## Calculus: Early Transcendentals 8th Edition

$\dfrac{23}{5}$
Integrate the given integral first with respect to $z$, then $x$, and then $y$ $\int_{0}^1 \int_{y}^{2y} \int_{0}^{x+y} 6xy dz dx dy= \int_{0}^1 \int_{y}^{2y} |6xyz|_{0}^{x+y} dx dy=\int_{0}^1\int_y^{2y} 6xy(x+y) dx dy$ This implies that $\int_0^1 \int_y^{2y} [6x^2y+6xy^2]dx dy=(6) \int^{0}_{1}[\dfrac{1}{3} 8y^4+\dfrac{4y^4}{2}] dy$ Hence, $\int_{0}^1 \int_{y}^{2y} \int_{0}^{x+y} 6xy dz dx dy=6 [\dfrac{8y^5}{15}+\dfrac{2y^5}{5}]_{0}^{1}=\dfrac{23}{5}$