Browse Questions

# Choose the correct answer in the area of the region bounded by the curve $y^2 = 4x$, $y$ - axis and the line $y = 3$ is

$(A) 2 \quad (B) \frac{9}{4} \quad (C) \frac{9}{3} \quad (D) \frac{9}{5}$

Toolbox:
• The area bounded by the curve g(y),y axis and the ordinate y=c,y=d is given by,$A=\int_c^dx\;dy=\int_c^dg(y)\;dy.$
Hence the area of the region bounded by the curve $y^2=4x$,y-axis and the line y=3 is the shaded portion as shown in the fig.
Hence $A=\int_0^3x\;dy.$
Here $x=\frac{y^2}{4}$
$A=\int_0^3\frac{y^2}{4}\;dy.$
On integrating we get,
$A=\frac{1}{4}\begin{bmatrix}\frac{y^3}{3}\end{bmatrix}_0^3$.
On applying limits we get,
$A=\frac{1}{12}[3^3-0]$
$\;\;\;\;=\frac{1}{12}\times 27=\frac{9}{4}$sq.units.
Hence B is the correct answer.