Calculate $\int_{∂B_R(0)} F \cdot ν dS_2$

Question

So I have this exercise where I want to check if I have done everything right.

Let $R > 0$ and $B_R(0) :=\{ x ∈ \mathbb{R}^3 : x_{1}^{2} + x_{2}^{2} + x_{3}^{2} < R\}$ be the open ball with radius R in $\mathbb{R}^3$ . Let $F : \mathbb{R}^3 → \mathbb{R}^3$ be given by $F(x1, x2, x3) := (x_1x_{2}^{2} ,x_{1}^{2}x_2 + e^{x_3} , x_{3}^{3}/3 + x_{2}^{4})$ .

1. State the exterior normal vector $ν(x)$ for $x ∈ ∂B_R(0)$ without proof.
2. Calculate: $$\int_{∂B_R(0)} F\cdot ν dS_2.$$

Hint: It does not need to be shown that $B_R(0)$ is a $C^1$−bounded set.

So what I have done is:

i)$v(x)=(x_2^2,x_1^2,x_3^2)$

From there for 2. we have: $$\int_{∂B_R(0)} F\cdot ν dS_2\overset{\text{Using Gauss}}{=} \\ x_2^2+x_1^2+x_3^3d \lambda_3(x)\overset{\text{Using polar coordinates}}{=}\\ \int_{0}^{R}\int_{0}^{2\pi}\int_{0}^{pi}r^4sin(\phi)d\phi, d\alpha, dr= \\ \int_{0}^{R}r^4 \int_{0}^{2\pi}\int_{0}^{pi}sin(\phi)d\phi, d\alpha, dr\\ =\int_{0}^{R}r^4 \int_{0}^{2\pi}[-cos(\phi)]_{0}^{pi}d\alpha, dr\\ =\int_{0}^{R}r^4 \int_{0}^{2\pi} 1d\alpha, dr=\\ 2\pi \int_{0}^{R}r^4 , dr=2\pi R/5$$

Is everything right or have I done any mistakes?