Mixing
Calculate the volume of points in a cube where a function f(x,y,z) satisfies given conditions
$begingroup$
A value for each point in my data-set is defined as $$v=xyz$$where $$ale xle b$$$$cle yle d$$$$ele zle f$$
for ease of visualisation, for now I will assume the following:
$$a=c=e=0$$$$b=d=f=1$$$$0le Lle 1$$
Treating $xyz$ as a cube, I am trying to calculate the volume of the cube where $v<L$. I have plotted a visualisation bellow as an aid in understanding my problem, where $L=0.2$.
For example, in the above cube I am trying to compute the volume of the yellow isovolume.
I thought that I might acheive this by setting $xyz=0.2$, and then triple integrating the function as follows:
$$int_a^bint_c^dint_e^f(xyz-0.2),dz,dy,dx$$
Which with my limits of $0$ and $1$ gives the following:
$$volume=frac{x^2y^2z^2}{8} - 0.2xyz$$
However, the result of this function is $-0.075$, which is both smaller than expected, and also negative. I am unsure if this behaviour is due to the asymptotes in the function?
Does anyone have any idea how I can acheive my goal, and also point out the errors in my method? It would be much appreciated.
Thank you in advance!
integration geometry limits volume
edited Feb 3 at 8:04
choco_addicted
8,08461947
asked Feb 2 at 22:51
CynicalBritCynicalBrit
85
$endgroup$
add a comment |
$begingroup$
A value for each point in my data-set is defined as $$v=xyz$$where $$ale xle b$$$$cle yle d$$$$ele zle f$$
for ease of visualisation, for now I will assume the following:
$$a=c=e=0$$$$b=d=f=1$$$$0le Lle 1$$
Treating $xyz$ as a cube, I am trying to calculate the volume of the cube where $v<L$. I have plotted a visualisation bellow as an aid in understanding my problem, where $L=0.2$.
For example, in the above cube I am trying to compute the volume of the yellow isovolume.
I thought that I might acheive this by setting $xyz=0.2$, and then triple integrating the function as follows:
$$int_a^bint_c^dint_e^f(xyz-0.2),dz,dy,dx$$
Which with my limits of $0$ and $1$ gives the following:
$$volume=frac{x^2y^2z^2}{8} - 0.2xyz$$
However, the result of this function is $-0.075$, which is both smaller than expected, and also negative. I am unsure if this behaviour is due to the asymptotes in the function?
Does anyone have any idea how I can acheive my goal, and also point out the errors in my method? It would be much appreciated.
Thank you in advance!
integration geometry limits volume
edited Feb 3 at 8:04
choco_addicted
8,08461947
asked Feb 2 at 22:51
CynicalBritCynicalBrit
85
$endgroup$
$begingroup$
You are subtracting from your cube's 'volume' another one with constant value $0.2$ at each point. There's no reason a subtraction can't yield a negative result. What's $v<L$ supposed to mean?
$endgroup$
– lightxbulb
Feb 2 at 22:56
$begingroup$
$L$ is an arbitrary number between $0$ and $1$ sorry I should have listed that in my question. I gave the example value of 0.2 for use in the attached image. With the equation given that may be true, but I would not expect it to be possible to calculate a negative volume? I am not sure my method is correct.
$endgroup$
– CynicalBrit
Feb 2 at 23:26
$begingroup$
It's possible to get a negative number, the reason however why you get a negative number here is that you subtract volumes. If I understood correctly what you want to do, then subtracting is wrong. You seem to simply want $v<L$, in which case this should transfer over to the limits. Note that $f(x,y,z)= xyz - L = 0$ defines a surface. You want to integrate below that surface, that is your integral limits should be dependent.
$endgroup$
– lightxbulb
Feb 3 at 0:05
$begingroup$
Yes that's it exactly. Forgive me, what little calculus I do know is very rusty. Please could you explain how I would insert the $v<L$ into the limits of my integration. I thought that the limits of my integrals with respect to $x$, $y$, and $z$ were setting the bounding box within which I want to calculate the volume below the surface?
$endgroup$
– CynicalBrit
Feb 3 at 0:35
add a comment |
$begingroup$
A value for each point in my data-set is defined as $$v=xyz$$where $$ale xle b$$$$cle yle d$$$$ele zle f$$
for ease of visualisation, for now I will assume the following:
$$a=c=e=0$$$$b=d=f=1$$$$0le Lle 1$$
Treating $xyz$ as a cube, I am trying to calculate the volume of the cube where $v<L$. I have plotted a visualisation bellow as an aid in understanding my problem, where $L=0.2$.
For example, in the above cube I am trying to compute the volume of the yellow isovolume.
I thought that I might acheive this by setting $xyz=0.2$, and then triple integrating the function as follows:
$$int_a^bint_c^dint_e^f(xyz-0.2),dz,dy,dx$$
Which with my limits of $0$ and $1$ gives the following:
$$volume=frac{x^2y^2z^2}{8} - 0.2xyz$$
However, the result of this function is $-0.075$, which is both smaller than expected, and also negative. I am unsure if this behaviour is due to the asymptotes in the function?
