Given linear transformation: $$T: \mathbb R^3 \to \mathbb R^3, T(x,y,x)= (3y+4z, 2x-3z, x+3y) $$ We need to evaluate the triple integral : $$\int \int \int_ {T(C)} (2x+y-2z) dx dy dz $$ where $C= \{{(x,y,z): 0 \le x \le 1, 0 \le y \le 1, 0 \le z\le 1}\}$
I understand that the unit cube gets transformed to a slanted parallelopiped, but how can I find the limits of integration in this case?
Kindly help. Thanks in advance.
Let $f(x,y,z)=2x+y-2z$. Then\begin{align}\iiint_{T(C)}2x-y-2z\,\mathrm dx\,\mathrm dy\,\mathrm dz&=\iiint_{T(C)}f(x,y,z)\,\mathrm dx\,\mathrm dy\,\mathrm dz\\&=\iiint_Cf\bigl(T(x,y,z)\bigr)|\det T|\,\mathrm dx\,\mathrm dy\,\mathrm dz\\&=15\int_0^1\int_0^1\int_0^15z\,\mathrm dx\,\mathrm dy\,\mathrm dz\\&=\frac{75}2.\end{align}