Check Divergence Theorem on Unit Cube

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SUMMARY

The discussion centers on verifying the Divergence Theorem using the vector field \(\bold{v} = \langle y^2, 2xy + z^2, 2yz \rangle\) within a unit cube. The user initially calculates the divergence as \((\nabla \cdot \bold{v}) = 2y + 2x + 2y\), which is incorrect. The correct divergence, as pointed out by another user, is \(2(x + y)\), highlighting the importance of accurately applying the formula \(\nabla \cdot \bold{v} = \frac{\partial v_x}{\partial x} + \frac{\partial v_y}{\partial y} + \frac{\partial v_z}{\partial z}\).

PREREQUISITES
  • Understanding of vector calculus, specifically the Divergence Theorem.
  • Familiarity with partial derivatives and their application in vector fields.
  • Knowledge of the notation and operations involving vector fields.
  • Basic proficiency in evaluating integrals over volume and surface.
NEXT STEPS
  • Review the Divergence Theorem and its applications in vector calculus.
  • Practice calculating divergences of various vector fields.
  • Explore examples of applying the Divergence Theorem to different geometric shapes.
  • Study the implications of divergence in physical contexts, such as fluid dynamics.
USEFUL FOR

Students studying vector calculus, educators teaching the Divergence Theorem, and anyone seeking to deepen their understanding of vector fields and their properties.

Saladsamurai
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Homework Statement


Check the Divergence Theorem \int_V(\nabla\cdot\bold{v})\,d\tau=\oint_S\bold{v}\cdot d\bold{a}

using the function \bold{v}=<y^2, 2xy+z^2, 2yz> and the unit cube below.
Photo1-1.jpg


Now when I calculate the divergence I get
(\nabla\cdot\bold{v})=2y+2x+2y

but Griffith's says that it is 2(x+y)

before I continue, I need to know what the heck I am missing?
 
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Saladsamurai said:
Now when I calculate the divergence I get
(\nabla\cdot\bold{v})=2y+2x+2y
How did you get that? You should have \frac{\partial v_x}{\partial x} + \frac{\partial v_y}{\partial y} + \frac{\partial v_z}{\partial z}. Check your first term \frac{\partial v_x}{\partial x}.
 
Defennder said:
How did you get that? You should have \frac{\partial v_x}{\partial x} + \frac{\partial v_y}{\partial y} + \frac{\partial v_z}{\partial z}. Check your first term \frac{\partial v_x}{\partial x}.

Oh man. Thanks Defennder. I am a retard :smile:
 

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