Does anyone have any idea how I can acheive my goal, and also point out the errors in my method? It would be much appreciated.
Thank you in advance!
integration geometry limits volume
edited Feb 3 at 8:04
choco_addicted
8,08461947
asked Feb 2 at 22:51
CynicalBritCynicalBrit
85
$endgroup$
A value for each point in my data-set is defined as $$v=xyz$$where $$ale xle b$$$$cle yle d$$$$ele zle f$$
for ease of visualisation, for now I will assume the following:
$$a=c=e=0$$$$b=d=f=1$$$$0le Lle 1$$
Treating $xyz$ as a cube, I am trying to calculate the volume of the cube where $v<L$. I have plotted a visualisation bellow as an aid in understanding my problem, where $L=0.2$.
For example, in the above cube I am trying to compute the volume of the yellow isovolume.
I thought that I might acheive this by setting $xyz=0.2$, and then triple integrating the function as follows:
$$int_a^bint_c^dint_e^f(xyz-0.2),dz,dy,dx$$
Which with my limits of $0$ and $1$ gives the following:
$$volume=frac{x^2y^2z^2}{8} - 0.2xyz$$
However, the result of this function is $-0.075$, which is both smaller than expected, and also negative. I am unsure if this behaviour is due to the asymptotes in the function?
Does anyone have any idea how I can acheive my goal, and also point out the errors in my method? It would be much appreciated.
Thank you in advance!
integration geometry limits volume
integration geometry limits volume
edited Feb 3 at 8:04
choco_addicted
8,08461947
asked Feb 2 at 22:51
CynicalBritCynicalBrit
85
edited Feb 3 at 8:04
choco_addicted
8,08461947
asked Feb 2 at 22:51
CynicalBritCynicalBrit
85
edited Feb 3 at 8:04
choco_addicted
8,08461947
edited Feb 3 at 8:04
choco_addicted
8,08461947
edited Feb 3 at 8:04
choco_addicted
8,08461947
8,08461947
asked Feb 2 at 22:51
CynicalBritCynicalBrit
85
asked Feb 2 at 22:51
CynicalBritCynicalBrit
85
asked Feb 2 at 22:51
CynicalBritCynicalBrit
85
85
$begingroup$
You are subtracting from your cube's 'volume' another one with constant value $0.2$ at each point. There's no reason a subtraction can't yield a negative result. What's $v<L$ supposed to mean?
$endgroup$
– lightxbulb
Feb 2 at 22:56
$begingroup$
$L$ is an arbitrary number between $0$ and $1$ sorry I should have listed that in my question. I gave the example value of 0.2 for use in the attached image. With the equation given that may be true, but I would not expect it to be possible to calculate a negative volume? I am not sure my method is correct.
$endgroup$
– CynicalBrit
Feb 2 at 23:26
$begingroup$
It's possible to get a negative number, the reason however why you get a negative number here is that you subtract volumes. If I understood correctly what you want to do, then subtracting is wrong. You seem to simply want $v<L$, in which case this should transfer over to the limits. Note that $f(x,y,z)= xyz - L = 0$ defines a surface. You want to integrate below that surface, that is your integral limits should be dependent.
$endgroup$
– lightxbulb
Feb 3 at 0:05
$begingroup$
Yes that's it exactly. Forgive me, what little calculus I do know is very rusty. Please could you explain how I would insert the $v<L$ into the limits of my integration. I thought that the limits of my integrals with respect to $x$, $y$, and $z$ were setting the bounding box within which I want to calculate the volume below the surface?
$endgroup$
– CynicalBrit
Feb 3 at 0:35
add a comment |
$begingroup$
You are subtracting from your cube's 'volume' another one with constant value $0.2$ at each point. There's no reason a subtraction can't yield a negative result. What's $v<L$ supposed to mean?
$endgroup$
– lightxbulb
Feb 2 at 22:56
$begingroup$
$L$ is an arbitrary number between $0$ and $1$ sorry I should have listed that in my question. I gave the example value of 0.2 for use in the attached image. With the equation given that may be true, but I would not expect it to be possible to calculate a negative volume? I am not sure my method is correct.
$endgroup$
– CynicalBrit
Feb 2 at 23:26
$begingroup$
It's possible to get a negative number, the reason however why you get a negative number here is that you subtract volumes. If I understood correctly what you want to do, then subtracting is wrong. You seem to simply want $v<L$, in which case this should transfer over to the limits. Note that $f(x,y,z)= xyz - L = 0$ defines a surface. You want to integrate below that surface, that is your integral limits should be dependent.
