Check Divergence Theorem on Unit Cube

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The discussion revolves around verifying the Divergence Theorem using the vector function v = <y^2, 2xy + z^2, 2yz> over a unit cube. A participant initially calculates the divergence as (∇·v) = 2y + 2x + 2y, but another contributor points out that it should be 2(x + y). The confusion arises from a miscalculation in the partial derivatives, specifically in the first term. After clarification, the participant acknowledges the mistake. The conversation highlights the importance of accurately applying the divergence formula in vector calculus.
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}=&lt;y^2, 2xy+z^2, 2yz&gt; 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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