The out-flux of the vector field $F(x,y,z)=(-sin(2x)+ye^(3z),-(y+1)^2,2z(y+cos(2x)+3)$

Question

The out-flux of the vector field $F(x,y,z)=(-sin(2x)+ye^(3z),-(y+1)^2,2z(y+cos(2x)+3)$ from the domain $D = \{(x, y,z) \in \mathbb R^3\colon\ x^2+y^2+z^2≤4,x≤0,y≤0,z ≥ 0\}$. is equal to

$(A)16π/ 3\quad\ (B) -16π/3 \quad\ (C) -π/8\quad\ (D) π/8$

Picture of actual question is Picture of Real Question

Correct answer is 'A' but I am unable to get it. I have tried calculating the surface integral as well as using the divergence theorem. The calculations are difficult to be shown here as I am not acquainted with the Math Latex. Please help me solve the question

Answer 1

Given the vector field $\mathbf{F} : \mathbb{R}^3 \to \mathbb{R}^3$ of law:

$$ \mathbf{F}(x,\,y,\,z) := \left(-\sin(2x) + y\,e^{3z}, \; -(y+1)^2, \; 2z\left(y+\cos(2x)+3\right)\right) $$

and the domain:

$$ \Omega := \left\{ (x,\,y,\,z) \in \mathbb{R}^3 : x^2+y^2+z^2\le2^2, \; x \le 0, \; y \le 0, \; z \ge 0 \right\}, $$

by the divergence theorem, the outgoing flow of $\mathbf{F}$ from the boundary $\partial\Omega$ of $\Omega$ is calculated as:

$$ \Phi_{\partial\Omega}(\mathbf{F}) = \iiint\limits_{\Omega} \nabla \cdot \mathbf{F}\,\text{d}\omega = \iiint\limits_{\Omega} 4\,\text{d}x\,\text{d}y\,\text{d}z = 4 \cdot ||\Omega|| = 4 \cdot \frac{\frac{4}{3}\,\pi\,2^3}{8} = \frac{16}{3}\pi\,, $$

i.e. it's equal to four times the volume of an eighth of a ball of radius two.