Rewriting the domain of my integral using a function for calculating the center of mass

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$Omega:= y leq x^2 $

Calculate the median point, with $rho(x,y)=1$

Also the mass is: $M=2/3$

We first rewrite Omega.

I know that there is a nicer way to rewrite Omega in a way that we can easily integrate it, but I also though I could do it like this:

We consider the function $f(x)=x^2$ on the set $B:=(x,y)inmathbb R^2 $

We only calculate the y-component:

$y_s=frac1Mint_Omega yrho(x,y)dmu(x,y)=frac32int_B f(x)ydmu(x,y)=frac32int_-1^1int_0^1yx^2 dydx=frac32int_-1^1frac12x^2dx$

$=frac32frac12[frac13x^3]_-1^1=frac32frac16(1-(-1))=frac32frac26=1/2$

Question: The correct result would be 1/10. How come, my idea doesn't work?

integration

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asked Jul 20 at 11:47

xotix

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up vote
1
down vote

favorite

$Omega:= y leq x^2 $

Calculate the median point, with $rho(x,y)=1$

Also the mass is: $M=2/3$

We first rewrite Omega.

I know that there is a nicer way to rewrite Omega in a way that we can easily integrate it, but I also though I could do it like this:

We consider the function $f(x)=x^2$ on the set $B:=(x,y)inmathbb R^2 $

We only calculate the y-component:

$y_s=frac1Mint_Omega yrho(x,y)dmu(x,y)=frac32int_B f(x)ydmu(x,y)=frac32int_-1^1int_0^1yx^2 dydx=frac32int_-1^1frac12x^2dx$

$=frac32frac12[frac13x^3]_-1^1=frac32frac16(1-(-1))=frac32frac26=1/2$

Question: The correct result would be 1/10. How come, my idea doesn't work?

integration

share|cite|improve this question

asked Jul 20 at 11:47

xotix

30519

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

$Omega:= y leq x^2 $

Calculate the median point, with $rho(x,y)=1$

Also the mass is: $M=2/3$

We first rewrite Omega.

I know that there is a nicer way to rewrite Omega in a way that we can easily integrate it, but I also though I could do it like this:

We consider the function $f(x)=x^2$ on the set $B:=(x,y)inmathbb R^2 $

We only calculate the y-component:

$y_s=frac1Mint_Omega yrho(x,y)dmu(x,y)=frac32int_B f(x)ydmu(x,y)=frac32int_-1^1int_0^1yx^2 dydx=frac32int_-1^1frac12x^2dx$

$=frac32frac12[frac13x^3]_-1^1=frac32frac16(1-(-1))=frac32frac26=1/2$

Question: The correct result would be 1/10. How come, my idea doesn't work?

integration

share|cite|improve this question

asked Jul 20 at 11:47

xotix

30519

$Omega:= y leq x^2 $

Calculate the median point, with $rho(x,y)=1$

Also the mass is: $M=2/3$

We first rewrite Omega.

I know that there is a nicer way to rewrite Omega in a way that we can easily integrate it, but I also though I could do it like this:

We consider the function $f(x)=x^2$ on the set $B:=(x,y)inmathbb R^2 $

We only calculate the y-component:

$y_s=frac1Mint_Omega yrho(x,y)dmu(x,y)=frac32int_B f(x)ydmu(x,y)=frac32int_-1^1int_0^1yx^2 dydx=frac32int_-1^1frac12x^2dx$

$=frac32frac12[frac13x^3]_-1^1=frac32frac16(1-(-1))=frac32frac26=1/2$

Question: The correct result would be 1/10. How come, my idea doesn't work?

integration

share|cite|improve this question

asked Jul 20 at 11:47

xotix

30519

share|cite|improve this question

share|cite|improve this question

asked Jul 20 at 11:47

xotix

30519

asked Jul 20 at 11:47

xotix

30519

asked Jul 20 at 11:47

xotix

30519

30519

add a comment | 

add a comment | 

1 Answer
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We want to find the following integral over $Omega$ :
$$ y_mathrms = frac32int limits_Omega y , mathrmd mu (x,y) , .$$
We can also integrate over $B$ instead if we find a suitable function $f$ :
$$ y_mathrms = frac32 int limits_B y f(x,y) , mathrmd mu (x,y) , .$$
We need to make sure, however, that we only integrate over the part of $B$ on which $y leq x^2$ holds. We cannot do this by choosing $f(x,y) = x^2$ , since this would lead to a totally different integral which has nothing to do with the centre of mass.

Instead we can use the Heaviside step function $H$ and let $f(x,y) = H(x^2 - y)$ . Then $f(x,y) = 1$ holds for $(x,y) in Omega$ (except possibly for a set of measure zero) and $f$ vanishes on $B setminus Omega$ , so the two integrals are indeed equal. This is the correct way to incorporate the condition $y leq x^2$ into the integration over $B$.

We then end up with the integral
$$ y_mathrms = frac32 int limits_-1^1 int limits_0^1 y H(x^2-y) , mathrmd y , mathrmd x = frac32 int limits_-1^1 int limits_0^x^2 y , mathrmd y , mathrmd x , . $$
Note that the final result is $y_mathrms = frac310$ and not $y_mathrms = frac110$ though.

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answered Jul 20 at 15:55

ComplexYetTrivial

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1 Answer
1

active

oldest

votes

1 Answer
1

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
1
down vote



accepted

We want to find the following integral over $Omega$ :
$$ y_mathrms = frac32int limits_Omega y , mathrmd mu (x,y) , .$$
We can also integrate over $B$ instead if we find a suitable function $f$ :
$$ y_mathrms = frac32 int limits_B y f(x,y) , mathrmd mu (x,y) , .$$
We need to make sure, however, that we only integrate over the part of $B$ on which $y leq x^2$ holds. We cannot do this by choosing $f(x,y) = x^2$ , since this would lead to a totally different integral which has nothing to do with the centre of mass.

