A paperweight has a slanted top described by x + y + z = 2. Its edges are orthogonal to the $xy$–plane, and the bottom of the paperweight is formed by the triangle with vertices $(1, 0, 0)$, $(0, −1, 0)$ and $(0, 1, 0)$. Use a triple integral to find the volume of the paperweight.
From the base of the graph I got $$ \int_{-1}^{1} \int_{y+1}^{-y+1} \;dx\;dy $$
After taking into account the slanted top I get
$$ \int_{-1}^{1} \int_{y+1}^{-y+1}\int_{0}^{2-y-x} \;dz \;dx\;dy $$
Does this look correct?
Doesn't look right to me. Here's what I'm getting:
$(\displaystyle\int_{-1}^{0}\int_{0}^{y+1}\int_{0}^{2-y-x}dzdxdy) + (\int_{0}^{1}\int_{0}^{-y+1}\int_{0}^{2-y-x}dzdxdy)$
or:
$\displaystyle\int_{0}^{1}\int_{x-1}^{-x+1}\int_0^{2-y-x}dzdydx$
Say we start with y as the outer integral. Pick a y-value. Then along the cross section go from the left most boundary to the right boundary to get the x boundaries. The left most boundary is the y-axis ie: x=0. No problem there. But the right boundary is represented by two different lines hence two different functions. We can't have two functions as the limit in the integral. So we need to split the integral.
But starting with x as the outer integral avoids this problem