Chapter 15 - Section 15.5 - Surfaces and Area - Exercises - Page 873: 39

$4$

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: $S=\iint_{R} \dfrac{|\nabla f|}{|\nabla f \cdot p|} dA=\iint_{R} \dfrac{\sqrt {4x^2+4y^2+1}}{1} dx dy$ This implies that $S=\int_{-1}^{1} \int_{y^2}^{2-y^2} (\dfrac{3}{2} dx dy$ Thus, $S=\int_{-1}^{1}(3-3y^2) dy=4$