How to use the divergence theorem to solve this question

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SUMMARY

The divergence theorem was incorrectly applied in the discussion regarding the volume of a half-cylinder. The correct volume calculation is ##\frac{\pi a^2 h}{2}##, as the surface integral must account for only half the cylinder. The divergence theorem states that the flux through a closed surface equals the volume integral of the divergence over the region enclosed. The confusion arose from not recognizing that the surface integral should only consider the specified half-cylinder, leading to an overestimation of the volume.

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Leo Liu
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The correct answer is ##\frac{\pi a^2 h} 2## by using the standard approach. However when I tried using the divergence theorem to solve this problem, I got a different answer. My work is as follows:
$$\iint_S \vec F\cdot\hat n\, dS = \iiint_D \nabla\cdot\vec F\,dV$$
$$= \iiint_D \frac{\partial y}{\partial y}\, dV$$
$$=\iiint_D \, dV=V_{cylinder}=\pi a^2h$$
Where did I go wrong?
 
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Is the surface integral over a closed surface?
 
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Your volume is only half the cylinder.
(After you add the surfaces needed to close the surface where the integral contribution is zero.)
 
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PeroK said:
Is the surface integral over a closed surface?
Nope. But IMO the flux passing through the led of the cylinder and the bottom of the cylinder is 0.
 
Orodruin said:
Your volume is only half the cylinder.
(After you add the surfaces needed to close the surface where the integral contribution is zero.)
I am sorry. I don't get it.
 
Leo Liu said:
I am sorry. I don't get it.
You've calculated the volume of the whole cylinder. The surface is only half the cylinder.
 
PeroK said:
You've calculated the volume of the whole cylinder. The surface is only half the cylinder.
1626972009451.png

Okay I see -- I didn't see the condition that the surface is to the right of the xz plane. Thank you! :smile:
 

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