## Calculus: Early Transcendentals 8th Edition

$\dfrac{9}{2}$
The Divergence Theorem states that $\iint_S \overrightarrow{F}\cdot d\overrightarrow{S}=\iiint_Ediv \overrightarrow{F}dV$ Here, we have $\iint_S \overrightarrow{F}\cdot d\overrightarrow{S}=\iiint_Ediv \overrightarrow{F}dV =\int_0^1\int_0^1\int_0^1 (3x+3) dz dy dx$ $=\int_0^1\int_0^1 [3xz+3z]_0^1 dy dx$ $=\int_0^1\int_0^1 3x(1)+3(1)-0 dy dx$ $=\int_0^1\int_0^1 (3x+3) dy dx$ $=\int_0^1 (3x+3)dx$ $=[(3/2)x^2+3x]_0^1$ $=\dfrac{9}{2}$