$endgroup$
– lightxbulb
Feb 3 at 0:05
$begingroup$
Yes that's it exactly. Forgive me, what little calculus I do know is very rusty. Please could you explain how I would insert the $v<L$ into the limits of my integration. I thought that the limits of my integrals with respect to $x$, $y$, and $z$ were setting the bounding box within which I want to calculate the volume below the surface?
$endgroup$
– CynicalBrit
Feb 3 at 0:35
$begingroup$
You are subtracting from your cube's 'volume' another one with constant value $0.2$ at each point. There's no reason a subtraction can't yield a negative result. What's $v<L$ supposed to mean?
$endgroup$
– lightxbulb
Feb 2 at 22:56
$begingroup$
You are subtracting from your cube's 'volume' another one with constant value $0.2$ at each point. There's no reason a subtraction can't yield a negative result. What's $v<L$ supposed to mean?
$endgroup$
– lightxbulb
Feb 2 at 22:56
$begingroup$
$L$ is an arbitrary number between $0$ and $1$ sorry I should have listed that in my question. I gave the example value of 0.2 for use in the attached image. With the equation given that may be true, but I would not expect it to be possible to calculate a negative volume? I am not sure my method is correct.
$endgroup$
– CynicalBrit
Feb 2 at 23:26
$begingroup$
$L$ is an arbitrary number between $0$ and $1$ sorry I should have listed that in my question. I gave the example value of 0.2 for use in the attached image. With the equation given that may be true, but I would not expect it to be possible to calculate a negative volume? I am not sure my method is correct.
$endgroup$
– CynicalBrit
Feb 2 at 23:26
$begingroup$
It's possible to get a negative number, the reason however why you get a negative number here is that you subtract volumes. If I understood correctly what you want to do, then subtracting is wrong. You seem to simply want $v<L$, in which case this should transfer over to the limits. Note that $f(x,y,z)= xyz - L = 0$ defines a surface. You want to integrate below that surface, that is your integral limits should be dependent.
$endgroup$
– lightxbulb
Feb 3 at 0:05
$begingroup$
It's possible to get a negative number, the reason however why you get a negative number here is that you subtract volumes. If I understood correctly what you want to do, then subtracting is wrong. You seem to simply want $v<L$, in which case this should transfer over to the limits. Note that $f(x,y,z)= xyz - L = 0$ defines a surface. You want to integrate below that surface, that is your integral limits should be dependent.
$endgroup$
– lightxbulb
Feb 3 at 0:05
$begingroup$
Yes that's it exactly. Forgive me, what little calculus I do know is very rusty. Please could you explain how I would insert the $v<L$ into the limits of my integration. I thought that the limits of my integrals with respect to $x$, $y$, and $z$ were setting the bounding box within which I want to calculate the volume below the surface?
$endgroup$
– CynicalBrit
Feb 3 at 0:35
$begingroup$
Yes that's it exactly. Forgive me, what little calculus I do know is very rusty. Please could you explain how I would insert the $v<L$ into the limits of my integration. I thought that the limits of my integrals with respect to $x$, $y$, and $z$ were setting the bounding box within which I want to calculate the volume below the surface?
$endgroup$
– CynicalBrit
Feb 3 at 0:35
add a comment |
1 Answer
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$begingroup$
It's vague to me which quantity are you trying to compute.
We want to have $xyz le L$, that is upon fixing $x, y$ where they are positive, we want $z$ to satisfy $z le frac{L}{xy}$.
If you want to compute the regular volume.
begin{align}&int_0^1int_0^1int_0^{min(frac{L}{xy},1)}1,,, dzdydx \&=int_0^1int_0^1 cdot minleft(frac{L}{xy},1right),,dydx\&=int_0^1int_{min(1, frac{L}{x})}^1 minleft(frac{L}{xy},1right),,dydx + int_0^1int_0^{min(1,frac{L}{x})} minleft(frac{L}{xy},1right),,dydx\
&=int_0^1int_{min(1, frac{L}{x})}^1frac{L}{xy},,dydx + int_0^1int_0^{min(1,frac{L}{x})}1 ,,dydx
\&=int_0^1frac{L}{x}left(-lnleft(min(1,frac{L}{x} right) right),dx +int_0^1min(1,frac{L}{x}) ,dx\
&=int_L^1frac{L}{x}left(-lnleft(min(1,frac{L}{x} right) right),dx \&+int_0^Lmin(1,frac{L}{x}) ,dx+int_L^1min(1,frac{L}{x}) ,dx\
&=int_L^1frac{L}{x}left(-lnleft(frac{L}{x} right) right),dx +int_0^L1 ,dx+int_L^1frac{L}{x} ,dx\
&=Lleft[frac12ln^2 left(frac{L}x right) right]_L^1 +L+L(-ln L)\
&=Lleft( frac{ln^2 L}2-ln L+1 right)
end{align}
Suppose you want to integrate $xyz$ over the region.
