1. The problem involves evaluating the triple integral of the function $f(x,y,z) = 2z$ over the region defined by the limits for $z$, $y$, and $x$ given as $z$ from 2 to -2, $y$ from $-\sqrt{4-x^2}$ to $\sqrt{4-x^2}$, and $x$ from 3 to 0. 2. The triple integral is written as: $$\int_{x=3}^{0} \int_{y=-\sqrt{4-x^2}}^{\sqrt{4-x^2}} \int_{z=2}^{-2} 2z \, dz \, dy \, dx$$ 3. Important note: The limits for $z$ are from 2 to -2, which is reversed. We can switch the limits and change the sign: $$\int_{z=2}^{-2} 2z \, dz = -\int_{z=-2}^{2} 2z \, dz$$ 4. Evaluate the innermost integral: $$-\int_{-2}^{2} 2z \, dz = -\left[ z^2 \right]_{-2}^{2} = -\left(4 - 4\right) = 0$$ 5. Since the innermost integral evaluates to 0, the entire triple integral evaluates to 0 regardless of the other integrals. 6. Therefore, the value of the triple integral is: $$0$$