Chapter 16: Integrals and Vector Fields - Section 16.5 - Surfaces and Area - Exercises 16.5 - Page 990: 39

$$4$$

Apply cylindrical coordinates. $ x=r \cos \theta ;\\ y= r \sin \theta ;\\ z \gt 0$ We know that $ r(r, \theta)=xi+yj+zk $ or, $ r^2=x^2+y^2+z^2$ We have $ x=y^2$ and $ x=2-y^2$ $$ Surface \space Area =\iint_{R} \dfrac{|\nabla f|}{|\nabla f \cdot p|} \space dA \\= \dfrac{3}{2} \times \int_{-1}^{1} \int_{y^2}^{2-y^2} \space dx \space dy \\ = \int_{-1}^{1}(3-3y^2) \space dy \\=4$$