First off, What does $f(x,y,z) = 1$ mean? versus $f(x,y,z) = x$.
I have trouble with this as I am not sure choosing an initial $x$ factors into the function. Similarly with choosing a corresponding z. (The choices are related to the problem).
OK, so, I am trying to find the volume of the following shape. For a given x, $0 \leq x \leq 1$, $0 \leq z \leq x$. For a given z based on the initial choice of x, $0 \leq y \leq\sqrt{1-z^2}$.
So, is the volume $$\int_0^1 \int_0^x \int_0^\sqrt{1-z^2} x dydzdx?$$
Or is it $$\int_0^1 \int_0^x \int_0^\sqrt{1-z^2} 1 dydzdx?$$
I also worry about the arbitrary choice of $z$ after the original choice of $x$. Thanks!
HINTS
$f(x,y,z)=1$ means $f$ maps any vector in $\mathbb{R}^3$ to $1$. When you say $g(x,y,z)=x$, this means that $g$ maps any vector in $\mathbb{R}^3$ to its first coordinate.
Volume of the region $A$ is defined as $\iiint_A dV$.