Calculate the volume integral of $T = xyz^2$ over the prism defined by vertices $(0,0,0), (0,0,3),(1,0,0), (0,1,0),(1, 0, 3), (0, 1, 3)$.
The limits of integration taken in the given answer are :
for x, $0$ to $1- y$,
for y, $0$ to $1$
and for z, $0$ to $3$.
I don't understand why $1-y$ is the upper limit. Should not it be $1$ ? And how do they came up with $1- y$ not something else ?
1 Answer
When we look at the bottom of the prism as shown here,
we see that choosing both $x$ and $y$ in the interval $[0,1]$ then we get the whole square $[0,1]^2$. But we only want the triangle, therefore we parameterize it. We can either choose $x$ or $y$ to go from $0$ to $1$ but the other is dependent on the first value through a linear function, in this case $1-y$.
