lmiller6
Can somebody explain to me how we did problem number 2 on the exam review sheet? I know someone showed the class their own solution but I’m having some trouble understanding what was done to find the bounds of integration.
\int_0^5 \int_{0}^{3-3x/5} \int_{0}^{2-(2/3)y-(2/5)x} 1 dz dy dx
is the set up integral from class.