Query about finding an expression for dV Announcing the arrival of Valued Associate #679:...
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Query about finding an expression for dV
Announcing the arrival of Valued Associate #679: Cesar Manara
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I was trying to prove the hydrostatic equilibrium relationship, the only way I found that it works, and I saw this done by someone online too, was by saying that $dV=dA dr$. But if $V=Ar$ then surely $dV=rdA+Adr$.
How can we say $dV=dAdr$ here?
Source image
calculus
asked Mar 24 at 12:51
Vishal JainVishal Jain
62
$endgroup$
add a comment |
$begingroup$
I was trying to prove the hydrostatic equilibrium relationship, the only way I found that it works, and I saw this done by someone online too, was by saying that $dV=dA dr$. But if $V=Ar$ then surely $dV=rdA+Adr$.
How can we say $dV=dAdr$ here?
Source image
calculus
asked Mar 24 at 12:51
Vishal JainVishal Jain
62
$endgroup$
$begingroup$
$V=Ar$ is irrelevant. There is nothing in the diagram labeled $A$ or $V$. $dV = dA;dr$ is just "the volume of a cylinder = base area $times$ height".
$endgroup$
– alephzero
Mar 24 at 13:00
add a comment |
$begingroup$
I was trying to prove the hydrostatic equilibrium relationship, the only way I found that it works, and I saw this done by someone online too, was by saying that $dV=dA dr$. But if $V=Ar$ then surely $dV=rdA+Adr$.
How can we say $dV=dAdr$ here?
Source image
calculus
asked Mar 24 at 12:51
Vishal JainVishal Jain
62
$endgroup$
I was trying to prove the hydrostatic equilibrium relationship, the only way I found that it works, and I saw this done by someone online too, was by saying that $dV=dA dr$. But if $V=Ar$ then surely $dV=rdA+Adr$.
How can we say $dV=dAdr$ here?
Source image
calculus
calculus
asked Mar 24 at 12:51
Vishal JainVishal Jain
62
asked Mar 24 at 12:51
Vishal JainVishal Jain
62
asked Mar 24 at 12:51
Vishal JainVishal Jain
62
asked Mar 24 at 12:51
Vishal JainVishal Jain
62
asked Mar 24 at 12:51
Vishal JainVishal Jain
62
62
$begingroup$
$V=Ar$ is irrelevant. There is nothing in the diagram labeled $A$ or $V$. $dV = dA;dr$ is just "the volume of a cylinder = base area $times$ height".
$endgroup$
– alephzero
Mar 24 at 13:00
add a comment |
$begingroup$
$V=Ar$ is irrelevant. There is nothing in the diagram labeled $A$ or $V$. $dV = dA;dr$ is just "the volume of a cylinder = base area $times$ height".
$endgroup$
– alephzero
Mar 24 at 13:00
$begingroup$
$V=Ar$ is irrelevant. There is nothing in the diagram labeled $A$ or $V$. $dV = dA;dr$ is just "the volume of a cylinder = base area $times$ height".
$endgroup$
– alephzero
Mar 24 at 13:00
$begingroup$
$V=Ar$ is irrelevant. There is nothing in the diagram labeled $A$ or $V$. $dV = dA;dr$ is just "the volume of a cylinder = base area $times$ height".
$endgroup$
– alephzero
Mar 24 at 13:00
add a comment |
1 Answer
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$begingroup$
$$frac{mathrm{d}V}{mathrm{d}r}=frac{mathrm{d}A}{mathrm{d}r}r+Acdot1=A,$$
where $frac{mathrm{d}A}{mathrm{d}r}=0$ ($A$ stays constant over $r$!), so $$mathrm{d}V=A,mathrm{d}r$$
(or $mathrm{d}V=mathrm{d}A,mathrm{d}r$ if you call the area $mathrm{d}A$ instead of $A$, like in the image).
answered Mar 24 at 12:55
st.mathst.math
1,268115
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$begingroup$
$$frac{mathrm{d}V}{mathrm{d}r}=frac{mathrm{d}A}{mathrm{d}r}r+Acdot1=A,$$
where $frac{mathrm{d}A}{mathrm{d}r}=0$ ($A$ stays constant over $r$!), so $$mathrm{d}V=A,mathrm{d}r$$
(or $mathrm{d}V=mathrm{d}A,mathrm{d}r$ if you call the area $mathrm{d}A$ instead of $A$, like in the image).
answered Mar 24 at 12:55
st.mathst.math
1,268115
$endgroup$
add a comment |
$begingroup$
$$frac{mathrm{d}V}{mathrm{d}r}=frac{mathrm{d}A}{mathrm{d}r}r+Acdot1=A,$$
where $frac{mathrm{d}A}{mathrm{d}r}=0$ ($A$ stays constant over $r$!), so $$mathrm{d}V=A,mathrm{d}r$$
(or $mathrm{d}V=mathrm{d}A,mathrm{d}r$ if you call the area $mathrm{d}A$ instead of $A$, like in the image).
answered Mar 24 at 12:55
st.mathst.math
1,268115
$endgroup$
add a comment |
$begingroup$
$$frac{mathrm{d}V}{mathrm{d}r}=frac{mathrm{d}A}{mathrm{d}r}r+Acdot1=A,$$
where $frac{mathrm{d}A}{mathrm{d}r}=0$ ($A$ stays constant over $r$!), so $$mathrm{d}V=A,mathrm{d}r$$
(or $mathrm{d}V=mathrm{d}A,mathrm{d}r$ if you call the area $mathrm{d}A$ instead of $A$, like in the image).
answered Mar 24 at 12:55
st.mathst.math
1,268115
$endgroup$
$$frac{mathrm{d}V}{mathrm{d}r}=frac{mathrm{d}A}{mathrm{d}r}r+Acdot1=A,$$
where $frac{mathrm{d}A}{mathrm{d}r}=0$ ($A$ stays constant over $r$!), so $$mathrm{d}V=A,mathrm{d}r$$
(or $mathrm{d}V=mathrm{d}A,mathrm{d}r$ if you call the area $mathrm{d}A$ instead of $A$, like in the image).
answered Mar 24 at 12:55
st.mathst.math
1,268115
edited Mar 24 at 13:02
answered Mar 24 at 12:55
st.mathst.math
1,268115
answered Mar 24 at 12:55
st.mathst.math
1,268115
answered Mar 24 at 12:55
st.mathst.math
1,268115
1,268115
add a comment |
add a comment |
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$begingroup$
$V=Ar$ is irrelevant. There is nothing in the diagram labeled $A$ or $V$. $dV = dA;dr$ is just "the volume of a cylinder = base area $times$ height".
$endgroup$
– alephzero
Mar 24 at 13:00