Instead we can use the Heaviside step function $H$ and let $f(x,y) = H(x^2 - y)$ . Then $f(x,y) = 1$ holds for $(x,y) in Omega$ (except possibly for a set of measure zero) and $f$ vanishes on $B setminus Omega$ , so the two integrals are indeed equal. This is the correct way to incorporate the condition $y leq x^2$ into the integration over $B$.

We then end up with the integral
$$ y_mathrms = frac32 int limits_-1^1 int limits_0^1 y H(x^2-y) , mathrmd y , mathrmd x = frac32 int limits_-1^1 int limits_0^x^2 y , mathrmd y , mathrmd x , . $$
Note that the final result is $y_mathrms = frac310$ and not $y_mathrms = frac110$ though.

share|cite|improve this answer

answered Jul 20 at 15:55

ComplexYetTrivial

2,607624

add a comment | 

up vote
1
down vote



accepted

We want to find the following integral over $Omega$ :
$$ y_mathrms = frac32int limits_Omega y , mathrmd mu (x,y) , .$$
We can also integrate over $B$ instead if we find a suitable function $f$ :
$$ y_mathrms = frac32 int limits_B y f(x,y) , mathrmd mu (x,y) , .$$
We need to make sure, however, that we only integrate over the part of $B$ on which $y leq x^2$ holds. We cannot do this by choosing $f(x,y) = x^2$ , since this would lead to a totally different integral which has nothing to do with the centre of mass.

Instead we can use the Heaviside step function $H$ and let $f(x,y) = H(x^2 - y)$ . Then $f(x,y) = 1$ holds for $(x,y) in Omega$ (except possibly for a set of measure zero) and $f$ vanishes on $B setminus Omega$ , so the two integrals are indeed equal. This is the correct way to incorporate the condition $y leq x^2$ into the integration over $B$.

We then end up with the integral
$$ y_mathrms = frac32 int limits_-1^1 int limits_0^1 y H(x^2-y) , mathrmd y , mathrmd x = frac32 int limits_-1^1 int limits_0^x^2 y , mathrmd y , mathrmd x , . $$
Note that the final result is $y_mathrms = frac310$ and not $y_mathrms = frac110$ though.

share|cite|improve this answer

answered Jul 20 at 15:55

ComplexYetTrivial

2,607624

add a comment | 

up vote
1
down vote



accepted

up vote
1
down vote



accepted

We want to find the following integral over $Omega$ :
$$ y_mathrms = frac32int limits_Omega y , mathrmd mu (x,y) , .$$
We can also integrate over $B$ instead if we find a suitable function $f$ :
$$ y_mathrms = frac32 int limits_B y f(x,y) , mathrmd mu (x,y) , .$$
We need to make sure, however, that we only integrate over the part of $B$ on which $y leq x^2$ holds. We cannot do this by choosing $f(x,y) = x^2$ , since this would lead to a totally different integral which has nothing to do with the centre of mass.

Instead we can use the Heaviside step function $H$ and let $f(x,y) = H(x^2 - y)$ . Then $f(x,y) = 1$ holds for $(x,y) in Omega$ (except possibly for a set of measure zero) and $f$ vanishes on $B setminus Omega$ , so the two integrals are indeed equal. This is the correct way to incorporate the condition $y leq x^2$ into the integration over $B$.

We then end up with the integral
$$ y_mathrms = frac32 int limits_-1^1 int limits_0^1 y H(x^2-y) , mathrmd y , mathrmd x = frac32 int limits_-1^1 int limits_0^x^2 y , mathrmd y , mathrmd x , . $$
Note that the final result is $y_mathrms = frac310$ and not $y_mathrms = frac110$ though.

share|cite|improve this answer

answered Jul 20 at 15:55

ComplexYetTrivial

2,607624

We want to find the following integral over $Omega$ :
$$ y_mathrms = frac32int limits_Omega y , mathrmd mu (x,y) , .$$
We can also integrate over $B$ instead if we find a suitable function $f$ :
$$ y_mathrms = frac32 int limits_B y f(x,y) , mathrmd mu (x,y) , .$$
We need to make sure, however, that we only integrate over the part of $B$ on which $y leq x^2$ holds. We cannot do this by choosing $f(x,y) = x^2$ , since this would lead to a totally different integral which has nothing to do with the centre of mass.

Instead we can use the Heaviside step function $H$ and let $f(x,y) = H(x^2 - y)$ . Then $f(x,y) = 1$ holds for $(x,y) in Omega$ (except possibly for a set of measure zero) and $f$ vanishes on $B setminus Omega$ , so the two integrals are indeed equal. This is the correct way to incorporate the condition $y leq x^2$ into the integration over $B$.

We then end up with the integral
$$ y_mathrms = frac32 int limits_-1^1 int limits_0^1 y H(x^2-y) , mathrmd y , mathrmd x = frac32 int limits_-1^1 int limits_0^x^2 y , mathrmd y , mathrmd x , . $$
Note that the final result is $y_mathrms = frac310$ and not $y_mathrms = frac110$ though.

share|cite|improve this answer

answered Jul 20 at 15:55

ComplexYetTrivial

2,607624

share|cite|improve this answer

share|cite|improve this answer

answered Jul 20 at 15:55

ComplexYetTrivial

2,607624

answered Jul 20 at 15:55

ComplexYetTrivial

2,607624

answered Jul 20 at 15:55

ComplexYetTrivial

2,607624

2,607624

add a comment | 

add a comment | 

 

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