If the area enclosed between the curves $y=kx^2$ and $x=ky^2 (k > 0)$ is 1 square unit. Then, k is

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

The curves $y=kx^2$ and $x = ky^2$ intersect at $(\frac{1}{k},\frac{1}{k})$ and at the origin. It is given that the area enclosed between the given curves is 1 square unit.

$∴\int\limits_0^{1/k}(y_2-y_1)dx=1$

$⇒\int\limits_0^{1/k}\left(\sqrt{\frac{x}{k}}-kx^2\right)dx=1$

$⇒\left[\frac{1}{\sqrt{k}}\frac{2}{3}x^{3/2}-\frac{k}{3}x^3\right]_0^{1/k}=1$

$⇒\frac{2}{3\sqrt{k}}(\frac{1}{k})^{3/2}-\frac{k}{3}(\frac{1}{k})^3=1$

$⇒\frac{2}{3k^2}-\frac{1}{3k^2}=1⇒\frac{1}{3k^2}=1⇒k^2=\frac{1}{3}⇒k=\frac{1}{\sqrt{3}}$