Area Using Integration Solution5





In this page area using integration solution5 we are going to see solution of some practice questions.

(9) Find the common area enclosed by the parabolas 4 y² = 9 x and 3 x² = 16 y.

Solution:

4 y² = 9 x  ---(1)

3 x² = 16 y  ---- (2)

y = (3 √x)/2

y =  3 x²/16

y = y

(3 √x)/2 = 3 x²/16

48 √x = 6 x²

(48/6)√x = x²

8 √x = x²

√x = x²/8

taking squares on both sides

 x = x⁴/64

 64 = x⁴/x

 64 = x³

 4³ = x³

 x = 4

now we are going to apply the value of x in the first or second equation get the value of y

 y = 3 (4)²/16

 y = 3

point of intersection are (0,0) and (4,3)

To find the required area that is shaded portion we have to subtract the area which is open upward from the area which is open rightward.

                        4                     4

Required area = ∫ (3/2) √x dx - ∫ 3 x²/16 dx

                       0                     0 

                                4                             4   

                     = (3/2)∫ x^(1/2) dx - (3/16)∫x² dx

                               0                              0

                                                     4                   4

                     = (3/2) [x^(3/2)/(3/2)] - (3/16)[x³/3]

                                                     0                   0         

                     = (3/2)(2/3) [4^(3/2) - 0] - (3/16)(1/3)[4³-0]                     

                     = 4 √4 - (1/16)[64]                     

                     = 4 (2) - (1/16)[64]                     

                     = 8 - 4                     

                     = 4 square units.

Therefore the required area is 4 square units.


(10) Find the area of the circle whose radius is a

Solution:

Equation of circle :

            x² + y² = a²

                   y² = a² - x²

                   y = √(a² - x²)

                        b

Required area = ∫ y dx

                       a

by using the limits x = 0 and x = a we can find the area only lies above the x axis. To find the total area of circle we have to multiply it by 4.

                           a

Required area =  4 ∫ √(a² - x²) dx

                           0

                                                                                             a

                   = 4 [(x/2)√(a² - x²) + a²/2 sin⁻¹ (X/a)]       

                                                                                             0

                   = 4[(a/2)√(a² - a²) + (a²/2) sin⁻¹ (a/a)] - 0 - 0]

                   = 4[0 + (a²/2) sin⁻¹ (1)]

                   = 4[0 + (a²/2) (π/2)]

                   = 4[(πa²/4)]

                   = πa² square units

Therefore the required area of circle is πa² square units.

area using integration solution5 area using integration solution5