Double integral as a volume of a solid

Question

I gotta evaluate $$\int\int_{R} dA$$ where $R$ is the area between $x=\sqrt{3-y}$, $x+y+3=0$ and $y=2x$. I know this integral gives me the value of this area. I wonder if, the same integral (but viewing it in $\mathbb{R}^{3}$) could be the volume of a solid.

Answer 1

If you integrate $$A = \int\int_{R} dA$$ by itself then you are guaranteed to get the area of $R$ no matter how you look at it. You would get a volume if you integrated a function $f:\mathbb{R^2}\rightarrow \mathbb{R}$ over a two dimensional region like

$$V = \int\int_{R} f(x,y) \; dA \;.$$

Note that this is equivalent in Cartesian coordinates to the volume integral

$$V = \int\int\int_{\Omega} dV = \int\int_{R} \int_0^{f(x,y)} dz \; dA = \int\int_{R} f(x,y) dA$$

where $dA$ is taken to be $dxdy.$ $\;\;f$ is thought of as the "height" of the region $\Omega$ with $R$ as the "extruded area" that generates the volume. In other words, this is the volume bound by $R,$ the $x\text{-}y$ plane, and $f.$

Hope this makes sense, let me know if you need any more clarification.