If the area bounded by the curves y=ax^2 and | vedclass

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(B) Given curves are $y=ax^2$ $(i)$ and $x=ay^2$ $(ii)$.Substituting $y$ from $(ii)$ into $(i)$, we get $x = a(ax^2)^2 = a^3x^4$, which implies $x(a^3

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$\frac{1}{3}$

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(B) Given curves are $y=ax^2$ $(i)$ and $x=ay^2$ $(ii)$.Substituting $y$ from $(ii)$ into $(i)$, we get $x = a(ax^2)^2 = a^3x^4$, which implies $x(a^3x^3 - 1) = 0$.Thus, $x=0$ or $x=\frac{1}{a}$.For $x=0$, $y=0$, and for $x=\frac{1}{a}$, $y=\frac{1}{a}$.The area bounded by the curves is given by $\int_0^{1/a} (\sqrt{x/a} - ax^2) dx = 3$.$\int_0^{1/a} \frac{1}{\sqrt{a}} x^{1/2} dx - \int_0^{1/a} ax^2 dx = 3$.$\frac{1}{\sqrt{a}} [\frac{2}{3} x^{3/2}]_0^{1/a} - a [\frac{x^3}{3}]_0^{1/a} = 3$.$\frac{1}{\sqrt{a}} \cdot \frac{2}{3} \cdot (\frac{1}{a})^{3/2} - \frac{a}{3} \cdot (\frac{1}{a})^3 = 3$.$\frac{2}{3a^2} - \frac{1}{3a^2} = 3$.$\frac{1}{3a^2} = 3 \Rightarrow 9a^2 = 1 \Rightarrow a^2 = \frac{1}{9}$.Since $a>0$, we have $a = \frac{1}{3}$.

If the area bounded by the curves $y=ax^2$ and $x=ay^2$ $(a>0)$ is $3$ sq units, then the value of $a$ is

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