begin{align}&int_0^1int_0^1int_0^{min(frac{L}{xy},1)}{xyz},,, dzdydx \&=frac{1}2int_0^1int_0^1xy cdot minleft(frac{L^2}{x^2y^2},1right),,dydx\&=frac{1}2int_0^1int_{min(1, frac{L}{x})}^1xy cdot minleft(frac{L^2}{x^2y^2},1right),,dydx + frac{1}2int_0^1int_0^{min(1,frac{L}{x})}xy cdot minleft(frac{L^2}{x^2y^2},1right),,dydx\
&=frac{1}2int_0^1int_{min(1, frac{L}{x})}^1frac{L^2}{xy},,dydx + frac{1}2int_0^1int_0^{min(1,frac{L}{x})}xy ,,dydx
\&=frac{1}2int_0^1frac{L^2}{x}left(-lnleft(min(1,frac{L}{x} right) right),dx + frac{1}4int_0^1xmin(1,frac{L^2}{x^2}) ,dx\
&=-frac{L^2}2int_L^1frac{1}{x}left(lnleft(frac{L}{x} right) right),dx + frac{1}4int_L^1frac{L^2}{x} ,dx+frac{1}4int_0^Lx ,dx\
&=-frac{L^2}{2}left[-frac12 ln^2 left( frac{L}{x}right) right]_L^1 + frac{L^2}4(-ln L)+frac{L^2}8\
&=frac{(Lln L)^2}{4} - frac{L^2ln L}4+frac{L^2}8end{align}
answered Feb 3 at 5:21
Siong Thye GohSiong Thye Goh
104k1468120
$endgroup$
$begingroup$
Hi Siong. Thank you for taking the time to answer my question, sorry I have not articulated it very well! If you look at the image in my post (that some kind soul has turned into an inline image - thank you!) you will see a blue and yellow region. The yellow region is where the $x$, $y$, and $z$ values result in $xyz le L$ (which in the case of that image $L=0.2$). I believe that this means that the first integration you provided is what I am trying to acheive?
$endgroup$
– CynicalBrit
Feb 3 at 18:52
$begingroup$
ah, yes, that is the regular volume then.
$endgroup$
– Siong Thye Goh
Feb 3 at 18:59
$begingroup$
Thank you very much for your help Siong. I ran through the regular volume example by hand, and compared it to some of my 'empirical' data (used to generate the 3D plot in my question. If $L=0.01$ then the volume should be ~16.21, and for $L=0.1$ it should be ~59.5. I think that there may be a mistake, and that the final line should be $L(frac{ln^2L}{2}-lnL+1)$.
$endgroup$
– CynicalBrit
Feb 8 at 19:35
$begingroup$
ah, thanks for the correction.
$endgroup$
– Siong Thye Goh
Feb 8 at 19:56
add a comment |
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$begingroup$
It's vague to me which quantity are you trying to compute.
We want to have $xyz le L$, that is upon fixing $x, y$ where they are positive, we want $z$ to satisfy $z le frac{L}{xy}$.
If you want to compute the regular volume.
begin{align}&int_0^1int_0^1int_0^{min(frac{L}{xy},1)}1,,, dzdydx \&=int_0^1int_0^1 cdot minleft(frac{L}{xy},1right),,dydx\&=int_0^1int_{min(1, frac{L}{x})}^1 minleft(frac{L}{xy},1right),,dydx + int_0^1int_0^{min(1,frac{L}{x})} minleft(frac{L}{xy},1right),,dydx\
&=int_0^1int_{min(1, frac{L}{x})}^1frac{L}{xy},,dydx + int_0^1int_0^{min(1,frac{L}{x})}1 ,,dydx
\&=int_0^1frac{L}{x}left(-lnleft(min(1,frac{L}{x} right) right),dx +int_0^1min(1,frac{L}{x}) ,dx\
&=int_L^1frac{L}{x}left(-lnleft(min(1,frac{L}{x} right) right),dx \&+int_0^Lmin(1,frac{L}{x}) ,dx+int_L^1min(1,frac{L}{x}) ,dx\
&=int_L^1frac{L}{x}left(-lnleft(frac{L}{x} right) right),dx +int_0^L1 ,dx+int_L^1frac{L}{x} ,dx\
&=Lleft[frac12ln^2 left(frac{L}x right) right]_L^1 +L+L(-ln L)\
&=Lleft( frac{ln^2 L}2-ln L+1 right)
end{align}
Suppose you want to integrate $xyz$ over the region.
