# Triple integral positive octant [closed]

I need to evaluate this integral over the portion of the positive octant cut off by the plane $x + y + z = \pi$. $$\iiint \sin(x+y+z)\, dx\, dy\,dz$$ What boundaries do I use?

## closed as off-topic by Namaste, Simply Beautiful Art, Michael Lee, José Carlos Santos, B. GoddardAug 27 '17 at 22:38

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Namaste, Simply Beautiful Art, Michael Lee, José Carlos Santos, B. Goddard
If this question can be reworded to fit the rules in the help center, please edit the question.

As you are working in the positive octant $$x \geq 0, y \geq 0, z \geq 0$$

You're other condition is $$x + y + z \leq \pi$$

Rewriting it: $z \leq \pi - x - y$

Results in $z \leq \pi$

Similarly, for any $z$: $y \leq \pi - x - z \leq \pi - z$

And for any $z$ and $y$: $x \leq \pi - y - z$

Therefore your boundary conditions are: $$0 \leq z \leq \pi$$ $$0 \leq y \leq \pi - z$$ $$0 \leq x \leq \pi - y - z$$

You're integral then becomes: $$\int_0^\pi \int_0^{\pi-z} \int_0^{\pi -y-z} \sin (x+y+z)dxdydz$$

Which is equal to $$\frac{\pi ^2}2 - 2$$

The boundaries are

• $z=0$ and $z(x,y)=\pi-x-y$
• $y=0$ and $y(x)=\pi-x$
• $x=0$ and $x=\pi$