Mixing
Finding the Centroid of Solid G?
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I saw this problem on one of my assignments and had no idea how to do it, mostly because I missed the section where it was covered. Anyways it states: Find the centroid of solid G defined by the inequalities $sqrt{x^2+y^2}$ $le$ z $le$ $20-x^2-y^2$. Find the coordinates of the centroid of G.
How should I approach this problem and others that could be similar to it. Thanks
multivariable-calculus centroid
asked Jun 7 '17 at 2:05
GatorGator
82
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add a comment |
$begingroup$
I saw this problem on one of my assignments and had no idea how to do it, mostly because I missed the section where it was covered. Anyways it states: Find the centroid of solid G defined by the inequalities $sqrt{x^2+y^2}$ $le$ z $le$ $20-x^2-y^2$. Find the coordinates of the centroid of G.
How should I approach this problem and others that could be similar to it. Thanks
multivariable-calculus centroid
asked Jun 7 '17 at 2:05
GatorGator
82
$endgroup$
$begingroup$
You solid G is not well defined.
$endgroup$
– hamam_Abdallah
Jun 7 '17 at 2:42
add a comment |
$begingroup$
I saw this problem on one of my assignments and had no idea how to do it, mostly because I missed the section where it was covered. Anyways it states: Find the centroid of solid G defined by the inequalities $sqrt{x^2+y^2}$ $le$ z $le$ $20-x^2-y^2$. Find the coordinates of the centroid of G.
How should I approach this problem and others that could be similar to it. Thanks
multivariable-calculus centroid
asked Jun 7 '17 at 2:05
GatorGator
82
$endgroup$
I saw this problem on one of my assignments and had no idea how to do it, mostly because I missed the section where it was covered. Anyways it states: Find the centroid of solid G defined by the inequalities $sqrt{x^2+y^2}$ $le$ z $le$ $20-x^2-y^2$. Find the coordinates of the centroid of G.
How should I approach this problem and others that could be similar to it. Thanks
multivariable-calculus centroid
multivariable-calculus centroid
asked Jun 7 '17 at 2:05
GatorGator
82
asked Jun 7 '17 at 2:05
GatorGator
82
asked Jun 7 '17 at 2:05
GatorGator
82
asked Jun 7 '17 at 2:05
GatorGator
82
asked Jun 7 '17 at 2:05
GatorGator
82
82
$begingroup$
You solid G is not well defined.
$endgroup$
– hamam_Abdallah
Jun 7 '17 at 2:42
add a comment |
$begingroup$
You solid G is not well defined.
$endgroup$
– hamam_Abdallah
Jun 7 '17 at 2:42
$begingroup$
You solid G is not well defined.
$endgroup$
– hamam_Abdallah
Jun 7 '17 at 2:42
$begingroup$
You solid G is not well defined.
$endgroup$
– hamam_Abdallah
Jun 7 '17 at 2:42
add a comment |
2 Answers
2
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Hint: Supposing the density is uniform across the solid, the centroid is simply calculated by
$bar x = {{iiint x dV} over {iiint dV}}$
$bar y = {{iiint y dV} over {iiint dV}}$
$bar z = {{iiint z dV} over {iiint dV}}$
Now you are going to calculate these volume integrals.
P.S If the density is not a constant, i.e. , $d=f(x,y,z)$, you should place it before $dV$ in every integrand. For example,
$bar x = {{iiint x f(x,y,z) dV} over {iiint f(x,y,z) dV}}$
answered Jun 7 '17 at 3:05
HuangHuang
46635
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From Wikipedia (https://en.wikipedia.org/wiki/Center_of_mass): If the mass distribution is continuous with the density $ρ(r)$ within a volume $V$, then the integral of the weighted position coordinates of the points in this volume relative to the center of mass $R$ is zero, that is
$$intlimits_{V}{rho left( mathbf{r} right)left( mathbf{r-R} right)dV},=0 $$
Clearly, then, for a volume of uniform density
$$mathbf{R}=frac{1}{V}intlimits_{V}{mathbf{r}dV}=frac{1}{V}intintint(xmathbf{i}+ymathbf{j}+z mathbf{k}),dx,dy,dz $$
where, of course
$$V=intintint dx,dy,dz $$
answered Jun 7 '17 at 15:51
Cye WaldmanCye Waldman
4,2322623
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2 Answers
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2 Answers
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$begingroup$
Hint: Supposing the density is uniform across the solid, the centroid is simply calculated by
$bar x = {{iiint x dV} over {iiint dV}}$
$bar y = {{iiint y dV} over {iiint dV}}$
$bar z = {{iiint z dV} over {iiint dV}}$
Now you are going to calculate these volume integrals.
P.S If the density is not a constant, i.e. , $d=f(x,y,z)$, you should place it before $dV$ in every integrand. For example,
$bar x = {{iiint x f(x,y,z) dV} over {iiint f(x,y,z) dV}}$
answered Jun 7 '17 at 3:05
HuangHuang
46635
$endgroup$
add a comment |
$begingroup$
Hint: Supposing the density is uniform across the solid, the centroid is simply calculated by
$bar x = {{iiint x dV} over {iiint dV}}$
$bar y = {{iiint y dV} over {iiint dV}}$
$bar z = {{iiint z dV} over {iiint dV}}$
Now you are going to calculate these volume integrals.
