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

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(B) The curves are $y = ax^2$ and $x = ay^2$ (which is $y = \sqrt{x/a}$ for $y \ge 0$).To find the intersection point,substitute $y = ax^2$ into $x =

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$\frac{1}{\sqrt{3}}$

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(B) The curves are $y = ax^2$ and $x = ay^2$ (which is $y = \sqrt{x/a}$ for $y \ge 0$).To find the intersection point,substitute $y = ax^2$ into $x = ay^2$:$x = a(ax^2)^2 = a^3x^4$$x(a^3x^3 - 1) = 0$Since $a > 0$,the intersection points are $x = 0$ and $x = 1/a$.The area $A$ is given by:$A = \int_{0}^{1/a} (\sqrt{x/a} - ax^2) dx = 1$$\frac{1}{\sqrt{a}} \int_{0}^{1/a} x^{1/2} dx - a \int_{0}^{1/a} x^2 dx = 1$$\frac{1}{\sqrt{a}} [\frac{2}{3} x^{3/2}]_{0}^{1/a} - a [\frac{x^3}{3}]_{0}^{1/a} = 1$$\frac{1}{\sqrt{a}} \cdot \frac{2}{3} \cdot (\frac{1}{a})^{3/2} - \frac{a}{3} \cdot (\frac{1}{a})^3 = 1$$\frac{2}{3a^2} - \frac{1}{3a^2} = 1$$\frac{1}{3a^2} = 1$$a^2 = \frac{1}{3} \implies a = \frac{1}{\sqrt{3}}$ (since $a > 0$).

If the area bounded by $y = ax^2$ and $x = ay^2$ for $a > 0$ is $1$,then $a =$

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