Answer
$$\dfrac{(36 \sqrt 3- 32 \sqrt 2+4)}{15}$$
Work Step by Step
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$
$ r_x=\lt 1,0, \sqrt x \gt ; \\ r_{y}=\lt 0, 1, \sqrt {y} \gt $
Also, $|r_x\times r_{y}|=\sqrt {x+y+1}$
$$ Area=\int_0^{1} \int_0^{1} (\sqrt {x+y+1}) \space dy \space dx\\=\int_0^{1} [\dfrac{2}{3 } \times (x+y+1)^{3/2}]_0^{1} \\= \dfrac{(36 \sqrt 3- 32 \sqrt 2+4)}{15}$$
