Chapter 16: Integrals and Vector Fields - Section 16.5 - Surfaces and Area - Exercises 16.5 - Page 990: 40

$$=\dfrac{7}{3}$$

Apply cylindrical coordinates. $ x=r \cos \theta ;\\ y= r \sin \theta ;\\ z \gt 0$ We know that $ x=y^2$ and $ x=2-y^2$ $$ Surface \space Area =\iint_{R} \dfrac{|\nabla f|}{|\nabla f \cdot p|} \space dA\\=\int_{0}^{\sqrt 3} \int_{0}^{x}\sqrt {x^2+1} \space dx \space dy \\ =(1/3) \times [(x^2+1)^{3/2}]_{0}^{\sqrt 3} \space \\=\dfrac{7}{3}$$