Triple integral - wedge shaped solid
Find the volume of of the wedge shaped solid that lies above the xy plane, below the $z=x$ plane and within the cylinder $x^2+y^2 = 4$.
I'm having serious trouble picturing this. I think the z bounds are [0,x], but can't figure out any of the other ones.
integration multivariable-calculus volume
asked Mar 28 '14 at 13:01
user127778
118112
add a comment |
Find the volume of of the wedge shaped solid that lies above the xy plane, below the $z=x$ plane and within the cylinder $x^2+y^2 = 4$.
I'm having serious trouble picturing this. I think the z bounds are [0,x], but can't figure out any of the other ones.
integration multivariable-calculus volume
asked Mar 28 '14 at 13:01
user127778
118112
add a comment |
Find the volume of of the wedge shaped solid that lies above the xy plane, below the $z=x$ plane and within the cylinder $x^2+y^2 = 4$.
I'm having serious trouble picturing this. I think the z bounds are [0,x], but can't figure out any of the other ones.
integration multivariable-calculus volume
asked Mar 28 '14 at 13:01
user127778
118112
Find the volume of of the wedge shaped solid that lies above the xy plane, below the $z=x$ plane and within the cylinder $x^2+y^2 = 4$.
I'm having serious trouble picturing this. I think the z bounds are [0,x], but can't figure out any of the other ones.
integration multivariable-calculus volume
integration multivariable-calculus volume
asked Mar 28 '14 at 13:01
user127778
118112
asked Mar 28 '14 at 13:01
user127778
118112
asked Mar 28 '14 at 13:01
user127778
118112
asked Mar 28 '14 at 13:01
user127778
118112
asked Mar 28 '14 at 13:01
user127778
118112
118112
add a comment |
add a comment |
4 Answers
4
active
oldest
votes
looks like this should be the required vol
$$2int_{0}^{pi/2} int_{0}^{2} int_{0}^{rcostheta} rdzdrdtheta $$
answered Mar 28 '14 at 13:06
ketan
1,530822
why is it [$0,frac{pi}{2}$] and not [$pi, 2pi]$? The $dtheta$ bound is what I'm asking about, because I got all of the other ones. Also, why do you multiply by 2?
– user127778
Mar 28 '14 at 13:09
the region in xy plane, is of 1st and 4th quadrant, so $[pi, 2pi]$ doesn't look like being there.the volume is above 1st and 4th quadrant of xy plane, so you can adjust $dtheta$ accordingly.and since i chose $[0,frac{pi}{2}]$ so multiplied by 2
– ketan
Mar 28 '14 at 13:14
$2times$ is because of symmetry. Otherwise you'd have to integrate $int_{-pi/2}^{pi/2}$.
– orion
Mar 28 '14 at 13:18
add a comment |
I believe that you are describing a variation of the hoof of Archimedes (see figure below). If that's the case, I can give you a detailed solution and show you how to reduce the triple integral for volume to a single integration. From what you describe, the height would be $R$, rather than the $2R$ shown in the figure.
