Answer
$-6$
Work Step by Step
We have the triple integral:
$\int^{\sqrt{2}}_0 \int^{3y}_0 \int^{8-x^2-y^2}_{x^2+3y^2} dz dx dy=\int^{\sqrt{2}}_0 \int^{3y}_0(8-2x^2-4y^2) dx \space dy $
or, $=\int^{\sqrt{2}}_0 [8x-\dfrac{2}{3}-4xy^2]^{3y}_0 dy $
or, $=\int^{\sqrt{2}}_0 (24y-18y^3-12y^3)dy $
or, $= [12y^2-\frac{15}{2}y^4]^{\sqrt{2}}_0=-6$
Therefore, $\int^{\sqrt{2}}_0 \int^{3y}_0 \int^{8-x^2-y^2}_{x^2+3y^2} dz \space dx \space dy=-6$
