## Thomas' Calculus 13th Edition

$-32$
We have: $F \cdot n =2uv+3v^2 -12$ and $\iint_{S} F \cdot n \ d S=\iint_{S} (2uv+3v^2 -12) \ dS$ or, $=\int_{-2}^{2} \int_{0}^{1} (2uv+3v^2 -12) \ du \ dv$ or, $= \int_{-2}^{2} [u^2 v+3uv^2-12u]_{0}^{1} \ dv$ or, $=\int_{-2}^{2} (v+3v^2-12) \ dv$ or, $=[\dfrac{v^2}{2}+v^3 -12 v]_{-2}^2$ or, $=(\dfrac{(2)^2}{2}+(2)^3 -12 \times 2) -(\dfrac{(-2)^2}{2}+(-)^3 -12 \times (-2))$ or, $=-32$