Why is the volume integral of zero equal to zero?

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The discussion clarifies that the volume integral of zero equals zero due to the nature of integration types. Volume integrals are treated as iterated definite integrals, while the integral of zero refers to indefinite integration. The expression for the volume integral of zero, represented as ∭V 0 dV, simplifies to zero because zero is a constant that can be factored out. Thus, the result of the volume integral remains zero regardless of the volume V. Understanding these distinctions in integration types is crucial for grasping why the volume integral of zero is zero.
tomwilliam2
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If the integral of zero is a constant, then why is the volume integral of zero just zero?
 
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You're addressing two types of integration. <Volume integrals> can be expressed as interated definite integration, while <integral of 0 is a constant> means that you're speaking of indefinite integration.
 
tomwilliam2 said:
If the integral of zero is a constant, then why is the volume integral of zero just zero?

Consider the expression ##\displaystyle \iiint\limits_{V}0 \, dV##, which I believe you mean by "volume integral". Since 0 is, itself, a constant, we can pull it out front, getting ##\displaystyle 0\iiint\limits_{V} \, dV = 0V = 0##
 
Thanks, both of you.
 
Good morning I have been refreshing my memory about Leibniz differentiation of integrals and found some useful videos from digital-university.org on YouTube. Although the audio quality is poor and the speaker proceeds a bit slowly, the explanations and processes are clear. However, it seems that one video in the Leibniz rule series is missing. While the videos are still present on YouTube, the referring website no longer exists but is preserved on the internet archive...

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