The area of the surface described by $z = f(x,y)$ for $(x,y)\in R$ is given by $$\iint_R\sqrt{1 +[f_x(x,y)]^2+[f_y(x,y)]^2}dA$$ Find an approximation to the area of the surface on the hemisphere $x^2+y^2+z^2=9$, $z\geq 0$ that lies above the reegion in the plane described by $R = \{(x,y)|0\leq x \leq 1, 0 \leq y \leq 1\}$ using the Trapezoidal rule in both directions.
Why do they give only a double integral when we are working with 3 unknowns in $x$, $y$ and $z$? I am not certain how to visualize this with the prepositions "on" and "above".
The integrand represents the roof on which we have to find the area, while the region R defines the floor (with the four walls) below the roof.
In fact $f(x, y)$ in this case is $z$. Hence the integral is
$$\int_{x=0}^{1}\int_{y=0}^{1} \frac{3dydx}{\sqrt{9 - x^2 - y^2}}$$
This integral is rather difficult to get an analytical solution, therefore the question asks an approximation using trapezoidal rule in both x and y directions.