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

04 y dx + 48 y dx

= 2 04 √x dx + 48  √[42– (x-4)2]dx

Now take x-4 = t, then the above equation is written in the form 

= 2 04 √x dx + 0 √[42– t2]dx …. (1)

Now, integrate the functions.

04 √x dx = (2)(⅔) (x3/2)04

04  √x dx = 32/3  …..(2)

04  √[42– t2]dx = [(t/2)(√[42-t2] + (½)(42)(sin-1(t/4)]04

04 √[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).