# Outward flux of a vector field

• DottZakapa
In summary, the conversation discusses using the Gauss and divergence theorems to evaluate a triple integral over a given volume. The speaker suggests using spherical coordinates with specific boundaries and solving the integral using substitution. There is some confusion about the correct bounds for one of the variables.

#### 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?

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

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

##2z## ?

BvU said:
##\nabla\cdot\vec F = (\cos (2x)2+2y+2-2z ( y+\cos (2x)+3) )##

##2z## ?
Ops i did not notice
So
divF = -4

BvU said:
##\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##

BvU said:
If ##\theta \le 0 ## it can't be bigger than ##{3\over 2}\pi##
So ##{3\over 2}\pi##to ##{2\pi}##

BvU

## 1. What is the concept of outward flux in a vector field?

The outward flux of a vector field is a measure of the flow of a vector field through a closed surface. It represents the amount of "stuff" flowing out of the surface, and is usually expressed as a scalar value.

## 2. How is outward flux calculated?

Outward flux is calculated by taking the dot product of the vector field and the unit normal vector of the surface, and then integrating this product over the surface. This is represented by the surface integral ∫∫F⋅n dS, where F is the vector field and n is the unit normal vector.

## 3. What is the significance of outward flux in vector calculus?

Outward flux is an important concept in vector calculus as it allows us to quantify the flow of a vector field through a surface. It has applications in many areas such as fluid dynamics, electromagnetism, and thermodynamics.

## 4. How does the orientation of the surface affect the calculation of outward flux?

The orientation of the surface is crucial in the calculation of outward flux. If the surface is oriented in the same direction as the unit normal vector, the flux will be positive. However, if the surface is oriented in the opposite direction, the flux will be negative. It is important to pay attention to the orientation of the surface when setting up the integral for calculating outward flux.

## 5. Can outward flux ever be negative?

Yes, outward flux can be negative. This occurs when the surface is oriented in the opposite direction of the unit normal vector, resulting in a negative dot product. This indicates that the "stuff" is flowing into the surface, rather than out of it. It is important to consider the sign of the outward flux when interpreting the results of a calculation.