Answer
$\dfrac{7}{3}$
Work Step by Step
Here, we have $x=y^2$ and $x=2-y^2$
We know that $x=r \cos \theta, y=r \sin \theta $
The surface area is given by:
$S=\iint_{R} \dfrac{|\nabla f|}{|\nabla f \cdot p|} dA$
This implies that
$S=\int_{0}^{\sqrt 3} \int_{0}^{x}\sqrt {x^2+1} dx dy$
Thus, $S=\int_{0}^{\sqrt 3} x \sqrt {x^2+1} dx=\dfrac{7}{3}$
