Area using $\iint dxdy$

Yes, consider $f(x,y)=1$. The volume under this curve is $1$ (the height) times the area of the base, so the numerical value of volume coincides with the area: they're the same thing.


If $B$ is a measurable and bounded subset of $ \mathbb R^2$, then the area of $B$ is given by

$\iint_B 1 d(x,y).$


One of the most common ways to think of a double integral is to think of it as the volume under a surface created by a function of two variables $z=f(x,y)$ and bounded by a region $R$ (typically rectangular in shape) in the $xy$ plane. If the given function of two variables is $f(x,y)=1$, then what you're basically doing is you're finding the volume of a solid whose height is $1$ and the base area is the area of the given region $R$. The volume of a solid whose height is $1$ is numerically equal to its base area. That's why

$$\iint_{R}1\,dx\,dy=area\ of\ region\ R.$$