Question

The area enclosed by the curve $x=y^2+2$, ordinates y = 0 & y = 3 and the Y - axis is sq. units.

• A. 5
• B. 10
• C. 15

Answer 1

The correct option is C 15

We have seen that if $g\left(y\right)\ge 0$ for $yϵ$ [c, d] then the area bounded by curve x = g(y) and y-axis between abscissa y = c and y = d is ${\int }_c^{d}g\left(y\right)dy$ .
We'll use the same concept here. Here the curve given is
$x=y^2+2$ and c = 0 $\&d=3.So,g\left(y\right)\ge 0\forall x\in (0,3)$
Let the area enclosed be A.
$A={\int }_0^3(y^2+2)dy.$
$A=(\frac{y^3}{3}+2y){|}_0^{3}A=9+6-0A=15$