answered Jun 7 '17 at 17:04
Cye Waldman
4,0952523
add a comment |
Planes $x={rm const.}$ intersect this solid $B$ in rectangles of width $2sqrt{4-x^2}$ and height $x$ (see the figure in Cye Waldman's answer). One therefore obtains
$${rm vol}(B)=int_0^2 2sqrt{4-x^2}>x>dx=-{2over3}(4-x^2)^{3/2}biggr|_0^2={16over3} .$$
answered Jul 9 '17 at 9:08
Christian Blatter
172k7112325
add a comment |
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
V & = iiint_{mathbb{R}^{large 3}}bracks{z > 0}
bracks{z < x}bracks{x^{2} + y^{2} < 4}dd x,dd y,dd z
\[5mm] & =
int_{-infty}^{infty}int_{-infty}^{infty}
int_{-infty}^{infty}bracks{0 < z < rhocospars{phi}}
bracks{rho > 0}bracks{rho^{2} < 4}rho,ddrho,ddphi
,dd z
\[5mm] & =
int_{0}^{infty}int_{0}^{2pi}
int_{0}^{2}bracks{0 < z < rhocospars{phi}}
rho,ddrho,ddphi,dd z
\[5mm] & =
2int_{0}^{infty}int_{0}^{pi}
int_{0}^{2}bracks{0 < z < -rhocospars{phi}}
rho,ddrho,ddphi,dd z
\[5mm] & =
2int_{0}^{infty}int_{0}^{pi/2}
int_{0}^{2}bracks{0 < z < rhosinpars{phi}}
rho,ddrho,ddphi,dd z
\[5mm] & =
2int_{0}^{pi/2}
int_{0}^{2}rhosinpars{phi}rho,ddrho,ddphi =
2pars{int_{0}^{pi/2}sinpars{phi},ddphi}
pars{int_{0}^{2}rho^{2},ddrho}
\[5mm] & = bbx{16 over 3}
end{align}
answered Dec 6 at 20:23
Felix Marin
66.9k7107139
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
looks like this should be the required vol
$$2int_{0}^{pi/2} int_{0}^{2} int_{0}^{rcostheta} rdzdrdtheta $$
answered Mar 28 '14 at 13:06
ketan
1,530822
why is it [$0,frac{pi}{2}$] and not [$pi, 2pi]$? The $dtheta$ bound is what I'm asking about, because I got all of the other ones. Also, why do you multiply by 2?
– user127778
Mar 28 '14 at 13:09
the region in xy plane, is of 1st and 4th quadrant, so $[pi, 2pi]$ doesn't look like being there.the volume is above 1st and 4th quadrant of xy plane, so you can adjust $dtheta$ accordingly.and since i chose $[0,frac{pi}{2}]$ so multiplied by 2
– ketan
Mar 28 '14 at 13:14
$2times$ is because of symmetry. Otherwise you'd have to integrate $int_{-pi/2}^{pi/2}$.
– orion
Mar 28 '14 at 13:18
add a comment |
looks like this should be the required vol
$$2int_{0}^{pi/2} int_{0}^{2} int_{0}^{rcostheta} rdzdrdtheta $$
answered Mar 28 '14 at 13:06
ketan
1,530822
why is it [$0,frac{pi}{2}$] and not [$pi, 2pi]$? The $dtheta$ bound is what I'm asking about, because I got all of the other ones. Also, why do you multiply by 2?
– user127778
Mar 28 '14 at 13:09
the region in xy plane, is of 1st and 4th quadrant, so $[pi, 2pi]$ doesn't look like being there.the volume is above 1st and 4th quadrant of xy plane, so you can adjust $dtheta$ accordingly.and since i chose $[0,frac{pi}{2}]$ so multiplied by 2
– ketan
Mar 28 '14 at 13:14
$2times$ is because of symmetry. Otherwise you'd have to integrate $int_{-pi/2}^{pi/2}$.
– orion
Mar 28 '14 at 13:18
add a comment |
looks like this should be the required vol
$$2int_{0}^{pi/2} int_{0}^{2} int_{0}^{rcostheta} rdzdrdtheta $$
answered Mar 28 '14 at 13:06
ketan
1,530822
looks like this should be the required vol
$$2int_{0}^{pi/2} int_{0}^{2} int_{0}^{rcostheta} rdzdrdtheta $$
answered Mar 28 '14 at 13:06
ketan
1,530822
answered Mar 28 '14 at 13:06
ketan
1,530822
answered Mar 28 '14 at 13:06
ketan
1,530822
answered Mar 28 '14 at 13:06
ketan
1,530822
1,530822
why is it [$0,frac{pi}{2}$] and not [$pi, 2pi]$? The $dtheta$ bound is what I'm asking about, because I got all of the other ones. Also, why do you multiply by 2?
– user127778
Mar 28 '14 at 13:09
the region in xy plane, is of 1st and 4th quadrant, so $[pi, 2pi]$ doesn't look like being there.the volume is above 1st and 4th quadrant of xy plane, so you can adjust $dtheta$ accordingly.and since i chose $[0,frac{pi}{2}]$ so multiplied by 2
– ketan
Mar 28 '14 at 13:14
$2times$ is because of symmetry. Otherwise you'd have to integrate $int_{-pi/2}^{pi/2}$.
