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 1


When we look at the bottom of the prism as shown here,

enter image description 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 $1-y$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .