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Equation of the parabola is y^{2} = 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 = âˆ«_{y=a}^{y=b} f(y) dy

Height of strip = Function of parabola - Function of Y - axis = y^{2}/6 - x

= y^{2}/6 - 0 = y^{2}/6

âˆ´ y^{2}/6 = f(y)

Area of the region = âˆ«_{y=0}^{y=4} f(y) dy

= âˆ«_{y=0}^{y=4} y^{2}/6 dy

= 1/6 .âˆ«_{y=0}^{y=4} y^{2} dy = 1/6[y^{3}/3]_{0}^{4}

= 1/6[4^{3}/3 - 0/3]

= 1/6[4^{3}/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 = âˆ«_{x=a}^{x=b} f(x) dx

Height of strip = Function of line - Function of parabola

f(x) = 4 - âˆš(6x)

Area of the region = âˆ«_{x=a}^{x=b} (4 - âˆš(6x)) dx

Solving y^{2} = 6x and y = 4, we get

16 = 6x â‡’ x = 16/6 = 8/3 âˆ´ A = (8/3 , 4)

Area of the region = âˆ«_{0}^{8/3} (4 - âˆš(6x)) dx

= [4x - âˆš6 . x^{3/2}/(3/2)]_{0}^{8/3}

= [4 x 8/3 - 2/3 âˆš6 . (8/3)^{3/2}] - 0

= 32/3 - 64/9 = 32/9 sq. Units.

Equation of the parabola is y^{2} = 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 = âˆ«_{y=a}^{y=b} f(y) dy

Height of strip = Function of parabola - Function of Y - axis = y^{2}/6 - x

= y^{2}/6 - 0 = y^{2}/6

âˆ´ y^{2}/6 = f(y)

Area of the region = âˆ«_{y=0}^{y=4} f(y) dy

= âˆ«_{y=0}^{y=4} y^{2}/6 dy

= 1/6 .âˆ«_{y=0}^{y=4} y^{2} dy = 1/6[y^{3}/3]_{0}^{4}

= 1/6[4^{3}/3 - 0/3]

= 1/6[4^{3}/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 = âˆ«_{x=a}^{x=b} f(x) dx

Height of strip = Function of line - Function of parabola

f(x) = 4 - âˆš(6x)

Area of the region = âˆ«_{x=a}^{x=b} (4 - âˆš(6x)) dx

Solving y^{2} = 6x and y = 4, we get

16 = 6x â‡’ x = 16/6 = 8/3 âˆ´ A = (8/3 , 4)

Area of the region = âˆ«_{0}^{8/3} (4 - âˆš(6x)) dx

= [4x - âˆš6 . x^{3/2}/(3/2)]_{0}^{8/3}

= [4 x 8/3 - 2/3 âˆš6 . (8/3)^{3/2}] - 0

= 32/3 - 64/9 = 32/9 sq. Units.