By divergence theorem

$$\int\int \overline{N} \cdot \overline{F}ds = \int\int\limits_{V}\int \overline{V} \cdot \overline{F}dv$$

Now $\overline{F} = 2x^2yi - y^2 j + 4xz^2 k$

$$\overline{V} \cdot \overline{F} = 4xy - 2y + 8xz$$

For the limits on z axis as the radius is 3 thus z -> 0 to 3

For y upper limits can be used as equation of circle

$$\therefore \int\int\limits_{V}\int \overline{V} \cdot \overline{F}dv = \int\limits_{z = 0}^{3} \int\limits_{y = 0}^{\sqrt{9 - z^2}}\int\limits_{x = 0}^{z} (4xy - 2y + 8xz)dxdydz$$ $$= \int\limits_{0}^{3}\int\limits_{0}^{\sqrt{9 - z^2}}[4y + 16z]dydz$$

$$= \int\limits_{0}^{3}[2y^2 + 16yz]_{0}^{\sqrt{9 - z^2}}$$

$$= \int\limits_{0}^{2} 2(9 - z^2) + 16 \sqrt{9 - z^2} dz$$

$$ = \bigg[18z - \frac{2z^3}{3} + 16(9 - z^2)^{\frac{3}{2}} \bigg(\frac{2}{3} * \frac{-1}{2} \bigg) \bigg]_{0}^{3}$$ $$ = (18 * 3) - 2(3)^2 + 0 - 16(9)^{\frac{3}{2}} * \bigg(\frac{-1}{3} \bigg) $$

$$ = 18 * 2 + 16(9) $$ $$ = 180 $$