Calculate the volume of $V$ limited by planes: $$x+y=1, x+y=2, x=0, y=0, z=0, z=1$$
The projection on $xy$ is
I tried finding the volume without integrals first: $$V=Sh=\frac{1}{2}(2\cdot2-1\cdot1)\cdot (1-0)=\frac{3}{2}$$
And I got a completely different answer when calculating with integrals: $$\int_{0}^{1}dz\int_{0}^{2}dx\int_{1-x}^{2-x}dy=\int_{0}^{1}dz\int_{0}^{2}dx(2-x-1+x)=\int_{0}^{1}dz\cdot2=2$$
Where did I go wrong?
What you did using integrals is not correct. The part corresponding to $z$, yes, it is fine, but the part corresponding to $x$ and $y$ should be$$\int_0^1\int_{1-x}^{2-x}1\,\mathrm dy\,\mathrm dx+\int_1^2\int_0^{2-x}1\,\mathrm dy\,\mathrm dx=\frac32.$$