1. **State the problem:** We need to evaluate the triple integral $$\iiint_V xy^2 z^3 \, dx \, dz \, dy$$ over the region $$V = \{(x,y,z) : z^2 \leq x \leq y, 0 \leq y \leq 2, \sqrt{y} \leq z \leq 1\}.$$ 2. **Analyze the region and order of integration:** The integral is given as $$\int_0^2 \int_{\sqrt{y}}^1 \int_{z^2}^y xy^2 z^3 \, dx \, dz \, dy.$$ 3. **Integrate with respect to $x$ first:** Treating $y$ and $z$ as constants, integrate $$\int_{z^2}^y x y^2 z^3 \, dx = y^2 z^3 \int_{z^2}^y x \, dx = y^2 z^3 \left[ \frac{x^2}{2} \right]_{x=z^2}^{x=y} = y^2 z^3 \left( \frac{y^2}{2} - \frac{z^4}{2} \right) = \frac{y^2 z^3}{2} (y^2 - z^4).$$ 4. **Rewrite the integral:** Now the integral reduces to $$\int_0^2 \int_{\sqrt{y}}^1 \frac{y^2 z^3}{2} (y^2 - z^4) \, dz \, dy = \frac{1}{2} \int_0^2 y^2 \int_{\sqrt{y}}^1 z^3 (y^2 - z^4) \, dz \, dy.$$ 5. **Expand the inner integrand:** $$z^3 (y^2 - z^4) = y^2 z^3 - z^7.$$ 6. **Integrate with respect to $z$:** $$\int_{\sqrt{y}}^1 (y^2 z^3 - z^7) \, dz = y^2 \int_{\sqrt{y}}^1 z^3 \, dz - \int_{\sqrt{y}}^1 z^7 \, dz.$$ Calculate each integral: - $$\int_{\sqrt{y}}^1 z^3 \, dz = \left[ \frac{z^4}{4} \right]_{\sqrt{y}}^1 = \frac{1}{4} - \frac{(\sqrt{y})^4}{4} = \frac{1}{4} - \frac{y^2}{4}.$$ - $$\int_{\sqrt{y}}^1 z^7 \, dz = \left[ \frac{z^8}{8} \right]_{\sqrt{y}}^1 = \frac{1}{8} - \frac{(\sqrt{y})^8}{8} = \frac{1}{8} - \frac{y^4}{8}.$$ 7. **Substitute back:** $$y^2 \left( \frac{1}{4} - \frac{y^2}{4} \right) - \left( \frac{1}{8} - \frac{y^4}{8} \right) = \frac{y^2}{4} - \frac{y^4}{4} - \frac{1}{8} + \frac{y^4}{8} = \frac{y^2}{4} - \frac{y^4}{4} + \frac{y^4}{8} - \frac{1}{8} = \frac{y^2}{4} - \frac{y^4}{8} - \frac{1}{8}.$$ 8. **Now the integral is:** $$\frac{1}{2} \int_0^2 y^2 \left( \frac{y^2}{4} - \frac{y^4}{8} - \frac{1}{8} \right) dy = \frac{1}{2} \int_0^2 \left( \frac{y^4}{4} - \frac{y^6}{8} - \frac{y^2}{8} \right) dy.$$ 9. **Split the integral:** $$= \frac{1}{2} \left( \int_0^2 \frac{y^4}{4} dy - \int_0^2 \frac{y^6}{8} dy - \int_0^2 \frac{y^2}{8} dy \right) = \frac{1}{2} \left( \frac{1}{4} \int_0^2 y^4 dy - \frac{1}{8} \int_0^2 y^6 dy - \frac{1}{8} \int_0^2 y^2 dy \right).$$ 10. **Calculate each integral:** - $$\int_0^2 y^4 dy = \left[ \frac{y^5}{5} \right]_0^2 = \frac{2^5}{5} = \frac{32}{5}.$$ - $$\int_0^2 y^6 dy = \left[ \frac{y^7}{7} \right]_0^2 = \frac{2^7}{7} = \frac{128}{7}.$$ - $$\int_0^2 y^2 dy = \left[ \frac{y^3}{3} \right]_0^2 = \frac{8}{3}.$$ 11. **Substitute values:** $$\frac{1}{2} \left( \frac{1}{4} \cdot \frac{32}{5} - \frac{1}{8} \cdot \frac{128}{7} - \frac{1}{8} \cdot \frac{8}{3} \right) = \frac{1}{2} \left( \frac{32}{20} - \frac{128}{56} - \frac{8}{24} \right).$$ Simplify fractions: - $$\frac{32}{20} = \frac{8}{5} = 1.6,$$ - $$\frac{128}{56} = \frac{32}{14} = \frac{16}{7} \approx 2.2857,$$ - $$\frac{8}{24} = \frac{1}{3} \approx 0.3333.$$ 12. **Calculate inside parentheses:** $$1.6 - 2.2857 - 0.3333 = 1.6 - 2.619 = -1.019.$$ 13. **Multiply by $\frac{1}{2}$:** $$\frac{1}{2} \times (-1.019) = -0.5095.$$ 14. **Final answer:** $$\boxed{-\frac{179}{351} \approx -0.5095}.$$ (Exact fraction found by common denominator: $$\frac{8}{5} - \frac{16}{7} - \frac{1}{3} = \frac{168}{105} - \frac{240}{105} - \frac{35}{105} = \frac{168 - 240 - 35}{105} = \frac{-107}{105}.$$ Then multiply by $\frac{1}{2}$: $$\frac{-107}{105} \times \frac{1}{2} = \frac{-107}{210}.$$ Rechecking arithmetic carefully: Actually, redoing step 11 with exact fractions: $$\frac{1}{4} \cdot \frac{32}{5} = \frac{32}{20} = \frac{8}{5},$$ $$\frac{1}{8} \cdot \frac{128}{7} = \frac{128}{56} = \frac{32}{14} = \frac{16}{7},$$ $$\frac{1}{8} \cdot \frac{8}{3} = \frac{8}{24} = \frac{1}{3}.$$ So inside parentheses: $$\frac{8}{5} - \frac{16}{7} - \frac{1}{3} = \frac{8}{5} - \left( \frac{16}{7} + \frac{1}{3} \right).$$ Find common denominator for $\frac{16}{7} + \frac{1}{3}$: $$\frac{16}{7} + \frac{1}{3} = \frac{16 \times 3}{21} + \frac{1 \times 7}{21} = \frac{48}{21} + \frac{7}{21} = \frac{55}{21}.$$ Now subtract: $$\frac{8}{5} - \frac{55}{21} = \frac{8 \times 21}{105} - \frac{55 \times 5}{105} = \frac{168}{105} - \frac{275}{105} = \frac{-107}{105}.$$ Multiply by $\frac{1}{2}$: $$\frac{-107}{105} \times \frac{1}{2} = \frac{-107}{210}.$$ So the exact final answer is $$\boxed{-\frac{107}{210}}.$$