Let $A = \left \{ (x,y,z)\in [0, 1]^{3}| x + y + z \leq 1 \right \}$. Calculate the volume.
So I found the following boundaries for the integrals:
- $0 \leq z\leq 1$
- $0 \leq y\leq 1-z$
- $0 \leq x\leq 1-y-z$
The integral is $\int_{0}^{1}\int_{0}^{1-z}\int_{0}^{1-y-z}f(x,y,z)dx dy dz$. But what is f(x,y,z)?
You are integrating $f(x, y, z) =1$.