Q:

if f(x,y) = 1 for all (x,y) where R has a nice shape like a rectangle or triangle what is another geometric interpretation of....?

If $f(x,y)$ is greater than or equal to $0$ on region $R$ in the plane, then $\iint_R f(x,y)dA$ can be interpreted geometrically as the volume of the solid under the surface $z=f(x,y)$ and above $R$. If $f(x,y)=1$ for all $(x,y)\in R$, where $R$ has a nice shape such as a rectangle, triangle, etc what is another geometric interpretation of $\iint_R f(x,y) dA$ ?
Could the answer simply be a circle, since it seems to be asking for just another "nice shape" for which z always equals 1? I am a bit confused as to how to prove that it could also be a circle though. 
This is for my Multivariable Calculus class. 
Thank you!

A:

If $f(x,y)=1$ for all $x,y\in R$ then $$\iint_R f(x,y) \mathrm{d}A=\iint_R 1 \mathrm{d}A=A$$
i.e. the integration just returns the area of $R$  

