Volume under a surface $z=3-2x-y$ and bound by surfaces

Question

I'm doing a practice problem and the question is to find the volume under the surface $z=3-2x-y$ with a region enclosed by: $x$-axis, $y=2x$ and $x=2$. The answer given was $6.25$ cubic units. My initial assumption was to use triple integration and integrate with respect to $z$ first, from $0$ to $3-2x-y$, then with respect to y from $0$ to $2x$, and then with respect to $x$ from $0$ to $2$. I got the value $-4$ so my assumptions were incorrect. I tried to visualize the domain using graphs and drawings but it was too difficult and of no help.

Answer 1

No triple integration (you don't have to integrate wrt variable $z$), Only a double integration.

You need first to make a sketch of your domain of integration which is a right triangle with vertices $(0,0),(2,0),(2,4)$; you will realize then how to "sweep it" by vertical segments joining point $(x,0)$ to point $(x,2x)$. Otherwise said by considering, for a fixed $x$, points $(x,y)$ with $y$ ranging from $0$ to $2x$, giving the double integral :

$$\int_{x=0}^{x=2}\int_{y=0}^{y=2x} (3-2x-y) dy dx$$

$$=\int_{x=0}^{x=2}[(3-2x)y-\tfrac12 y^2]_{y=0}^{y=2x} dx=\int_{x=0}^{x=2}((3-2x)2x-2 x^2) dx$$

$$=\int_{x=0}^{x=2}(6x-6x^2) dx=[3x^2-2x^3]_{x=0}^{x=2}=12-16=\color{red}{-4}$$