# Chapter 15: Multiple Integrals - Section 15.5 - Triple Integrals in Rectangular Coordinates - Exercises 15.5 - Page 900: 8

-6

#### Work Step by Step

$\int^{\sqrt{2}}_0 \int^{3y}_0 \int^{8-x^2-y^2}_{x^2+3y^2}$ dz dx dy =$\int^{\sqrt{2}}_0 \int^{3y}_0(8-2x^2-4y^2)$ dx dy =$\int^{\sqrt{2}}_0 [8x-\frac{2}{3}-4xy^2]^{3y}_0$ dy =$\int^{\sqrt{2}}_0 (24y-18y^3-12y^3)dy$ =$[12y^2-\frac{15}{2}y^4]^{\sqrt{2}}_0$ =24-30 =-6

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