Approximating triple integrals over tiny regions












0












$begingroup$


As a way of learning the divergence theorem, we have considered microscopic flux of a vector field through a tiny region.



Consider:



$$iiint_D (partial_x^2 + partial_y^2 +partial_z^2)(u) dV$$



We don't know anything about u except that it is a function of (t,x,y,z).



I am told that if we make the region D very tiny, then we can approximate the integral above as :



$$(partial_x^2 + partial_y^2 +partial_z^2)(u) cdot volumespace of D$$



Can someone explain why this is so? Is it something to do with the 2nd order partials being constants, or something? I'm having a hard time visualising



Thanks






calculus







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  • $begingroup$
    You can write an integral $$iiint_D f dV = left( frac{1}{|D|} iiint_D f dV right) |D|$$ where $|D|$ stands for the volume of $D$. The first term represents an average of $f$, which can be replaced by the value of $f$ if the region is small enough - a continuous function can't vary too much over a sufficiently small region. This is just a version of the fundamental theorem of calculus, really - taking $|D| to 0$ corresponds to differentiating an integral.
    $endgroup$
    – T. Bongers
    Dec 2 '18 at 23:06


















0












$begingroup$


As a way of learning the divergence theorem, we have considered microscopic flux of a vector field through a tiny region.



Consider:



$$iiint_D (partial_x^2 + partial_y^2 +partial_z^2)(u) dV$$



We don't know anything about u except that it is a function of (t,x,y,z).



I am told that if we make the region D very tiny, then we can approximate the integral above as :



$$(partial_x^2 + partial_y^2 +partial_z^2)(u) cdot volumespace of D$$



Can someone explain why this is so? Is it something to do with the 2nd order partials being constants, or something? I'm having a hard time visualising



Thanks






calculus







share|cite|improve this question









$endgroup$












  • $begingroup$
    You can write an integral $$iiint_D f dV = left( frac{1}{|D|} iiint_D f dV right) |D|$$ where $|D|$ stands for the volume of $D$. The first term represents an average of $f$, which can be replaced by the value of $f$ if the region is small enough - a continuous function can't vary too much over a sufficiently small region. This is just a version of the fundamental theorem of calculus, really - taking $|D| to 0$ corresponds to differentiating an integral.
    $endgroup$
    – T. Bongers
    Dec 2 '18 at 23:06
















0












0








0





$begingroup$


As a way of learning the divergence theorem, we have considered microscopic flux of a vector field through a tiny region.



Consider:



$$iiint_D (partial_x^2 + partial_y^2 +partial_z^2)(u) dV$$



We don't know anything about u except that it is a function of (t,x,y,z).



I am told that if we make the region D very tiny, then we can approximate the integral above as :



$$(partial_x^2 + partial_y^2 +partial_z^2)(u) cdot volumespace of D$$



Can someone explain why this is so? Is it something to do with the 2nd order partials being constants, or something? I'm having a hard time visualising



Thanks






calculus







share|cite|improve this question









$endgroup$




As a way of learning the divergence theorem, we have considered microscopic flux of a vector field through a tiny region.



Consider:



$$iiint_D (partial_x^2 + partial_y^2 +partial_z^2)(u) dV$$



We don't know anything about u except that it is a function of (t,x,y,z).



I am told that if we make the region D very tiny, then we can approximate the integral above as :



$$(partial_x^2 + partial_y^2 +partial_z^2)(u) cdot volumespace of D$$



Can someone explain why this is so? Is it something to do with the 2nd order partials being constants, or something? I'm having a hard time visualising



Thanks






calculus




calculus






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asked Dec 2 '18 at 23:04









big daddybig daddy

434




434












  • $begingroup$
    You can write an integral $$iiint_D f dV = left( frac{1}{|D|} iiint_D f dV right) |D|$$ where $|D|$ stands for the volume of $D$. The first term represents an average of $f$, which can be replaced by the value of $f$ if the region is small enough - a continuous function can't vary too much over a sufficiently small region. This is just a version of the fundamental theorem of calculus, really - taking $|D| to 0$ corresponds to differentiating an integral.
    $endgroup$
    – T. Bongers
    Dec 2 '18 at 23:06




















  • $begingroup$
    You can write an integral $$iiint_D f dV = left( frac{1}{|D|} iiint_D f dV right) |D|$$ where $|D|$ stands for the volume of $D$. The first term represents an average of $f$, which can be replaced by the value of $f$ if the region is small enough - a continuous function can't vary too much over a sufficiently small region. This is just a version of the fundamental theorem of calculus, really - taking $|D| to 0$ corresponds to differentiating an integral.
    $endgroup$
    – T. Bongers
    Dec 2 '18 at 23:06


















$begingroup$
You can write an integral $$iiint_D f dV = left( frac{1}{|D|} iiint_D f dV right) |D|$$ where $|D|$ stands for the volume of $D$. The first term represents an average of $f$, which can be replaced by the value of $f$ if the region is small enough - a continuous function can't vary too much over a sufficiently small region. This is just a version of the fundamental theorem of calculus, really - taking $|D| to 0$ corresponds to differentiating an integral.
$endgroup$
– T. Bongers
Dec 2 '18 at 23:06






$begingroup$
You can write an integral $$iiint_D f dV = left( frac{1}{|D|} iiint_D f dV right) |D|$$ where $|D|$ stands for the volume of $D$. The first term represents an average of $f$, which can be replaced by the value of $f$ if the region is small enough - a continuous function can't vary too much over a sufficiently small region. This is just a version of the fundamental theorem of calculus, really - taking $|D| to 0$ corresponds to differentiating an integral.
$endgroup$
– T. Bongers
Dec 2 '18 at 23:06












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$begingroup$

Intuitively, for a small region,
$$int f mathrm{d}V approx f int mathrm{d}V=fV$$
If the function $f$ is "nice enough", because we can treat $f$ as a constant there.






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    $begingroup$

    Intuitively, for a small region,
    $$int f mathrm{d}V approx f int mathrm{d}V=fV$$
    If the function $f$ is "nice enough", because we can treat $f$ as a constant there.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Intuitively, for a small region,
      $$int f mathrm{d}V approx f int mathrm{d}V=fV$$
      If the function $f$ is "nice enough", because we can treat $f$ as a constant there.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Intuitively, for a small region,
        $$int f mathrm{d}V approx f int mathrm{d}V=fV$$
        If the function $f$ is "nice enough", because we can treat $f$ as a constant there.






        share|cite|improve this answer









        $endgroup$



        Intuitively, for a small region,
        $$int f mathrm{d}V approx f int mathrm{d}V=fV$$
        If the function $f$ is "nice enough", because we can treat $f$ as a constant there.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 2 '18 at 23:08









        BotondBotond

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