If the definitions of type I and type II regions is the same as in the link provided, then as a type I region the integration domain is the set
[tex]R_{\rm I} = \left\{(x,y) \mid 0 \le x \le 3 \text{ and } \sqrt{x+1} \le y \le 2\right\}[/tex]
and as a type II region,
[tex]R_{\rm II} = \left\{(x,y) \mid 0 \le x \le y^2-1 \text{ and } 1 \le y \le 2\right\}[/tex]
where we solve y = √(x + 1) for x to get x as a function of y.
A. The area of the type I region is
[tex]\displaystyle \iint_{R_{\rm I}} dA = \int_0^3 \int_{\sqrt{x+1}}^2 dy \, dx = \int_0^3 (2 - \sqrt{x+1}) \, dx = \boxed{\frac43}[/tex]
B. The area of the type II region is of course also
[tex]\displaystyle \iint_{R_{\rm II}} dA = \int_1^2 \int_0^{y^2-1} dx \, dy = \int_1^2 (y^2-1) \, dy = \boxed{\frac43}[/tex]
I've attached a plot of the type II region to give an idea of how it was determined. The black arrows indicate the domain of x as it varies from the line x = 0 (y-axis) to the curve y = √(x + 1).
