## Calculus 8th Edition

$\dfrac{8}{5}$
Let us consider that $V=\int_{-1}^1 \int_{x^2}^{1} \int_{0}^{1-y} dz dy dx= \int_{-1}^1 \int_{-x^2}^{1} [z]_{0}^{1-y} dy dx$ This implies that $\int_{-1}^{1} \int_{x^2}^{1} (1-y) dy dx=\int_{-1}^1[y-(\dfrac{1}{2})y^2]_{x^2}^1 dx$ or, $\int_{-1}^{1} \dfrac{1}{2}-x^2+\dfrac{x^4}{2} dx=[\dfrac{1}{2}x-\dfrac{1}{3}x^3+\dfrac{1}{(2)(5)}x^5]_{-1}^1$ Thus, $V= \dfrac{8}{5}$