"A parallelepiped is described by the vectors $(2,2,3),(2,4,3)$ and $(0,1,5)$
given that the density $= 2x+2y$, find the mass of the parallelepiped."
I can find the volume just fine, but setting up the integral to find the mass is giving me a lot of trouble.
this is my best guess so far...
$$20 \int_0^2 \int_0^4 \int_0^5 (2x+2y) dzdxdy$$
The vectors do stem from the origin
You’re starting out with the right idea: you need to compute the triple integral $$\iiint 2x+2y\,dV$$ over the paralellepiped, but since the edges aren’t parallel to the coordinate axes, finding the right bounds with $dV=dx\,dy\,dz$ is a bit of a nuisance. So, perform a change of variables to a coordinate system in which the edges are parallel to the coordinate axes. One possibility is the transformation $$(x,y,z) = (2,2,3)u+(2,4,3)v+(0,1,5)w = (2u+2v,2u+4v+w,3u+3v+5w)$$ so that each variable falls in the range $[0,1]$. The integral then becomes $$\int_0^1\int_0^1\int_0^1 2(2u+2v)+2(2u+4v+w)\,\left|{\partial(x,y,z)\over\partial(u,v,w)}\right|\,du\,dv\,dw.$$ Since the coordinate transformation is linear, its Jacobian is just the determinant of its matrix, which in turn is equal to the volume $V$ of the paralellepiped, so the integral becomes $$2V \int_0^1\int_0^1\int_0^1 4u+6v+w\,du\,dv\,dw$$ which I’m sure you can evaluate.