Answer
$\dfrac{65}{28}$
Work Step by Step
$\iiint_E 6xy dV= \int_{0}^1 \int_{0}^{\sqrt x} \int_{0}^{1+x+y} (6xy) dz dy dx$
$= 6 \int_{0}^1\int_{0}^{\sqrt x} [xyz]_{0}^{1+x+y} dy dx$
$= 6 \int_{0}^1\int_{0}^{\sqrt x} xy(1+x+y) dy dx$
$=6 \int_{0}^1\int_{0}^{\sqrt x} [xy+x^2y+xy^2] dy dx$
$=\int_0^1 [3xy^2+3x^2y^2+2xy^3]_{0}^{\sqrt x}dx$
$= \int^{0}_{1}3x^2+3x^3+2x^{5/2} dx$
$= [x^3+\dfrac{3x^4}{4}+(2)\dfrac{2x^{7/2}}{7}]^{0}_{1}$
$= [(0-1)^3+\dfrac{3(0-1)^4}{4}+(2)\dfrac{2(0-1)^{7/2}}{7}]$
$=\dfrac{65}{28}$
