Mixing
Calculate the volume between $2$ surfaces
$begingroup$
Calculate the volume between $x^2+y^2+z^2=8$ and $x^2+y^2-2z=0$. I don't know how to approach this but I still tried something:
I rewrote the second equation as: $x^2+y^2+(z-1)^2=z^2+1$ and then combined it with the first one and got $2(x^2+y^2)+(z-1)^2=9$ and then parametrized this with the regular spheric parametrization which is:
$$x=frac {1}{sqrt{2}}rsin theta cos phi$$
$$y=frac 1{sqrt{2}}sinthetasinphi$$
$$z=rcostheta + 1$$
And of course the volume formula:
$$V(Omega)=intintint_{Omega} dxdydz$$
But that led me to a wrong answer.. what should I do?
Else, I tried parametrizing like this: $x=rcos t$, $y=rsin t$. then $r^2+z^2=8$ and $r^2-2z=0$ giving the only 'good' solutions $r=2, z=2$ then $rin[0,2]$ and $z=[frac {r^2}2,sqrt{8-r^2}]$ positive root because it's in the plane $z=2$.
giving $int_{0}^{2pi}int_0^2int_{frac {r^2}2}^{sqrt{8-r^2}}rdzdrdt.$ But still i god the wrong answer..
integration volume
asked Jan 27 at 12:17
C. CristiC. Cristi
1,639218
$endgroup$
add a comment |
$begingroup$
Calculate the volume between $x^2+y^2+z^2=8$ and $x^2+y^2-2z=0$. I don't know how to approach this but I still tried something:
I rewrote the second equation as: $x^2+y^2+(z-1)^2=z^2+1$ and then combined it with the first one and got $2(x^2+y^2)+(z-1)^2=9$ and then parametrized this with the regular spheric parametrization which is:
$$x=frac {1}{sqrt{2}}rsin theta cos phi$$
$$y=frac 1{sqrt{2}}sinthetasinphi$$
$$z=rcostheta + 1$$
And of course the volume formula:
$$V(Omega)=intintint_{Omega} dxdydz$$
But that led me to a wrong answer.. what should I do?
Else, I tried parametrizing like this: $x=rcos t$, $y=rsin t$. then $r^2+z^2=8$ and $r^2-2z=0$ giving the only 'good' solutions $r=2, z=2$ then $rin[0,2]$ and $z=[frac {r^2}2,sqrt{8-r^2}]$ positive root because it's in the plane $z=2$.
giving $int_{0}^{2pi}int_0^2int_{frac {r^2}2}^{sqrt{8-r^2}}rdzdrdt.$ But still i god the wrong answer..
integration volume
asked Jan 27 at 12:17
C. CristiC. Cristi
1,639218
$endgroup$
$begingroup$
There are two surfaces. Your approach was to find a consequence of both surface equations and then parametrize that, which gives a single surface. But that doesn't seem related to getting volume between the two starting surfaces.
$endgroup$
– coffeemath
Jan 27 at 12:31
$begingroup$
@coffeemath Why wouldn't it be related to getting the volume between the 2 of them? It really made sense... actually.. I think you're right, then what approach should I be using?
$endgroup$
– C. Cristi
Jan 27 at 12:34
$begingroup$
You have an almost identical exercise here: math.stackexchange.com/questions/3065012/…
$endgroup$
– Rafa Budría
Jan 27 at 12:37
$begingroup$
I tried to follow that example but still get the wrong answer: I will update in the question!
$endgroup$
– C. Cristi
Jan 27 at 12:47
add a comment |
$begingroup$
Calculate the volume between $x^2+y^2+z^2=8$ and $x^2+y^2-2z=0$. I don't know how to approach this but I still tried something:
I rewrote the second equation as: $x^2+y^2+(z-1)^2=z^2+1$ and then combined it with the first one and got $2(x^2+y^2)+(z-1)^2=9$ and then parametrized this with the regular spheric parametrization which is:
$$x=frac {1}{sqrt{2}}rsin theta cos phi$$
$$y=frac 1{sqrt{2}}sinthetasinphi$$
$$z=rcostheta + 1$$
And of course the volume formula:
$$V(Omega)=intintint_{Omega} dxdydz$$
But that led me to a wrong answer.. what should I do?
Else, I tried parametrizing like this: $x=rcos t$, $y=rsin t$. then $r^2+z^2=8$ and $r^2-2z=0$ giving the only 'good' solutions $r=2, z=2$ then $rin[0,2]$ and $z=[frac {r^2}2,sqrt{8-r^2}]$ positive root because it's in the plane $z=2$.
giving $int_{0}^{2pi}int_0^2int_{frac {r^2}2}^{sqrt{8-r^2}}rdzdrdt.$ But still i god the wrong answer..
integration volume
asked Jan 27 at 12:17
C. CristiC. Cristi
1,639218
$endgroup$
Calculate the volume between $x^2+y^2+z^2=8$ and $x^2+y^2-2z=0$. I don't know how to approach this but I still tried something:
I rewrote the second equation as: $x^2+y^2+(z-1)^2=z^2+1$ and then combined it with the first one and got $2(x^2+y^2)+(z-1)^2=9$ and then parametrized this with the regular spheric parametrization which is:
$$x=frac {1}{sqrt{2}}rsin theta cos phi$$
$$y=frac 1{sqrt{2}}sinthetasinphi$$
$$z=rcostheta + 1$$
And of course the volume formula:
$$V(Omega)=intintint_{Omega} dxdydz$$
But that led me to a wrong answer.. what should I do?
