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.

Thanks in advance.

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

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


enter image description here 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. $A(0,0)$, $B(0,1)$ $C(1,0)$.

  • $\begingroup$ 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 $yz$? $\endgroup$ – Mevve Nov 14 '19 at 6:18
  • $\begingroup$ because you have $x>0$and $y>0$ $\endgroup$ – AppoopanThaadi Nov 14 '19 at 6:22

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.