begin{align}&int_0^1int_0^1int_0^{min(frac{L}{xy},1)}{xyz},,, dzdydx \&=frac{1}2int_0^1int_0^1xy cdot minleft(frac{L^2}{x^2y^2},1right),,dydx\&=frac{1}2int_0^1int_{min(1, frac{L}{x})}^1xy cdot minleft(frac{L^2}{x^2y^2},1right),,dydx + frac{1}2int_0^1int_0^{min(1,frac{L}{x})}xy cdot minleft(frac{L^2}{x^2y^2},1right),,dydx\
&=frac{1}2int_0^1int_{min(1, frac{L}{x})}^1frac{L^2}{xy},,dydx + frac{1}2int_0^1int_0^{min(1,frac{L}{x})}xy ,,dydx
\&=frac{1}2int_0^1frac{L^2}{x}left(-lnleft(min(1,frac{L}{x} right) right),dx + frac{1}4int_0^1xmin(1,frac{L^2}{x^2}) ,dx\
&=-frac{L^2}2int_L^1frac{1}{x}left(lnleft(frac{L}{x} right) right),dx + frac{1}4int_L^1frac{L^2}{x} ,dx+frac{1}4int_0^Lx ,dx\
&=-frac{L^2}{2}left[-frac12 ln^2 left( frac{L}{x}right) right]_L^1 + frac{L^2}4(-ln L)+frac{L^2}8\
&=frac{(Lln L)^2}{4} - frac{L^2ln L}4+frac{L^2}8end{align}
answered Feb 3 at 5:21
Siong Thye GohSiong Thye Goh
104k1468120
$endgroup$
$begingroup$
Hi Siong. Thank you for taking the time to answer my question, sorry I have not articulated it very well! If you look at the image in my post (that some kind soul has turned into an inline image - thank you!) you will see a blue and yellow region. The yellow region is where the $x$, $y$, and $z$ values result in $xyz le L$ (which in the case of that image $L=0.2$). I believe that this means that the first integration you provided is what I am trying to acheive?
$endgroup$
– CynicalBrit
Feb 3 at 18:52
$begingroup$
ah, yes, that is the regular volume then.
$endgroup$
– Siong Thye Goh
Feb 3 at 18:59
$begingroup$
Thank you very much for your help Siong. I ran through the regular volume example by hand, and compared it to some of my 'empirical' data (used to generate the 3D plot in my question. If $L=0.01$ then the volume should be ~16.21, and for $L=0.1$ it should be ~59.5. I think that there may be a mistake, and that the final line should be $L(frac{ln^2L}{2}-lnL+1)$.
$endgroup$
– CynicalBrit
Feb 8 at 19:35
$begingroup$
ah, thanks for the correction.
$endgroup$
– Siong Thye Goh
Feb 8 at 19:56
add a comment |
$begingroup$
It's vague to me which quantity are you trying to compute.
We want to have $xyz le L$, that is upon fixing $x, y$ where they are positive, we want $z$ to satisfy $z le frac{L}{xy}$.
If you want to compute the regular volume.
begin{align}&int_0^1int_0^1int_0^{min(frac{L}{xy},1)}1,,, dzdydx \&=int_0^1int_0^1 cdot minleft(frac{L}{xy},1right),,dydx\&=int_0^1int_{min(1, frac{L}{x})}^1 minleft(frac{L}{xy},1right),,dydx + int_0^1int_0^{min(1,frac{L}{x})} minleft(frac{L}{xy},1right),,dydx\
&=int_0^1int_{min(1, frac{L}{x})}^1frac{L}{xy},,dydx + int_0^1int_0^{min(1,frac{L}{x})}1 ,,dydx
\&=int_0^1frac{L}{x}left(-lnleft(min(1,frac{L}{x} right) right),dx +int_0^1min(1,frac{L}{x}) ,dx\
&=int_L^1frac{L}{x}left(-lnleft(min(1,frac{L}{x} right) right),dx \&+int_0^Lmin(1,frac{L}{x}) ,dx+int_L^1min(1,frac{L}{x}) ,dx\
&=int_L^1frac{L}{x}left(-lnleft(frac{L}{x} right) right),dx +int_0^L1 ,dx+int_L^1frac{L}{x} ,dx\
&=Lleft[frac12ln^2 left(frac{L}x right) right]_L^1 +L+L(-ln L)\
&=Lleft( frac{ln^2 L}2-ln L+1 right)
end{align}
Suppose you want to integrate $xyz$ over the region.
begin{align}&int_0^1int_0^1int_0^{min(frac{L}{xy},1)}{xyz},,, dzdydx \&=frac{1}2int_0^1int_0^1xy cdot minleft(frac{L^2}{x^2y^2},1right),,dydx\&=frac{1}2int_0^1int_{min(1, frac{L}{x})}^1xy cdot minleft(frac{L^2}{x^2y^2},1right),,dydx + frac{1}2int_0^1int_0^{min(1,frac{L}{x})}xy cdot minleft(frac{L^2}{x^2y^2},1right),,dydx\
&=frac{1}2int_0^1int_{min(1, frac{L}{x})}^1frac{L^2}{xy},,dydx + frac{1}2int_0^1int_0^{min(1,frac{L}{x})}xy ,,dydx
\&=frac{1}2int_0^1frac{L^2}{x}left(-lnleft(min(1,frac{L}{x} right) right),dx + frac{1}4int_0^1xmin(1,frac{L^2}{x^2}) ,dx\
&=-frac{L^2}2int_L^1frac{1}{x}left(lnleft(frac{L}{x} right) right),dx + frac{1}4int_L^1frac{L^2}{x} ,dx+frac{1}4int_0^Lx ,dx\
&=-frac{L^2}{2}left[-frac12 ln^2 left( frac{L}{x}right) right]_L^1 + frac{L^2}4(-ln L)+frac{L^2}8\
&=frac{(Lln L)^2}{4} - frac{L^2ln L}4+frac{L^2}8end{align}
answered Feb 3 at 5:21
Siong Thye GohSiong Thye Goh
104k1468120
$endgroup$
$begingroup$
Hi Siong. Thank you for taking the time to answer my question, sorry I have not articulated it very well! If you look at the image in my post (that some kind soul has turned into an inline image - thank you!) you will see a blue and yellow region. The yellow region is where the $x$, $y$, and $z$ values result in $xyz le L$ (which in the case of that image $L=0.2$). I believe that this means that the first integration you provided is what I am trying to acheive?
