# Triple Integral $\int \int \int_D x^2yz dxdydz$ over a strange area

Fairly simple triple integral $$\int \int \int_D (x^2yz) dxdydz$$ 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: $$\int_0^1 \int_{-z}^0 \int_0^1 (x^2yz) dxdydz$$

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.

• Shouldn't the limit of $z$ would be $[x+y,1]$ – Naman Jain Nov 14 '19 at 5:38
• I made an error writing the volume, fixed now. – Mevve Nov 14 '19 at 6:12

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

$$\int_0^1 \int_{-z}^0 \int_0^{y+z} x^2yz dxdydz = -\frac{1}{420}$$ 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. $$A(0,0)$$, $$B(0,1)$$ $$C(1,0)$$.

• Hi, I made an error writing down the volume limits. However I dont understand how you came to that conclusion. I'm guessing you get the surface plane from $0 \leq x+y \leq z$ but from where do you get that the sides cut by $xz$ and $yz$? – Mevve Nov 14 '19 at 6:18
• because you have $x>0$and $y>0$ – AppoopanThaadi Nov 14 '19 at 6:22