Area bounded by the parabolas y^2=4x and x^2 =4y is :

The correct answer is option (1) → $\frac{16}{3}$

finding points of intersection

$x^2=4y⇒x^4=4×4y^2$

$⇒x^4=16×4x$

so $x=0$ or $x=4$

$y=0$ or $y=4$

area = $\int\limits_0^42\sqrt{x}-\frac{x^2}{4}dx=\frac{16}{3}$ sq. units