$endgroup$
– CynicalBrit
Feb 3 at 18:52
$begingroup$
ah, yes, that is the regular volume then.
$endgroup$
– Siong Thye Goh
Feb 3 at 18:59
$begingroup$
Thank you very much for your help Siong. I ran through the regular volume example by hand, and compared it to some of my 'empirical' data (used to generate the 3D plot in my question. If $L=0.01$ then the volume should be ~16.21, and for $L=0.1$ it should be ~59.5. I think that there may be a mistake, and that the final line should be $L(frac{ln^2L}{2}-lnL+1)$.
$endgroup$
– CynicalBrit
Feb 8 at 19:35
$begingroup$
ah, thanks for the correction.
$endgroup$
– Siong Thye Goh
Feb 8 at 19:56
add a comment |
$begingroup$
It's vague to me which quantity are you trying to compute.
We want to have $xyz le L$, that is upon fixing $x, y$ where they are positive, we want $z$ to satisfy $z le frac{L}{xy}$.
If you want to compute the regular volume.
begin{align}&int_0^1int_0^1int_0^{min(frac{L}{xy},1)}1,,, dzdydx \&=int_0^1int_0^1 cdot minleft(frac{L}{xy},1right),,dydx\&=int_0^1int_{min(1, frac{L}{x})}^1 minleft(frac{L}{xy},1right),,dydx + int_0^1int_0^{min(1,frac{L}{x})} minleft(frac{L}{xy},1right),,dydx\
&=int_0^1int_{min(1, frac{L}{x})}^1frac{L}{xy},,dydx + int_0^1int_0^{min(1,frac{L}{x})}1 ,,dydx
\&=int_0^1frac{L}{x}left(-lnleft(min(1,frac{L}{x} right) right),dx +int_0^1min(1,frac{L}{x}) ,dx\
&=int_L^1frac{L}{x}left(-lnleft(min(1,frac{L}{x} right) right),dx \&+int_0^Lmin(1,frac{L}{x}) ,dx+int_L^1min(1,frac{L}{x}) ,dx\
&=int_L^1frac{L}{x}left(-lnleft(frac{L}{x} right) right),dx +int_0^L1 ,dx+int_L^1frac{L}{x} ,dx\
&=Lleft[frac12ln^2 left(frac{L}x right) right]_L^1 +L+L(-ln L)\
&=Lleft( frac{ln^2 L}2-ln L+1 right)
end{align}
Suppose you want to integrate $xyz$ over the region.
begin{align}&int_0^1int_0^1int_0^{min(frac{L}{xy},1)}{xyz},,, dzdydx \&=frac{1}2int_0^1int_0^1xy cdot minleft(frac{L^2}{x^2y^2},1right),,dydx\&=frac{1}2int_0^1int_{min(1, frac{L}{x})}^1xy cdot minleft(frac{L^2}{x^2y^2},1right),,dydx + frac{1}2int_0^1int_0^{min(1,frac{L}{x})}xy cdot minleft(frac{L^2}{x^2y^2},1right),,dydx\
&=frac{1}2int_0^1int_{min(1, frac{L}{x})}^1frac{L^2}{xy},,dydx + frac{1}2int_0^1int_0^{min(1,frac{L}{x})}xy ,,dydx
\&=frac{1}2int_0^1frac{L^2}{x}left(-lnleft(min(1,frac{L}{x} right) right),dx + frac{1}4int_0^1xmin(1,frac{L^2}{x^2}) ,dx\
&=-frac{L^2}2int_L^1frac{1}{x}left(lnleft(frac{L}{x} right) right),dx + frac{1}4int_L^1frac{L^2}{x} ,dx+frac{1}4int_0^Lx ,dx\
&=-frac{L^2}{2}left[-frac12 ln^2 left( frac{L}{x}right) right]_L^1 + frac{L^2}4(-ln L)+frac{L^2}8\
&=frac{(Lln L)^2}{4} - frac{L^2ln L}4+frac{L^2}8end{align}
answered Feb 3 at 5:21
Siong Thye GohSiong Thye Goh
104k1468120
$endgroup$
It's vague to me which quantity are you trying to compute.
