Answer
$\dfrac{4}{15}(3^{5 / 2}-2^{7 / 2}+1)$
Work Step by Step
We write the surface area of the part $z=f(x,y)$ as follows: $A(S)=\iint_{D} \sqrt {1+(f_x)^2+(f_y)^2} dx \ dy$
and, $\iint_{D} dA$ defines the projection of the surface on the xy-plane.
Our aim is to calculate the area of the given surface.
$A(S)=\iint_{D} \sqrt{(\sqrt{x})^{2}+(\sqrt{y})^{2}+1} d A =\int_{0}^{1} \int_{0}^{1} \sqrt{x+y+1} d y d x $
or, $=\int_{0}^{1}\left[\dfrac{2}{3}(x+y+1)^{3 / 2}\right]_{y=0}^{y=1} d x $
or, $ =\dfrac{2}{3}\times \int_{0}^{1}\left[(x+2)^{3 / 2}-(x+1)^{3 / 2}\right] \ d x$
or, $=\dfrac{2}{3} \times [\dfrac{2}{5}(x+2)^{5 / 2}-\dfrac{2}{5}(x+1)^{5 / 2}]_{0}^{1}$
or, $=\dfrac{4}{15}(3^{5 / 2}-2^{7 / 2}+1)$
