Class 12 Maths Application of Integrals Area between Two Curves

Area between Two Curves

The most general class of problems in calculating the area of bounded regions is the one involving the regions between two curves. Let we are given two curves represented by y = f(x), y = g(x), where f(x) ≥ g(x) in [a, b] as shown in figure.

         Class_12_Maths_Applications_Of_Integrals_Area_Between_Two_Curves                                       

Here the points of intersection of these two curves are given by x = a and x = b obtained by taking common values of y from the given equation of two curves.

Elementary strip has height f(x) – g(x) and width dx so that the elementary area

dA = [f(x) – g(x)] dx

Total area A can be calculated as A = ab [f(x) – g(x)]

Alternatively,

A = [area bounded by y = f (x), x-axis and the lines x = a, x = b]

– [area bounded by y = g (x), x-axis and the lines x = a, x = b]

A = ab f(x) = ab g(x) = ab [f(x) – g(x)], where f(x) ≥ g(x) in [a, b]

If f (x) ≥ g (x) in [a, c] and f (x) ≤ g (x) in [c, b], where a < c < b as shown in the figure, then the area of the regions bounded by curves can be calculated as

Total Area = Area of the region ACBDA + Area of the region BPRQB.

                   = ac [f(x) – g(x)] + cb [g(x) – f(x)]  

Class_12_Maths_Applications_Of_Integrals_Area_Between_Two_Curves_1

                            

Example:

Find the area of the circle 4x2 + 4y2 = 9 which is interior to the parabola x2 = 4y.

Solution:

The required area is represented by the shaded area OBCDO.

    Class_12_Maths_Applications_Of_Integrals_Area_Between_Two_Curves_2                                              

Solving the given equation of circle, 4x2 + 4y2 = 9, and parabola, x2 = 4y, we get the point of intersection as B(√2, 1/2) and D(-√2, 1/2).

It can be observed that the required area is symmetrical about y-axis.

So, Area OBCDO = 2 * Area OBCO

Now, draw BM perpendicular to OA.

Therefore, the coordinates of M is (√2, 0).

Therefore, Area OBCO = Area OMBCO – Area OMBO

                                         = ʃ0√2  {√(9 – 4x2)/4} dx - ʃ0√2 x2/4 dx

                                         = (1/2)ʃ0√2 √(9 – 4x2) dx – (1/4)ʃ0√2 x2 dx

                                        = (1/4)[x√(9 – 4x2) + (9/2) * sin-1 2x/3]0√2  - (1/4)[x3/3]0√2            

                                        = (1/4)[√2 * √(9 – 8) + (9/2) * sin-1 2√2/3] - (1/12)(√2)3

                                        = (√2/4) + (9/8) * sin-1 2√2/3 - √2/6

                                        = (√2/12) + (9/8) * sin-1 2√2/3

                                        = (1/2)[√2/6 + (9/4) * sin-1 2√2/3]

Therefore, the required area OBCDO = 2 * (1/2)[√2/6 + (9/4) * sin-1 2√2/3]

                                                                   = [√2/6 + (9/4) * sin-1 2√2/3] units

Example:

Find the area between the curves y = x and y = x2

Solution:

The required area is represented by the shaded area OBAO as

The points of intersection of the curves, y = x and y = x2 is A (1, 1).

Now, draw AC perpendicular to x-axis.

Class_12_Maths_Applications_Of_Integrals_Area_Between_Two_Curves_3

So, Area (OBAO) = Area (∆OCA) – Area (OCABO)

                              = ʃ01 x dx - ʃ01 x2 dx

                              = [x2/2]01 – [x3/3]01

                              = 1/2 - 1/3

                              = 1/6 units

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