We want to have $xyz le L$, that is upon fixing $x, y$ where they are positive, we want $z$ to satisfy $z le frac{L}{xy}$.
If you want to compute the regular volume.
begin{align}&int_0^1int_0^1int_0^{min(frac{L}{xy},1)}1,,, dzdydx \&=int_0^1int_0^1 cdot minleft(frac{L}{xy},1right),,dydx\&=int_0^1int_{min(1, frac{L}{x})}^1 minleft(frac{L}{xy},1right),,dydx + int_0^1int_0^{min(1,frac{L}{x})} minleft(frac{L}{xy},1right),,dydx\
&=int_0^1int_{min(1, frac{L}{x})}^1frac{L}{xy},,dydx + int_0^1int_0^{min(1,frac{L}{x})}1 ,,dydx
\&=int_0^1frac{L}{x}left(-lnleft(min(1,frac{L}{x} right) right),dx +int_0^1min(1,frac{L}{x}) ,dx\
&=int_L^1frac{L}{x}left(-lnleft(min(1,frac{L}{x} right) right),dx \&+int_0^Lmin(1,frac{L}{x}) ,dx+int_L^1min(1,frac{L}{x}) ,dx\
&=int_L^1frac{L}{x}left(-lnleft(frac{L}{x} right) right),dx +int_0^L1 ,dx+int_L^1frac{L}{x} ,dx\
&=Lleft[frac12ln^2 left(frac{L}x right) right]_L^1 +L+L(-ln L)\
&=Lleft( frac{ln^2 L}2-ln L+1 right)
end{align}
Suppose you want to integrate $xyz$ over the region.
begin{align}&int_0^1int_0^1int_0^{min(frac{L}{xy},1)}{xyz},,, dzdydx \&=frac{1}2int_0^1int_0^1xy cdot minleft(frac{L^2}{x^2y^2},1right),,dydx\&=frac{1}2int_0^1int_{min(1, frac{L}{x})}^1xy cdot minleft(frac{L^2}{x^2y^2},1right),,dydx + frac{1}2int_0^1int_0^{min(1,frac{L}{x})}xy cdot minleft(frac{L^2}{x^2y^2},1right),,dydx\
&=frac{1}2int_0^1int_{min(1, frac{L}{x})}^1frac{L^2}{xy},,dydx + frac{1}2int_0^1int_0^{min(1,frac{L}{x})}xy ,,dydx
\&=frac{1}2int_0^1frac{L^2}{x}left(-lnleft(min(1,frac{L}{x} right) right),dx + frac{1}4int_0^1xmin(1,frac{L^2}{x^2}) ,dx\
&=-frac{L^2}2int_L^1frac{1}{x}left(lnleft(frac{L}{x} right) right),dx + frac{1}4int_L^1frac{L^2}{x} ,dx+frac{1}4int_0^Lx ,dx\
&=-frac{L^2}{2}left[-frac12 ln^2 left( frac{L}{x}right) right]_L^1 + frac{L^2}4(-ln L)+frac{L^2}8\
&=frac{(Lln L)^2}{4} - frac{L^2ln L}4+frac{L^2}8end{align}
answered Feb 3 at 5:21
Siong Thye GohSiong Thye Goh
104k1468120
edited Feb 8 at 19:56
answered Feb 3 at 5:21
Siong Thye GohSiong Thye Goh
104k1468120
answered Feb 3 at 5:21
Siong Thye GohSiong Thye Goh
104k1468120
answered Feb 3 at 5:21
Siong Thye GohSiong Thye Goh
104k1468120
104k1468120
$begingroup$
Hi Siong. Thank you for taking the time to answer my question, sorry I have not articulated it very well! If you look at the image in my post (that some kind soul has turned into an inline image - thank you!) you will see a blue and yellow region. The yellow region is where the $x$, $y$, and $z$ values result in $xyz le L$ (which in the case of that image $L=0.2$). I believe that this means that the first integration you provided is what I am trying to acheive?
$endgroup$
– CynicalBrit
Feb 3 at 18:52
$begingroup$
ah, yes, that is the regular volume then.
$endgroup$
– Siong Thye Goh
Feb 3 at 18:59
$begingroup$
Thank you very much for your help Siong. I ran through the regular volume example by hand, and compared it to some of my 'empirical' data (used to generate the 3D plot in my question. If $L=0.01$ then the volume should be ~16.21, and for $L=0.1$ it should be ~59.5. I think that there may be a mistake, and that the final line should be $L(frac{ln^2L}{2}-lnL+1)$.
$endgroup$
– CynicalBrit
Feb 8 at 19:35
$begingroup$
ah, thanks for the correction.
