$\small{~=~2 \sqrt{a} \left[\frac{x^{3/2}}{3/2} \right]_0^a~=~4 \sqrt{a} \left[\frac{x^{3/2}}{3} \right]_0^a~=~4 \sqrt{a} \left[\frac{a^{3/2}}{3} \right]}$
$\small{~=~4 \sqrt{a} \left[\frac{a \sqrt a}{3} \right]~=~\frac{4 a^2}{3}}$ sq.units
4. Twice the blue area = $\small{\frac{8 a^2}{3}}$ sq.units
Solved example 24.18
Prove that the curves y2 = 4x and x2 = 4y divide the area of the square bounded by x = 0, x = 4, y = 4 and y = 0 into three equal parts.
Solution:
1. First we represent the given curves as functions:
$\small{y~=~f(x)~=~\pm 2 \sqrt x~~~~\rm{and}~~~~y~=~g(x)~=~\frac{x^2}{4}}$
•
Solving the two equations, we see that, they intersect at (0,0), (4,4) and (4,−4).
•
We are interested in (0,0) and (4,4) because, they are in the I quadrant. They are marked as O and A in the fig.24.24 below:
 |
| Fig.24.24 |
•
The horizontal line can be represented as: y = h(x) = 4
2. The blue area can be obtained as:
$\small{\int_0^4{\left[h(x) - f(x) \right]dx}~=~\int_0^4{\left[4 - 2 \sqrt x \right]dx}~=~\frac{16}{3}}$ sq.units
(The reader may write all steps related to the integration process)
3. The magenta area can be obtained as:
$\small{\int_0^4{\left[f(x) - g(x) \right]dx}~=~\int_0^4{\left[2 \sqrt x - \frac{x^2}{4} \right]dx}~=~\frac{16}{3}}$ sq.units
(The reader may write all steps related to the integration process)
4. The violet area can be obtained as:
$\small{\int_0^4{\left[g(x) \right]dx}~=~\int_0^4{\left[\frac{x^2}{4} \right]dx}~=~\frac{16}{3}}$ sq.units
(The reader may write all steps related to the integration process)
5. One third of the area of the square ABOC
$\small{~=~\frac{1}{3} (4 \times 4)~=~\frac{16}{3}}$ sq.units
6. From steps (2), (3), (4) and (5), we see that:
Area of blue = Area of magenta = Area of violet
= one third of the area of the square.
• That means, the curves divide the square into three equal parts.
Solved example 24.19
Find the area of the region
$\small{\{(x,y):0 \le y\le x^2 + 1,~~0 \le y\le x + 1,~~0 \le x \le 2 \}}$
Solution:
•
In this problem, a region is given as a set.
♦ That set contains three inequalities.
♦ Any ordered pair (x,y) which satisfies all the three inequalities, is eligible to be a member of the set.
♦ For any inequality, there will be a corresponding region on the x-y plane. That is., any point (x,y) in that region will satisfy that inequality.
♦ In our present case, there will be three regions. The points in the intersection of the three regions, will satisfy all three inequalities.
♦ We are asked to find the area of that intersection region.
1. In the fig.24.25 below, the red curve represents $\small{y~=~f(x)~=~x^2 + 1}$
 |
| Fig.24.25 |
• The inequality: $\small{0 \le y\le x^2 + 1}$ will be represented by a region. This region is hatched with vertical lines. That means, if any region has vertical lines, that region will be part of the inequality.
• Note that all the region below the red curve has vertical lines.
• However, the vertical lines do not extend below the x-axis. This is because, it is specified that 0 ≤ y
2. In the fig.24.25 above, the green line represents $\small{y~=~g(x)~=~x + 1}$
• The inequality: $\small{0 \le y\le x + 1}$ will be represented by a region. This region is hatched with slanting lines. That means, if any region has slanting lines, that region will be part of the inequality.
• Note that all the region below the green line has slanting lines.
• However, the slanting lines do not extend below the x-axis. This is because, it is specified that 0 ≤ y
3. In the fig.24.25 above, the magenta line represents $\small{x~=~2}$
• The inequality: $\small{0 \le x\le 2}$ will be represented by a region. This region is hatched with horizontal lines. That means, if any region has horizontal lines, that region will be part of the inequality.
• Note that all the region to the left of the magenta line has horizontal lines.
• However, the horizontal lines do not extend to the left of the y-axis. This is because, it is specified that 0 ≤ x
4. We see that, a region has all the three type:
Vertical, slanting and horizontal lines.
• This region is the intersection of all the three regions that we saw above. Such a region will satisfy all three inequalities. We are asked to find the area of that region.
