Chapter 15 - Section 15.5 - Surfaces and Area - Exercises - Page 873: 48

$ \dfrac{1}{15}(36 \sqrt 3- 32 \sqrt 2+4)$

Here, we have $r_x=\lt 1,0, \sqrt x \gt$ and $r_{y}=\lt 0, 1, \sqrt {y} \gt$ Also, $|r_x\times r_{y}|=\sqrt {x+y+1}$ Thus, the area is given by $A=\int_0^{1} \int_0^{1} (\sqrt {x+y+1}) dy dx$ or, $A=\int_0^{1} [\dfrac{2}{3}(x+y+1)^{3/2}]_0^{1}= \dfrac{1}{15}(36 \sqrt 3- 32 \sqrt 2+4)$