P.S If the density is not a constant, i.e. , $d=f(x,y,z)$, you should place it before $dV$ in every integrand. For example,
$bar x = {{iiint x f(x,y,z) dV} over {iiint f(x,y,z) dV}}$
answered Jun 7 '17 at 3:05
HuangHuang
46635
$endgroup$
add a comment |
$begingroup$
Hint: Supposing the density is uniform across the solid, the centroid is simply calculated by
$bar x = {{iiint x dV} over {iiint dV}}$
$bar y = {{iiint y dV} over {iiint dV}}$
$bar z = {{iiint z dV} over {iiint dV}}$
Now you are going to calculate these volume integrals.
P.S If the density is not a constant, i.e. , $d=f(x,y,z)$, you should place it before $dV$ in every integrand. For example,
$bar x = {{iiint x f(x,y,z) dV} over {iiint f(x,y,z) dV}}$
answered Jun 7 '17 at 3:05
HuangHuang
46635
$endgroup$
Hint: Supposing the density is uniform across the solid, the centroid is simply calculated by
$bar x = {{iiint x dV} over {iiint dV}}$
$bar y = {{iiint y dV} over {iiint dV}}$
$bar z = {{iiint z dV} over {iiint dV}}$
Now you are going to calculate these volume integrals.
P.S If the density is not a constant, i.e. , $d=f(x,y,z)$, you should place it before $dV$ in every integrand. For example,
$bar x = {{iiint x f(x,y,z) dV} over {iiint f(x,y,z) dV}}$
answered Jun 7 '17 at 3:05
HuangHuang
46635
answered Jun 7 '17 at 3:05
HuangHuang
46635
answered Jun 7 '17 at 3:05
HuangHuang
46635
answered Jun 7 '17 at 3:05
HuangHuang
46635
46635
add a comment |
add a comment |
$begingroup$
From Wikipedia (https://en.wikipedia.org/wiki/Center_of_mass): If the mass distribution is continuous with the density $ρ(r)$ within a volume $V$, then the integral of the weighted position coordinates of the points in this volume relative to the center of mass $R$ is zero, that is
$$intlimits_{V}{rho left( mathbf{r} right)left( mathbf{r-R} right)dV},=0 $$
Clearly, then, for a volume of uniform density
$$mathbf{R}=frac{1}{V}intlimits_{V}{mathbf{r}dV}=frac{1}{V}intintint(xmathbf{i}+ymathbf{j}+z mathbf{k}),dx,dy,dz $$
where, of course
$$V=intintint dx,dy,dz $$
answered Jun 7 '17 at 15:51
Cye WaldmanCye Waldman
4,2322623
$endgroup$
add a comment |
$begingroup$
From Wikipedia (https://en.wikipedia.org/wiki/Center_of_mass): If the mass distribution is continuous with the density $ρ(r)$ within a volume $V$, then the integral of the weighted position coordinates of the points in this volume relative to the center of mass $R$ is zero, that is
$$intlimits_{V}{rho left( mathbf{r} right)left( mathbf{r-R} right)dV},=0 $$
Clearly, then, for a volume of uniform density
$$mathbf{R}=frac{1}{V}intlimits_{V}{mathbf{r}dV}=frac{1}{V}intintint(xmathbf{i}+ymathbf{j}+z mathbf{k}),dx,dy,dz $$
where, of course
$$V=intintint dx,dy,dz $$
answered Jun 7 '17 at 15:51
Cye WaldmanCye Waldman
4,2322623
$endgroup$
add a comment |
$begingroup$
From Wikipedia (https://en.wikipedia.org/wiki/Center_of_mass): If the mass distribution is continuous with the density $ρ(r)$ within a volume $V$, then the integral of the weighted position coordinates of the points in this volume relative to the center of mass $R$ is zero, that is
$$intlimits_{V}{rho left( mathbf{r} right)left( mathbf{r-R} right)dV},=0 $$
Clearly, then, for a volume of uniform density
$$mathbf{R}=frac{1}{V}intlimits_{V}{mathbf{r}dV}=frac{1}{V}intintint(xmathbf{i}+ymathbf{j}+z mathbf{k}),dx,dy,dz $$
where, of course
$$V=intintint dx,dy,dz $$
answered Jun 7 '17 at 15:51
Cye WaldmanCye Waldman
4,2322623
$endgroup$
From Wikipedia (https://en.wikipedia.org/wiki/Center_of_mass): If the mass distribution is continuous with the density $ρ(r)$ within a volume $V$, then the integral of the weighted position coordinates of the points in this volume relative to the center of mass $R$ is zero, that is
$$intlimits_{V}{rho left( mathbf{r} right)left( mathbf{r-R} right)dV},=0 $$
Clearly, then, for a volume of uniform density
$$mathbf{R}=frac{1}{V}intlimits_{V}{mathbf{r}dV}=frac{1}{V}intintint(xmathbf{i}+ymathbf{j}+z mathbf{k}),dx,dy,dz $$
where, of course
$$V=intintint dx,dy,dz $$
answered Jun 7 '17 at 15:51
Cye WaldmanCye Waldman
4,2322623
answered Jun 7 '17 at 15:51
Cye WaldmanCye Waldman
4,2322623
answered Jun 7 '17 at 15:51
Cye WaldmanCye Waldman
4,2322623
answered Jun 7 '17 at 15:51
Cye WaldmanCye Waldman
4,2322623
4,2322623
add a comment |
add a comment |
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$begingroup$
You solid G is not well defined.
$endgroup$
– hamam_Abdallah
Jun 7 '17 at 2:42