– orion
Mar 28 '14 at 13:18
add a comment |
why is it [$0,frac{pi}{2}$] and not [$pi, 2pi]$? The $dtheta$ bound is what I'm asking about, because I got all of the other ones. Also, why do you multiply by 2?
– user127778
Mar 28 '14 at 13:09
the region in xy plane, is of 1st and 4th quadrant, so $[pi, 2pi]$ doesn't look like being there.the volume is above 1st and 4th quadrant of xy plane, so you can adjust $dtheta$ accordingly.and since i chose $[0,frac{pi}{2}]$ so multiplied by 2
– ketan
Mar 28 '14 at 13:14
$2times$ is because of symmetry. Otherwise you'd have to integrate $int_{-pi/2}^{pi/2}$.
– orion
Mar 28 '14 at 13:18
why is it [$0,frac{pi}{2}$] and not [$pi, 2pi]$? The $dtheta$ bound is what I'm asking about, because I got all of the other ones. Also, why do you multiply by 2?
– user127778
Mar 28 '14 at 13:09
why is it [$0,frac{pi}{2}$] and not [$pi, 2pi]$? The $dtheta$ bound is what I'm asking about, because I got all of the other ones. Also, why do you multiply by 2?
– user127778
Mar 28 '14 at 13:09
the region in xy plane, is of 1st and 4th quadrant, so $[pi, 2pi]$ doesn't look like being there.the volume is above 1st and 4th quadrant of xy plane, so you can adjust $dtheta$ accordingly.and since i chose $[0,frac{pi}{2}]$ so multiplied by 2
– ketan
Mar 28 '14 at 13:14
the region in xy plane, is of 1st and 4th quadrant, so $[pi, 2pi]$ doesn't look like being there.the volume is above 1st and 4th quadrant of xy plane, so you can adjust $dtheta$ accordingly.and since i chose $[0,frac{pi}{2}]$ so multiplied by 2
– ketan
Mar 28 '14 at 13:14
$2times$ is because of symmetry. Otherwise you'd have to integrate $int_{-pi/2}^{pi/2}$.
– orion
Mar 28 '14 at 13:18
$2times$ is because of symmetry. Otherwise you'd have to integrate $int_{-pi/2}^{pi/2}$.
– orion
Mar 28 '14 at 13:18
add a comment |
I believe that you are describing a variation of the hoof of Archimedes (see figure below). If that's the case, I can give you a detailed solution and show you how to reduce the triple integral for volume to a single integration. From what you describe, the height would be $R$, rather than the $2R$ shown in the figure.
answered Jun 7 '17 at 17:04
Cye Waldman
4,0952523
add a comment |
I believe that you are describing a variation of the hoof of Archimedes (see figure below). If that's the case, I can give you a detailed solution and show you how to reduce the triple integral for volume to a single integration. From what you describe, the height would be $R$, rather than the $2R$ shown in the figure.
answered Jun 7 '17 at 17:04
Cye Waldman
4,0952523
add a comment |
I believe that you are describing a variation of the hoof of Archimedes (see figure below). If that's the case, I can give you a detailed solution and show you how to reduce the triple integral for volume to a single integration. From what you describe, the height would be $R$, rather than the $2R$ shown in the figure.
answered Jun 7 '17 at 17:04
Cye Waldman
4,0952523
I believe that you are describing a variation of the hoof of Archimedes (see figure below). If that's the case, I can give you a detailed solution and show you how to reduce the triple integral for volume to a single integration. From what you describe, the height would be $R$, rather than the $2R$ shown in the figure.
