Notes On Area of a Region Bounded by a Curve and a Line - CBSE Class 12 Maths
Equation of the parabola is y2 = 6x and equation of the line is y = 4. The entire shaded region if we move it horizontally from y =0 to y = 4. Area of the region = ∫y=ay=b f(y) dy Height of strip = Function of parabola - Function of Y - axis = y2/6 - x                        = y2/6 - 0 = y2/6                       ∴ y2/6 = f(y) Area of the region = ∫y=0y=4 f(y) dy                           = ∫y=0y=4 y2/6 dy                           = 1/6 .∫y=0y=4 y2 dy = 1/6[y3/3]04                          = 1/6[43/3  - 0/3]                          = 1/6[43/3] = 64 / 18 = 32 / 9 sq.units Therefore, the area of the required region is 32/9 square units. Alternatively, the area of the shaded region can also be found by considering a strip of width dx parallel to the y-axis. Area of the region = ∫x=ax=b f(x) dx Height of strip = Function of line - Function of parabola f(x) = 4 - √(6x) Area of the region = ∫x=ax=b (4 - √(6x)) dx Solving y2 = 6x and y = 4, we get 16 = 6x        ⇒ x = 16/6 = 8/3          ∴ A = (8/3 , 4) Area of the region = ∫08/3 (4 - √(6x)) dx                            = [4x - √6 . x3/2/(3/2)]08/3                            = [4 x 8/3 - 2/3 √6 . (8/3)3/2] - 0                            = 32/3 - 64/9 = 32/9 sq. Units.

#### Summary

Equation of the parabola is y2 = 6x and equation of the line is y = 4. The entire shaded region if we move it horizontally from y =0 to y = 4. Area of the region = ∫y=ay=b f(y) dy Height of strip = Function of parabola - Function of Y - axis = y2/6 - x                        = y2/6 - 0 = y2/6                       ∴ y2/6 = f(y) Area of the region = ∫y=0y=4 f(y) dy                           = ∫y=0y=4 y2/6 dy                           = 1/6 .∫y=0y=4 y2 dy = 1/6[y3/3]04                          = 1/6[43/3  - 0/3]                          = 1/6[43/3] = 64 / 18 = 32 / 9 sq.units Therefore, the area of the required region is 32/9 square units. Alternatively, the area of the shaded region can also be found by considering a strip of width dx parallel to the y-axis. Area of the region = ∫x=ax=b f(x) dx Height of strip = Function of line - Function of parabola f(x) = 4 - √(6x) Area of the region = ∫x=ax=b (4 - √(6x)) dx Solving y2 = 6x and y = 4, we get 16 = 6x        ⇒ x = 16/6 = 8/3          ∴ A = (8/3 , 4) Area of the region = ∫08/3 (4 - √(6x)) dx                            = [4x - √6 . x3/2/(3/2)]08/3                            = [4 x 8/3 - 2/3 √6 . (8/3)3/2] - 0                            = 32/3 - 64/9 = 32/9 sq. Units.

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