1. The problem involves evaluating the triple integral with limits for $x$ from 2 to -2, for $y$ from $\sqrt{4-x}$ to $\sqrt{4-x^2}$, and for $z$ from 3 to 0, integrating the function $2z$ with respect to $dz$, $dy$, and $dx$. 2. The triple integral is written as: $$\int_{x=2}^{-2} \int_{y=\sqrt{4-x}}^{\sqrt{4-x^2}} \int_{z=3}^{0} 2z \, dz \, dy \, dx$$ 3. First, integrate with respect to $z$: $$\int_3^0 2z \, dz = \left[z^2\right]_3^0 = 0^2 - 3^2 = -9$$ 4. Now the integral reduces to: $$\int_{2}^{-2} \int_{\sqrt{4-x}}^{\sqrt{4-x^2}} -9 \, dy \, dx$$ 5. Integrate with respect to $y$: $$\int_{\sqrt{4-x}}^{\sqrt{4-x^2}} -9 \, dy = -9 \left( \sqrt{4-x^2} - \sqrt{4-x} \right)$$ 6. The integral now is: $$\int_2^{-2} -9 \left( \sqrt{4-x^2} - \sqrt{4-x} \right) \, dx$$ 7. Since the limits for $x$ are from 2 to -2 (upper limit less than lower), reverse limits and change sign: $$= 9 \int_{-2}^2 \left( \sqrt{4-x^2} - \sqrt{4-x} \right) \, dx$$ 8. This integral can be split: $$9 \left( \int_{-2}^2 \sqrt{4-x^2} \, dx - \int_{-2}^2 \sqrt{4-x} \, dx \right)$$ 9. The first integral $\int_{-2}^2 \sqrt{4-x^2} \, dx$ represents the area of a semicircle of radius 2: $$= \frac{\pi \times 2^2}{2} = 2\pi$$ 10. The second integral $\int_{-2}^2 \sqrt{4-x} \, dx$ is improper because $\sqrt{4-x}$ is real only for $x \leq 4$, but the interval is valid. We split it into two parts: - From $-2$ to $2$: $$\int_{-2}^2 \sqrt{4-x} \, dx$$ 11. Substitute $u = 4 - x$, then $du = -dx$, when $x=-2$, $u=6$, when $x=2$, $u=2$: $$\int_{u=6}^{2} \sqrt{u} (-du) = \int_2^6 u^{1/2} du = \left[ \frac{2}{3} u^{3/2} \right]_2^6 = \frac{2}{3} (6^{3/2} - 2^{3/2})$$ 12. Calculate numeric values: $$6^{3/2} = 6 \times \sqrt{6} \approx 6 \times 2.4495 = 14.697$$ $$2^{3/2} = 2 \times \sqrt{2} \approx 2 \times 1.4142 = 2.828$$ 13. So the integral is: $$\frac{2}{3} (14.697 - 2.828) = \frac{2}{3} \times 11.869 = 7.913$$ 14. Finally, the value of the triple integral is: $$9 (2\pi - 7.913) = 9 (6.283 - 7.913) = 9 (-1.63) = -14.67$$ 15. Therefore, the value of the triple integral is approximately $-14.67$.