answered Jun 7 '17 at 17:04
Cye Waldman
4,0952523
answered Jun 7 '17 at 17:04
Cye Waldman
4,0952523
answered Jun 7 '17 at 17:04
Cye Waldman
4,0952523
answered Jun 7 '17 at 17:04
Cye Waldman
4,0952523
4,0952523
add a comment |
add a comment |
Planes $x={rm const.}$ intersect this solid $B$ in rectangles of width $2sqrt{4-x^2}$ and height $x$ (see the figure in Cye Waldman's answer). One therefore obtains
$${rm vol}(B)=int_0^2 2sqrt{4-x^2}>x>dx=-{2over3}(4-x^2)^{3/2}biggr|_0^2={16over3} .$$
answered Jul 9 '17 at 9:08
Christian Blatter
172k7112325
add a comment |
Planes $x={rm const.}$ intersect this solid $B$ in rectangles of width $2sqrt{4-x^2}$ and height $x$ (see the figure in Cye Waldman's answer). One therefore obtains
$${rm vol}(B)=int_0^2 2sqrt{4-x^2}>x>dx=-{2over3}(4-x^2)^{3/2}biggr|_0^2={16over3} .$$
answered Jul 9 '17 at 9:08
Christian Blatter
172k7112325
add a comment |
Planes $x={rm const.}$ intersect this solid $B$ in rectangles of width $2sqrt{4-x^2}$ and height $x$ (see the figure in Cye Waldman's answer). One therefore obtains
$${rm vol}(B)=int_0^2 2sqrt{4-x^2}>x>dx=-{2over3}(4-x^2)^{3/2}biggr|_0^2={16over3} .$$
answered Jul 9 '17 at 9:08
Christian Blatter
172k7112325
Planes $x={rm const.}$ intersect this solid $B$ in rectangles of width $2sqrt{4-x^2}$ and height $x$ (see the figure in Cye Waldman's answer). One therefore obtains
$${rm vol}(B)=int_0^2 2sqrt{4-x^2}>x>dx=-{2over3}(4-x^2)^{3/2}biggr|_0^2={16over3} .$$
answered Jul 9 '17 at 9:08
Christian Blatter
172k7112325
answered Jul 9 '17 at 9:08
Christian Blatter
172k7112325
answered Jul 9 '17 at 9:08
Christian Blatter
172k7112325
answered Jul 9 '17 at 9:08
Christian Blatter
172k7112325
172k7112325
add a comment |
add a comment |
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
V & = iiint_{mathbb{R}^{large 3}}bracks{z > 0}
bracks{z < x}bracks{x^{2} + y^{2} < 4}dd x,dd y,dd z
\[5mm] & =
int_{-infty}^{infty}int_{-infty}^{infty}
int_{-infty}^{infty}bracks{0 < z < rhocospars{phi}}
bracks{rho > 0}bracks{rho^{2} < 4}rho,ddrho,ddphi
,dd z
\[5mm] & =
int_{0}^{infty}int_{0}^{2pi}
int_{0}^{2}bracks{0 < z < rhocospars{phi}}
rho,ddrho,ddphi,dd z
\[5mm] & =
2int_{0}^{infty}int_{0}^{pi}
int_{0}^{2}bracks{0 < z < -rhocospars{phi}}
rho,ddrho,ddphi,dd z
\[5mm] & =
2int_{0}^{infty}int_{0}^{pi/2}
int_{0}^{2}bracks{0 < z < rhosinpars{phi}}
rho,ddrho,ddphi,dd z
\[5mm] & =
2int_{0}^{pi/2}
int_{0}^{2}rhosinpars{phi}rho,ddrho,ddphi =
2pars{int_{0}^{pi/2}sinpars{phi},ddphi}
pars{int_{0}^{2}rho^{2},ddrho}
\[5mm] & = bbx{16 over 3}
end{align}
answered Dec 6 at 20:23
Felix Marin
66.9k7107139
add a comment |
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
V & = iiint_{mathbb{R}^{large 3}}bracks{z > 0}
