- Books Name
- Mathmatics Book Based on NCERT
- Publication
- KRISHNA PUBLICATIONS
- Course
- CBSE Class 12
- Subject
- Mathmatics
To find the area of the region bounded by a curve and a line
Let’s have a look at the example to understand how to find the area of the region bounded by a curve and a line.
Example 2:
Find the area of the region bounded between the line x = 2 and the parabola y2 = 8x.
Solution:
Given equation of parabola is y2 = 8x.
Equation of line is x = 2.
Here, y2 = 8x as a right handed parabola having its vertex at the origin and x = 2 is the line which is parallel to y-axis at x = 2 units distance
Similarly,
y2 = 8x has only even power of y and is symmetrical about x-axis.
So, the required area = Area of OAC + Area of OAB
= 2 (Area of OAB)
= 2 ∫02 y dx
Substituting the value of y, i.e. y2 = 8x and y = √(8x) = 2 √2 √x, we get;
= 2 ∫02 (2 √2 √x) dx
= 4√2 ∫02 (√x) dx
= 4√2 [x3/2/ (3/2)]02
By applying the limits,
= 4√2 {[23/2/ (3/2)] – 0}
= (8√2/3) × 2√2
= (16 × √2 × √2)
= 32/3
Go through the example given below to learn how to find the area between two curves.
Example 3:
Determine the area which lies above the x-axis and included between the circle and parabola, where the circle equation is given as x2+y2 = 8x, and parabola equation is y2 = 4x.
Solution:
The circle equation x2+y2 = 8x can be written as (x-4)2+y2=16. Hence, the centre of the circle is (4, 0), and the radius is 4 units. The intersection of the circle with the parabola y2 = 4x is as follows:
Now, substitute y2 = 4x in the given circle equation,
x2+4x = 8x
x2– 4x = 0
On solving the above equation, we get
x=0 and x=4
Therefore, the point of intersection of the circle and the parabola above the x-axis is obtained as O(0,0) and P(4,4).
Hence, from the above figure, the area of the region OPQCO included between these two curves above the x-axis is written as
= Area of OCPO + Area of PCQP
= 0∫4 y dx + 4∫8 y dx
= 2 0∫4 √x dx + 4∫8 √[42– (x-4)2]dx
Now take x-4 = t, then the above equation is written in the form
= 2 0∫4 √x dx + 0∫4 √[42– t2]dx …. (1)
Now, integrate the functions.
2 0∫4 √x dx = (2)(⅔) (x3/2)04
2 0∫4 √x dx = 32/3 …..(2)
0∫4 √[42– t2]dx = [(t/2)(√[42-t2] + (½)(42)(sin-1(t/4)]04
0∫4 √[42– t2]dx = 4π …..(3)
Now, substitute (2) and (3) in (1), we get
= (32/3) + 4π
= (4/3) (8+3π)
Therefore, the area of the region that lies above the x-axis, and included between the circle and parabola is (4/3) (8+3π).