$endgroup$
– Siong Thye Goh
Feb 8 at 19:56
add a comment |
$begingroup$
Hi Siong. Thank you for taking the time to answer my question, sorry I have not articulated it very well! If you look at the image in my post (that some kind soul has turned into an inline image - thank you!) you will see a blue and yellow region. The yellow region is where the $x$, $y$, and $z$ values result in $xyz le L$ (which in the case of that image $L=0.2$). I believe that this means that the first integration you provided is what I am trying to acheive?
$endgroup$
– CynicalBrit
Feb 3 at 18:52
$begingroup$
ah, yes, that is the regular volume then.
$endgroup$
– Siong Thye Goh
Feb 3 at 18:59
$begingroup$
Thank you very much for your help Siong. I ran through the regular volume example by hand, and compared it to some of my 'empirical' data (used to generate the 3D plot in my question. If $L=0.01$ then the volume should be ~16.21, and for $L=0.1$ it should be ~59.5. I think that there may be a mistake, and that the final line should be $L(frac{ln^2L}{2}-lnL+1)$.
$endgroup$
– CynicalBrit
Feb 8 at 19:35
$begingroup$
ah, thanks for the correction.
$endgroup$
– Siong Thye Goh
Feb 8 at 19:56
$begingroup$
Hi Siong. Thank you for taking the time to answer my question, sorry I have not articulated it very well! If you look at the image in my post (that some kind soul has turned into an inline image - thank you!) you will see a blue and yellow region. The yellow region is where the $x$, $y$, and $z$ values result in $xyz le L$ (which in the case of that image $L=0.2$). I believe that this means that the first integration you provided is what I am trying to acheive?
$endgroup$
– CynicalBrit
Feb 3 at 18:52
$begingroup$
Hi Siong. Thank you for taking the time to answer my question, sorry I have not articulated it very well! If you look at the image in my post (that some kind soul has turned into an inline image - thank you!) you will see a blue and yellow region. The yellow region is where the $x$, $y$, and $z$ values result in $xyz le L$ (which in the case of that image $L=0.2$). I believe that this means that the first integration you provided is what I am trying to acheive?
$endgroup$
– CynicalBrit
Feb 3 at 18:52
$begingroup$
ah, yes, that is the regular volume then.
$endgroup$
– Siong Thye Goh
Feb 3 at 18:59
$begingroup$
ah, yes, that is the regular volume then.
$endgroup$
– Siong Thye Goh
Feb 3 at 18:59
$begingroup$
Thank you very much for your help Siong. I ran through the regular volume example by hand, and compared it to some of my 'empirical' data (used to generate the 3D plot in my question. If $L=0.01$ then the volume should be ~16.21, and for $L=0.1$ it should be ~59.5. I think that there may be a mistake, and that the final line should be $L(frac{ln^2L}{2}-lnL+1)$.
$endgroup$
– CynicalBrit
Feb 8 at 19:35
$begingroup$
Thank you very much for your help Siong. I ran through the regular volume example by hand, and compared it to some of my 'empirical' data (used to generate the 3D plot in my question. If $L=0.01$ then the volume should be ~16.21, and for $L=0.1$ it should be ~59.5. I think that there may be a mistake, and that the final line should be $L(frac{ln^2L}{2}-lnL+1)$.
$endgroup$
– CynicalBrit
Feb 8 at 19:35
$begingroup$
ah, thanks for the correction.
$endgroup$
– Siong Thye Goh
Feb 8 at 19:56
$begingroup$
ah, thanks for the correction.
$endgroup$
– Siong Thye Goh
Feb 8 at 19:56
add a comment |
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$begingroup$
You are subtracting from your cube's 'volume' another one with constant value $0.2$ at each point. There's no reason a subtraction can't yield a negative result. What's $v<L$ supposed to mean?
$endgroup$
– lightxbulb
Feb 2 at 22:56
$begingroup$
$L$ is an arbitrary number between $0$ and $1$ sorry I should have listed that in my question. I gave the example value of 0.2 for use in the attached image. With the equation given that may be true, but I would not expect it to be possible to calculate a negative volume? I am not sure my method is correct.
$endgroup$
– CynicalBrit
Feb 2 at 23:26
$begingroup$
It's possible to get a negative number, the reason however why you get a negative number here is that you subtract volumes. If I understood correctly what you want to do, then subtracting is wrong. You seem to simply want $v<L$, in which case this should transfer over to the limits. Note that $f(x,y,z)= xyz - L = 0$ defines a surface. You want to integrate below that surface, that is your integral limits should be dependent.
$endgroup$
– lightxbulb
Feb 3 at 0:05
$begingroup$
Yes that's it exactly. Forgive me, what little calculus I do know is very rusty. Please could you explain how I would insert the $v<L$ into the limits of my integration. I thought that the limits of my integrals with respect to $x$, $y$, and $z$ were setting the bounding box within which I want to calculate the volume below the surface?
$endgroup$
– CynicalBrit
Feb 3 at 0:35