Triple integral $\iiint_D x^2yz \,dx\,dy\,dz$ over a strange area

Question

Fairly simple triple integral $$\iiint_D x^2yz \,dx\,dy\,dz$$ over the area $D = \{(x,y,z):0 \leq x \leq y+z \leq z \leq 1\}$.

I'm not sure how to interpret this area, this is what I have done so far:

Since the area is strictly positive we get from $0 \leq x \leq y+z \leq z \leq 1$ $$\begin{align} 0 &\leq x \leq 1 \\ -z &\leq y \leq 0 \qquad \text{and} \\ 0 &\leq z \leq 1\end{align}$$

Which gives me the integral: $$\int_0^1 \int_{-z}^0 \int_0^1 (x^2yz) \,dx\,dy\,dz$$

This I can fairly easily calculate, giving me the final answer $\frac{1}{24}$, (I dont have the key).

I'm not sure my integration limits are correct, if not any pointers to how I can figure them out would be greatly appreciated.

Thanks in advance.

Answer 1

You should be suspicious of your first bounds because they are constants, but the inequalities for $x$ are not bounded by constants. Let's look at the inequalities and choose to do $x$ first.

$$ 0 \leq x \leq y+z$$

Next, after the $x$ is gone, we have the inequalities

$$ 0 \leq y+z \leq z \implies -z \leq y \leq 0 $$

Lastly, with our $y$ gone, the inequalities now read

$$ 0 \leq z \leq 1$$

leaving us with the integral

$$\int_0^1 \int_{-z}^0 \int_0^{y+z} x^2yz dxdydz = -\frac{1}{420}$$

Answer 2

It is not strange area!(actually volume), let us say $f(x,y,z)=x^2yz$ is the formula for finding calories at a location in this cake piece and we want to find out total calories, that is the problem statement. The base of cake is $xy$ plane and surface is the plane $x+y-z=0$and sides are cut by $xz$ and $yz$ planes. $A(0,0)$, $B(0,1)$ $C(1,0)$.