Chapter 15 - Multiple Integrals - 15.7 Exercises - Page 1049: 21

$ \dfrac{8}{15}$

Here, we have $V=\int_{-1}^1 \int_{x^2}^{1} \int_{0}^{1-y} dz dy dx$ $= \int_{-1}^1 \int_{x^2}^{1} [z]_{0}^{1-y} dy dx $ $=\int_{-1}^{1} \int_{x^2}^{1} [1-y] dy dx$ $=\int_{-1}^1[y-\dfrac{y^2}{2}]_{x^2}^1 dx $ $=[\dfrac{x}{2}-\dfrac{x^3}{3}+\dfrac{x^5}{10}]_{-1}^1$ $= \dfrac{8}{15}$