# 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?

Homework Helper
##\frac {3\pi} {2}\leq\theta\leq 0## ?

Homework Helper
##\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

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