Area of the region bounded by the curve y 2 = 4 x , y -axis and the line y = 3 is A. 2 B. C. D.
We have to find the area bounded by the parabola y 2 = 4 x and line y = 3 . Draw the graph of these equations.
Figure (1)
Solve the equations to find the point of intersection.
y 2 = 4 x ( 3 ) 2 = 4 x x = 9 4
To find the area of OABO, we take a horizontal strip in the region with infinitely small width d y and integrate for the area of the strip.
Area of the region OABO = ∫ 0 3 x d y (1)
From the equation of parabola, find the value of x in terms of y and substitute in equation (1).
y 2 = 4 x x = y 2 4
Substitute y 2 4 for x in equation (1) and integrate.
Area of the region OABO = ∫ 0 3 y 2 4 d y = 1 4 ∫ 0 3 y 2 d y = 1 4 [ y 3 3 ] 0 3 = 1 4 [ 3 3 3 − 0 3 3 ]
Simplify further,
Area of the region OABO = 1 4 [ 27 3 ] = 1 4 [ 9 ] = 9 4 sq units
The area calculated for the region OABO is 9 4 sq units .
Thus, out of all the four options, option (B) is correct.
We have to find the area bounded by the parabola
Figure (1)
Solve the equations to find the point of intersection.
To find the area of OABO, we take a horizontal strip in the region with infinitely small width
From the equation of parabola, find the value of x in terms of y and substitute in equation (1).
Substitute
Simplify further,
The area calculated for the region OABO is
Thus, out of all the four options, option (B) is correct.