• This region of intersection is shown in fig.24.26 below:
 |
| Fig.24.26 |
•
Coordinates of A, B and C can be obtained by solving appropriate pairs of equations
5. Area of pink region
$\small{~=~\int_0^1{\left[f(x) \right]dx}~=~\int_0^1{\left[x^2 + 1 \right]dx}~=~\frac{4}{3}}$ sq.units
(The reader may write all steps related to the integration process)
6. Area of blue region
$\small{~=~\int_1^2{\left[g(x) \right]dx}~=~\int_1^2{\left[x + 1 \right]dx}~=~\frac{5}{2}}$ sq.units
(The reader may write all steps related to the integration process)
7. So we get:
Pink + Blue = $\small{\frac{4}{3}~+~\frac{5}{2}~=~\frac{23}{6}}$ sq.units
Solved example 24.20
Find the area of the region
$\small{\{(x,y): y^2 \le 4x,~~4x^2 + 4y^2 \le 9 \}}$
Solution:
•
In this problem, a region is given as a set.
♦ That set contains two inequalities.
♦ Any ordered pair (x,y) which satisfies both the inequalities, is eligible to be a member of the set.
♦ For any inequality, there will be a corresponding region on the x-y
plane. That is., any point (x,y) in that region will satisfy that
inequality.
♦ In our present case, there will be two regions.
The points in the intersection of the two regions, will satisfy both the inequalities.
♦ We are asked to find the area of that intersection region.
1. In the fig.24.27 below, the red curve represents $\small{y~=~f(x)~=~\pm 2 \sqrt x}$
 |
| Fig.24.27 |
•
The inequality: $\small{y^2 \le 4x}$ will be represented by a region.
This region is hatched with vertical lines. That means, if any region
has vertical lines, that region will be part of the inequality.
• Note that all the region inside the red curve has vertical lines.
2. In the fig.24.27 above, the green curve represents $\small{y~=~g(x)~=~\pm \sqrt{\frac{9}{4}~-~x^2}}$
•
The inequality: $\small{4x^2 + 4y^2 \le 9}$ will be represented by a
region. This region is hatched with horizontal lines. That means, if
any region has horizontal lines, that region will be part of the
inequality.
• Note that all the region inside the green circle has horizontal lines.
3. We see that, a particular region has both the types:
Vertical and horizontal lines.
•
This region is the intersection of all the two regions that we saw
above. Such a region will satisfy both the inequalities. We are asked to
find the area of that region.
• This region of intersection is shown in fig.24.28 below:
 |
| Fig.24.28 |
• Coordinates of A and B can be obtained by solving the two equations
4. Area of blue region
$\small{~=~\int_0^{0.5}{\left[g(x)
- f(x)
\right]dx}~=~\int_0^{0.5}{\left[\sqrt{\frac{9}{4}~-~x^2}~-~\left(2 \sqrt
x \right) \right]dx}}$ sq.units
$\small{~=~\left[\frac{27}{24} \sin^{-1}\left(\frac{1}{3} \right)~-~\frac{\sqrt 2}{12} \right]}$ sq.units
(The reader may write all steps related to the integration process)
[Here we use $\small{\sqrt{\frac{9}{4}~-~x^2}}$
Instead of $\small{-\sqrt{\frac{9}{4}~-~x^2}}$
This is because, in the first quadrant, y values are +ve]
6. Blue + Magenta gives one fourth the total area of the circle.
• So area of magenta region
$\small{~=~\Bigg[\frac{\pi
(3/2)^2}{4}~-~\left[\frac{27}{24} \sin^{-1}\left(\frac{1}{3}
\right)~-~\frac{\sqrt 2}{12} \right]\Bigg]}$ sq.units
$\small{~=~\Bigg[\frac{9 \pi }{16}~-~\frac{27}{24} \sin^{-1}\left(\frac{1}{3} \right)~+~\frac{\sqrt 2}{12} \Bigg]}$ sq.units
7. The area of intersection, is twice the magenta area. So we can write:
Required
area $\small{~=~\Bigg[\frac{9 \pi }{8}~-~\frac{27}{12}
\sin^{-1}\left(\frac{1}{3} \right)~+~\frac{\sqrt 2}{6} \Bigg]}$ sq.units
$\small{~=~\Bigg[\frac{9 \pi }{8}~-~\frac{9}{4} \sin^{-1}\left(\frac{1}{3} \right)~+~\frac{1}{3 \sqrt 2} \Bigg]}$ sq.units
Solved example 24.21
Find the area bounded by the curves
$\small{\{(x,y): y \ge x^2,~~\rm{and}~~y~=~|x| \}}$
Solution:
•
In this problem, the set contains an inequality.
♦ For any inequality, there will be a corresponding region on the x-y
plane. That is., any point (x,y) in that region will satisfy that
inequality.
♦ The above mentioned region should be bounded by the line $\small{y~=~|x|}$.
♦ Any point (x,y), in that bounded region is eligible to be a member of the set.
