Question

The area bounded by the curve $x^2=ky,k>0$ and the line $y=3$ is 12 unit${}^2$. Then $k$ is:

• A. $3$
• B. $3\sqrt{3}$
• C. $\frac{3}{4}$
• D. none of these

Answer 1

The correct option is A $3$

From the figure, the area enclosed
$2.{\int }_0^3\sqrt{ky}dy=12$
Integrating and solving,
$=2.2/3\sqrt{k{3}^3}=12$
$\Rightarrow k=3$