Find the surface area of $z=2-(x^2+y^2)$ cut off by the lines $x=0,y=1,y=x$
S.A.$=\int \int_{R} \sqrt{1+4(x^2+y^2)}dxdy$
Where we know that $R$ is the region bounded by $y=1,x=0,y=x$ and $x^2+y^2=(\sqrt{2})^2$
This is essentially the area bounded by $y=1,x=0,y=x$ as $\sqrt{2}>1$
The final integral i am getting is:
$\int_{y=0}^1 \int_{x=0}^y \sqrt{1+4(x^2+y^2)} dxdy$
Since I am facing some problem to solve this. Is this okay?