Approximating triple integrals over tiny regions
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
asked Dec 2 '18 at 23:04
big daddy
434
add a comment |
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
asked Dec 2 '18 at 23:04
big daddy
434
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.
– T. Bongers
Dec 2 '18 at 23:06
add a comment |
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
asked Dec 2 '18 at 23:04
big daddy
434
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
asked Dec 2 '18 at 23:04
big daddy
434
asked Dec 2 '18 at 23:04
big daddy
434
asked Dec 2 '18 at 23:04
big daddy
434
asked Dec 2 '18 at 23:04
big daddy
434
asked Dec 2 '18 at 23:04
big daddy
434
434
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.
– T. Bongers
Dec 2 '18 at 23:06
add a comment |
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.
– T. Bongers
Dec 2 '18 at 23:06
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.
– T. Bongers
Dec 2 '18 at 23:06
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.
– T. Bongers
Dec 2 '18 at 23:06
add a comment |
1 Answer
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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.
answered Dec 2 '18 at 23:08
Botond
5,5532732
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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.
answered Dec 2 '18 at 23:08
Botond
5,5532732
add a comment |
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.
answered Dec 2 '18 at 23:08
Botond
5,5532732
add a comment |
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.
answered Dec 2 '18 at 23:08
Botond
5,5532732
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.
answered Dec 2 '18 at 23:08
Botond
5,5532732
answered Dec 2 '18 at 23:08
Botond
5,5532732
answered Dec 2 '18 at 23:08
Botond
5,5532732
answered Dec 2 '18 at 23:08
Botond
5,5532732
5,5532732
add a comment |
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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.
– T. Bongers
Dec 2 '18 at 23:06