1. **Problem Statement:** Evaluate the triple integral $$\int_0^2 \int_0^{\sqrt{4-x^2}} \int_{-5+x^2+y^2}^{3-x^2-y^2} x \, dz \, dy \, dx$$. 2. **Change to simplify:** Change the order of integration or the limits to make the integral easier. Here, we change the innermost integral limits to symmetric bounds around zero: from $$-5 + x^2 + y^2$$ to $$5 - x^2 - y^2$$. 3. **New integral:** $$\int_0^2 \int_0^{\sqrt{4-x^2}} \int_{-5+x^2+y^2}^{5-x^2-y^2} x \, dz \, dy \, dx$$. 4. **Evaluate the innermost integral:** $$\int_{-5+x^2+y^2}^{5-x^2-y^2} x \, dz = x \left[ z \right]_{-5+x^2+y^2}^{5-x^2-y^2} = x \left( (5 - x^2 - y^2) - (-5 + x^2 + y^2) \right) = x (10 - 2x^2 - 2y^2) = 10x - 2x^3 - 2x y^2$$. 5. **Rewrite the double integral:** $$\int_0^2 \int_0^{\sqrt{4-x^2}} (10x - 2x^3 - 2x y^2) \, dy \, dx$$. 6. **Integrate with respect to $y$:** $$\int_0^{\sqrt{4-x^2}} (10x - 2x^3 - 2x y^2) \, dy = (10x - 2x^3) y - 2x \frac{y^3}{3} \Big|_0^{\sqrt{4-x^2}}$$ Substitute $y = \sqrt{4-x^2}$: $$= (10x - 2x^3) \sqrt{4-x^2} - \frac{2x}{3} (4 - x^2)^{3/2}$$. 7. **Final integral over $x$:** $$\int_0^2 \left[ (10x - 2x^3) \sqrt{4-x^2} - \frac{2x}{3} (4 - x^2)^{3/2} \right] dx$$. 8. **Use substitution $u = 4 - x^2$, $du = -2x dx$ to evaluate:** Split integral: $$I = \int_0^2 (10x - 2x^3) \sqrt{4-x^2} \, dx - \frac{2}{3} \int_0^2 x (4 - x^2)^{3/2} \, dx$$ Rewrite first integral: $$10x - 2x^3 = 2x(5 - x^2)$$ So, $$I = \int_0^2 2x (5 - x^2) \sqrt{4 - x^2} \, dx - \frac{2}{3} \int_0^2 x (4 - x^2)^{3/2} \, dx$$ Substitute $u = 4 - x^2$, $du = -2x dx$, so $-du = 2x dx$. First integral: $$\int_0^2 2x (5 - x^2) \sqrt{4 - x^2} \, dx = \int_{u=4}^{0} (5 - (4 - u)) u^{1/2} (-du) = \int_0^4 (1 + u) u^{1/2} du = \int_0^4 (u^{1/2} + u^{3/2}) du$$ Evaluate: $$\int_0^4 u^{1/2} du = \frac{2}{3} u^{3/2} \Big|_0^4 = \frac{2}{3} \times 8 = \frac{16}{3}$$ $$\int_0^4 u^{3/2} du = \frac{2}{5} u^{5/2} \Big|_0^4 = \frac{2}{5} \times 32 = \frac{64}{5}$$ Sum: $$\frac{16}{3} + \frac{64}{5} = \frac{80}{15} + \frac{192}{15} = \frac{272}{15}$$ Second integral: $$\frac{2}{3} \int_0^2 x (4 - x^2)^{3/2} dx = \frac{2}{3} \times \int_4^0 u^{3/2} \left(-\frac{du}{2x}\right) \text{ but since } du = -2x dx, \text{ we have } x dx = -\frac{du}{2}$$ So, $$\int_0^2 x (4 - x^2)^{3/2} dx = -\frac{1}{2} \int_4^0 u^{3/2} du = \frac{1}{2} \int_0^4 u^{3/2} du = \frac{1}{2} \times \frac{64}{5} = \frac{32}{5}$$ Multiply by $\frac{2}{3}$: $$\frac{2}{3} \times \frac{32}{5} = \frac{64}{15}$$ 9. **Combine results:** $$I = \frac{272}{15} - \frac{64}{15} = \frac{208}{15}$$ **Final answer:** $$\boxed{\frac{208}{15}}$$