Outward flux of a vector field

[DottZakapa]

Homework Statement:

The out-flux of the vector field
F(x,y,z) = (sin(2x) + ye3z,(y + 1)2,−2z(y + cos(2x) + 3)
from the domain
D = {(x, y, z) ∈ R^3 : x^2 + y^2 + z^2 ≤ 1, x ≥ 0, y ≤ 0, z ≥ 0}

Relevant Equations:

flux of a vector field

My idea is to evaluate it using gauss theorem/divergence theorem.

so the divergence would be

$divF = (\cos (2x)2+2y+2-2z ( y+\cos (2x)+3) )$

is it correct?
In this way i'ma able to compute a triple integral on the volume given by the domain

$D = \left\{ (x, y, z) ∈ R^3 : x^2 + y^2 + z^2 ≤ 1, x ≥ 0, y ≤ 0, z ≥ 0 \right\}$
then using spherical coordinates
$x= r cos\theta \sin\phi\\ y= r\sin\theta\sin\phi\\ z=r\sin\phi$
with the following boundaries

$0\leq\phi\leq\frac \pi 2$

$\frac {3\pi} {2}\leq\theta\leq 0$

then i do the substitution in the divergence and solve the integral

$\iiint_V divF r^2\sin\phi dr d\theta d\phi$

am i doing it correctly or is there anything wrong?

 

[BvU]

$\frac {3\pi} {2}\leq\theta\leq 0$ ?

 

[BvU]

$\nabla\cdot\vec F = (\cos (2x)2+2y+2-2z ( y+\cos (2x)+3) )$

$2z$ ?

 

[DottZakapa]

$\nabla\cdot\vec F = (\cos (2x)2+2y+2-2z ( y+\cos (2x)+3) )$

$2z$ ?

Ops i did not notice
So
divF = -4

 

[DottZakapa]

$\frac {3\pi} {2}\leq\theta\leq 0$ ?

If that is not a correct bound then i did not understend how to find it.
If someone would be so kind to explain it

 

[BvU]

If $\theta \le 0$ it can't be bigger than ${3\over 2}\pi$

 

[DottZakapa]

If $\theta \le 0$ it can't be bigger than ${3\over 2}\pi$

So ${3\over 2}\pi$to ${2\pi}$