Answer
$\dfrac{4}{7}$
Work Step by Step
Our aim is to integrate the triple integral as follows:
$$ V=2\int^1_0 \int^{1-y^2}_0 \int^{x^2+y^2}_0 \space dz \space dx \space dy \\=2\int^2_0 \int^{1-y^2}_0 (x^2+y^2) dx \space dy \\=2\int^1_0 [\dfrac{x^3}{3}+xy^2]^{1-y^2}_0 \space dy \\= 2\int^1_0 (1-y^2)[\dfrac{1}{3}(1-y^2)+y^2] \space dy \\= 2\int^1_0 (1-y^2)(\dfrac{1}{3}+\dfrac{1}{3}y^2+\dfrac{1}{3}y^4) \space dy \\= \dfrac{2}{3}\int^1_0 (1-y^6) \space dy \\=\dfrac{2}{3}[y-\dfrac{1}{7} \times y^{7}] \\ =\dfrac{4}{7}$$