bracks{z < x}bracks{x^{2} + y^{2} < 4}dd x,dd y,dd z
\[5mm] & =
int_{-infty}^{infty}int_{-infty}^{infty}
int_{-infty}^{infty}bracks{0 < z < rhocospars{phi}}
bracks{rho > 0}bracks{rho^{2} < 4}rho,ddrho,ddphi
,dd z
\[5mm] & =
int_{0}^{infty}int_{0}^{2pi}
int_{0}^{2}bracks{0 < z < rhocospars{phi}}
rho,ddrho,ddphi,dd z
\[5mm] & =
2int_{0}^{infty}int_{0}^{pi}
int_{0}^{2}bracks{0 < z < -rhocospars{phi}}
rho,ddrho,ddphi,dd z
\[5mm] & =
2int_{0}^{infty}int_{0}^{pi/2}
int_{0}^{2}bracks{0 < z < rhosinpars{phi}}
rho,ddrho,ddphi,dd z
\[5mm] & =
2int_{0}^{pi/2}
int_{0}^{2}rhosinpars{phi}rho,ddrho,ddphi =
2pars{int_{0}^{pi/2}sinpars{phi},ddphi}
pars{int_{0}^{2}rho^{2},ddrho}
\[5mm] & = bbx{16 over 3}
end{align}
answered Dec 6 at 20:23
Felix Marin
66.9k7107139
add a comment |
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
V & = iiint_{mathbb{R}^{large 3}}bracks{z > 0}
bracks{z < x}bracks{x^{2} + y^{2} < 4}dd x,dd y,dd z
\[5mm] & =
int_{-infty}^{infty}int_{-infty}^{infty}
int_{-infty}^{infty}bracks{0 < z < rhocospars{phi}}
bracks{rho > 0}bracks{rho^{2} < 4}rho,ddrho,ddphi
,dd z
\[5mm] & =
int_{0}^{infty}int_{0}^{2pi}
int_{0}^{2}bracks{0 < z < rhocospars{phi}}
rho,ddrho,ddphi,dd z
\[5mm] & =
2int_{0}^{infty}int_{0}^{pi}
int_{0}^{2}bracks{0 < z < -rhocospars{phi}}
rho,ddrho,ddphi,dd z
\[5mm] & =
2int_{0}^{infty}int_{0}^{pi/2}
int_{0}^{2}bracks{0 < z < rhosinpars{phi}}
rho,ddrho,ddphi,dd z
\[5mm] & =
2int_{0}^{pi/2}
int_{0}^{2}rhosinpars{phi}rho,ddrho,ddphi =
2pars{int_{0}^{pi/2}sinpars{phi},ddphi}
pars{int_{0}^{2}rho^{2},ddrho}
\[5mm] & = bbx{16 over 3}
end{align}
answered Dec 6 at 20:23
Felix Marin
66.9k7107139
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
V & = iiint_{mathbb{R}^{large 3}}bracks{z > 0}
bracks{z < x}bracks{x^{2} + y^{2} < 4}dd x,dd y,dd z
\[5mm] & =
int_{-infty}^{infty}int_{-infty}^{infty}
int_{-infty}^{infty}bracks{0 < z < rhocospars{phi}}
bracks{rho > 0}bracks{rho^{2} < 4}rho,ddrho,ddphi
,dd z
\[5mm] & =
int_{0}^{infty}int_{0}^{2pi}
int_{0}^{2}bracks{0 < z < rhocospars{phi}}
rho,ddrho,ddphi,dd z
\[5mm] & =
2int_{0}^{infty}int_{0}^{pi}
int_{0}^{2}bracks{0 < z < -rhocospars{phi}}
rho,ddrho,ddphi,dd z
\[5mm] & =
2int_{0}^{infty}int_{0}^{pi/2}
int_{0}^{2}bracks{0 < z < rhosinpars{phi}}
rho,ddrho,ddphi,dd z
\[5mm] & =
2int_{0}^{pi/2}
int_{0}^{2}rhosinpars{phi}rho,ddrho,ddphi =
2pars{int_{0}^{pi/2}sinpars{phi},ddphi}
pars{int_{0}^{2}rho^{2},ddrho}
\[5mm] & = bbx{16 over 3}
end{align}
answered Dec 6 at 20:23
Felix Marin
66.9k7107139
answered Dec 6 at 20:23
Felix Marin
66.9k7107139
answered Dec 6 at 20:23
Felix Marin
66.9k7107139
answered Dec 6 at 20:23
Felix Marin
66.9k7107139
66.9k7107139
add a comment |
add a comment |
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