Triple integral volume problem

Question

I recently learned triple integrals. This is my first problem solving one with my own skills.

Let $D$ be the region bounded by $x=0$, $y=0$, $z=0$ and $x+y+z=1$. Compute $$\iiint_D \frac{dx\,dy\,dz}{(x+y+z+1)^3}$$

So far I found the points where $x+y+z=1$ meets the $x$-, $y$- and $z$-axes: $$x=0,~y=0,~z=1 \text{ or } x=0,~z=0,~y=1 \text{ or } z=0,~y=0,~x=1$$

And this is the function I came up with,

$$\int_0^1\int_0^1\int_0^{1-(y+z)} \frac{dx\,dy\,dz}{(x+y+z+1)^3}$$

Is this the correct way to do it? Or is there something wrong with my limits?

Answer 1

With your limits, both $y$ and $z$ can be simultaneously $1$, which is incorrect.

The correct limits are $0\le z \le 1$, $0 \le y \le 1-z$, and $0 \le x \le 1-y-z$. So you have to evaluate $$\int_{0}^{1}\int_{0}^{1-z}\int_{0}^{1-y-z} \frac{1}{(x+y+z+1)^3}\,dx\,dy\,dz$$

Answer 2

The projection of the body on the $xy$-plane isn't $[0,1]\times[0,1]$. Hint: what is $x + y + z = 1$ if $z = 0$?

Answer 3

close

$$\int_0^1\int_0^{1-z}\int_0^{1-y-z} \frac {1}{(x+y+z+1)^3}\,dx\,dy\,dz$$