Area Using Integration Solution4





In this page area using integration solution4 we are going to see solution of some practice questions.

(7) Find the area of the region bounded by the ellipse (x²/9) + (y²/5) = 1 between the two latus rectum.

Solution:

From the given equation we come to know that the given ellipse is symmetric about x axis.

a² = 9

b² = 5

e = √[1-(b²/a²)]

e = √[1-(5/9)]

   = √[(9-5)/9]

   = √(4/9)

e = 2/3

equation of latus rectum x = ± ae

a = 3  e = 2/3

 ae = 3 (2/3)

 ae = 2

Equation of latus rectum x = ± 2

                        b

Required area = ∫ y dx

                       a

(x²/9) + (y²/5) = 1

(y²/5) = 1 - (x²/9)

y²/5 = (9 - x²)/9

y²= (5/9) (9 - x²)

y = √(5/9) (9 - x²)

y = √5/3 √(9 - x²)

By using the limits x = 0 and x = 2 we can find area above the x-axis.To find the total shaded area we have to multiply the above area by 4.

                                2     

                   = 4 ∫ √5/3 √(9 - x²) dx

                         0       

                                          2     

                   = 4√5/3  ∫ √(9 - x²) dx

                                 0                          

                                                                                                   2

                   = 4√5/3 [(x/2)√(9 - x²) + 9/2 sin⁻¹ (X/3)]       

                                                                                                  0

                     
= 4√5/3 [(2/2)√(9-2²)+(9/2)sin⁻¹ (2/3)]-[(2/2)√(9-0²)+9/2 sin⁻¹ (0/3)]

= 4√5/3 [√5+(9/2)sin⁻¹ (2/3)]-[0+9/2 sin⁻¹ (0)]

= 4√5/3 [√5+(9/2)sin⁻¹ (2/3)]-[0+9/2 (0)]

= 4√5/3 [√5+(9/2)sin⁻¹ (2/3)] square units


(8) Find the area of the region bounded by the parabola y² = 4 x and the line 2 x - y = 4.

Solution:

To find the point of intersection we have to solve both equations.

  x = y²/4   ------ (1)

  x = (4 + y)/2 ------ (2)

 (1) = (2)

y²/4 = (4 + y)/2

2 y² = 4 (4 + y)

2 y² = 16 + 4 y

 2 y² - 4 y- 16 = 0

now we are going to divide the whole equation by 2

 y² - 2 y- 8 = 0

 (y - 4) (y + 2) = 0

 y - 4 = 0          y + 2 = 0

  y = 4                   y = -2

 point of intersection of two curves are (0,4) (0,-2)

                        b

Required area = ∫ x dy

                       a

                             4      

                   =  ∫ [(4 + y)/2]-[y²/4] dy

                      -2        

                                                                              4

                   =  (1/2)[4 y + y²/2] - [y³/12]      

                                                                             -2

                                                                        4

                   = {[(8y + y²)/4] - [y³/12]}      

                                                                     - 2

                   = [ (32 + 16)/4 - (64/12) ] - [ (-16 + 4)/4 - (-8/12) ]

                   = [(48/4) - (16/3)] - [(-12/4) - (-8/12)]

                   = [12-(16/3)] - [-3 + (2/3)]

                   = [(36-16)/3)] - [(-9+2)/3]

                   = [20/3] - [-7/3]

                   = (20/3) + (7/3)

                   = (20+7)/3

                   = 27/3

                   = 9 square units    

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