The area enclosed between the curves $y=ax^2$ and $x=ay^2$ ($a>0$) is 1 sq. unit. Then the value of $a$ is
Step-by-Step Solution
Key Concept: Curves intersect at $(0,0)$ and $(1/a,1/a)$. Area $=\int_0^{1/a}\!\left(\sqrt{x/a}-ax^2\right)dx=\dfrac{1}{3a^2}$. Set $=1\Rightarrow a=1/\sqrt3$.
Area $=1/(3a^2)=1\Rightarrow a=1/\sqrt3$.
Correct Answer: (A) $\dfrac{1}{\sqrt3}$