♦ We are asked to find the area of that bounded region.
1. In the fig.24.29 below, the red curve represents $\small{y~=~f(x)~=~x^2}$. It is a parabola.
 |
| Fig.24.29 |
The interior of the parabola is hatched with vertical lines. Any point (x,y) in the interior will satisfy the inequality $\small{y \ge x^2}$. It is an infinite region.
2. The above mentioned infinite region is bounded by the green line and pink line.
♦ The green line represent $\small{y~=~g(x)~=~x}$
♦ The pink line represent $\small{y~=~h(x)~=~-x}$
♦ The two lines together represent $\small{y~=~|x|}$
•
The bounded region is hatched with both vertical and horizontal lines. We want the area of this hatched region.
3. The equations of the curves can be solved to find the points of intersection.
♦ Solving red and green, we get: A(1,1)
♦ Solving red and pink, we get: B(−1,1)
4. The area AOA
$\small{~=~\int_0^{1}{\left[g(x) - f(x) \right]dx}~=~\int_0^{1}{\left[x~-~x^2 \right]dx}~=~\frac{1}{6}}$ sq.units
(The reader may write all steps related to the integration process)
5. Since the graph is symmetrical, we can write:
Required area = $\small{2\left(\frac{1}{6} \right)~=~\frac{1}{3}}$ sq.units
Solved example 24.22
Using the method of integration find the area bounded by the curve $\small{|x| + |y| = 1}$
[Hint: The required region is bounded by lines $\small{x+y=1,~x-y=1,~-x+y=1,~\rm{and}~-x-y=1}$]
Solution:
1. Consider the equation $\small{|x| + |y| = 1}$
•
For this equation, four cases are possible:
(i) x is +ve and y is also +ve
Then |x| is x. |y| is y
The equation becomes $\small{x+y=1}$
(ii) x is +ve and y is −ve
Then |x| is x. |y| is −y
The equation becomes $\small{x-y=1}$
(iii) x is −ve and y is +ve
Then |x| is −x. |y| is y
The equation becomes $\small{-x+y=1}$
(iv) x is −ve and y is also −ve
Then |x| is −x. |y| is −y
The equation becomes $\small{-x-y=1}$
2. Now we can plot the graphs:
•
Case (i) gives:
$\small{y = f(x) = 1-x}$
This is the red line in fig.24.30 below:
 |
| Fig.24.30 |
•
Case (ii) gives:
$\small{y = g(x) = x-1}$
This is the green line in fig.24.30 above.
•
Case (iii) gives:
$\small{y = h(x) = 1+x}$
This is the pink line in fig.24.30 above.
•
Case (iv) gives:
$\small{y = j(x) = -x-1}$
This is the blue line in fig.24.30 above.
3. We are asked to find the area of ABCD
•
This area is the sum of the areas of OAB, OBC, OCD and ODA
•
OAB is shaded with magenta color. It's area
$\small{~=~\int_0^{1}{\left[f(x) \right]dx}~=~\int_0^{1}{\left[1-x \right]dx}~=~\frac{1}{2}}$ sq.units
4. Due to symmetry, all four areas are equal.
Therefore, total area of ABCD = $\small{4\left(\frac{1}{2} \right)~=~2}$ sq.units
Solved example 24.23
Sketch the graph of y = |x+3| and evaluate $\small{\int_{-6}^{0}{\left[\left|x+3 \right| \right]dx}}$
Solution:
1. Consider the equation $\small{y~=~\left|x+3 \right|}$
• For this equation, two cases are possible:
(i) x is less than −3
Then (x+3) is −ve. So |x+3| is −(x+3)
The equation becomes $\small{y=-x-3}$
(ii) x is greater than −3
Then (x+3) is +ve. So |x+3| is +(x+3)
The equation becomes $\small{y=x+3}$
2. Now we can plot the graphs:
• Case (i) gives:
$\small{y = f(x) = -x-3}$
This is the red line in fig.24.31 below:
 |
| Fig.24.31 |
• Case (ii) gives:
$\small{y = g(x) = x+3}$
This is the green line in fig.24.31 above.
The two lines together represent y = |x+3|
3. We are asked to find the area of magenta + blue
• Magenta area
$\small{~=~\int_{-6}^{-3}{\left[f(x) \right]dx}~=~\int_{-6}^{-3}{\left[-x-3 \right]dx}~=~\frac{9}{2}}$ sq.units
• Blue area
$\small{~=~\int_{-3}^{0}{\left[f(x) \right]dx}~=~\int_{-3}^{0}{\left[x+3 \right]dx}~=~\frac{9}{2}}$ sq.units
4. Therefore, the required area = $\small{2\left(\frac{9}{2} \right)~=~9}$ sq.units.
In the next section, we will see a few more miscellaneous examples.
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