Else, I tried parametrizing like this: $x=rcos t$, $y=rsin t$. then $r^2+z^2=8$ and $r^2-2z=0$ giving the only 'good' solutions $r=2, z=2$ then $rin[0,2]$ and $z=[frac {r^2}2,sqrt{8-r^2}]$ positive root because it's in the plane $z=2$.
giving $int_{0}^{2pi}int_0^2int_{frac {r^2}2}^{sqrt{8-r^2}}rdzdrdt.$ But still i god the wrong answer..
integration volume
integration volume
asked Jan 27 at 12:17
C. CristiC. Cristi
1,639218
asked Jan 27 at 12:17
C. CristiC. Cristi
1,639218
edited Jan 27 at 12:50
C. Cristi
asked Jan 27 at 12:17
C. CristiC. Cristi
1,639218
asked Jan 27 at 12:17
C. CristiC. Cristi
1,639218
asked Jan 27 at 12:17
C. CristiC. Cristi
1,639218
1,639218
$begingroup$
There are two surfaces. Your approach was to find a consequence of both surface equations and then parametrize that, which gives a single surface. But that doesn't seem related to getting volume between the two starting surfaces.
$endgroup$
– coffeemath
Jan 27 at 12:31
$begingroup$
@coffeemath Why wouldn't it be related to getting the volume between the 2 of them? It really made sense... actually.. I think you're right, then what approach should I be using?
$endgroup$
– C. Cristi
Jan 27 at 12:34
$begingroup$
You have an almost identical exercise here: math.stackexchange.com/questions/3065012/…
$endgroup$
– Rafa Budría
Jan 27 at 12:37
$begingroup$
I tried to follow that example but still get the wrong answer: I will update in the question!
$endgroup$
– C. Cristi
Jan 27 at 12:47
add a comment |
$begingroup$
There are two surfaces. Your approach was to find a consequence of both surface equations and then parametrize that, which gives a single surface. But that doesn't seem related to getting volume between the two starting surfaces.
$endgroup$
– coffeemath
Jan 27 at 12:31
$begingroup$
@coffeemath Why wouldn't it be related to getting the volume between the 2 of them? It really made sense... actually.. I think you're right, then what approach should I be using?
$endgroup$
– C. Cristi
Jan 27 at 12:34
$begingroup$
You have an almost identical exercise here: math.stackexchange.com/questions/3065012/…
$endgroup$
– Rafa Budría
Jan 27 at 12:37
$begingroup$
I tried to follow that example but still get the wrong answer: I will update in the question!
$endgroup$
– C. Cristi
Jan 27 at 12:47
$begingroup$
There are two surfaces. Your approach was to find a consequence of both surface equations and then parametrize that, which gives a single surface. But that doesn't seem related to getting volume between the two starting surfaces.
$endgroup$
– coffeemath
Jan 27 at 12:31
$begingroup$
There are two surfaces. Your approach was to find a consequence of both surface equations and then parametrize that, which gives a single surface. But that doesn't seem related to getting volume between the two starting surfaces.
$endgroup$
– coffeemath
Jan 27 at 12:31
$begingroup$
@coffeemath Why wouldn't it be related to getting the volume between the 2 of them? It really made sense... actually.. I think you're right, then what approach should I be using?
$endgroup$
– C. Cristi
Jan 27 at 12:34
$begingroup$
@coffeemath Why wouldn't it be related to getting the volume between the 2 of them? It really made sense... actually.. I think you're right, then what approach should I be using?
$endgroup$
– C. Cristi
Jan 27 at 12:34
$begingroup$
You have an almost identical exercise here: math.stackexchange.com/questions/3065012/…
$endgroup$
– Rafa Budría
Jan 27 at 12:37
$begingroup$
You have an almost identical exercise here: math.stackexchange.com/questions/3065012/…
$endgroup$
– Rafa Budría
Jan 27 at 12:37
$begingroup$
I tried to follow that example but still get the wrong answer: I will update in the question!
$endgroup$
– C. Cristi
Jan 27 at 12:47
$begingroup$
I tried to follow that example but still get the wrong answer: I will update in the question!
$endgroup$
– C. Cristi
Jan 27 at 12:47
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
V & stackrel{mrm{def.}}{equiv} iiint_{mathbb{R}^{large 3}}bracks{x^{2} + y^{2} + z^{2} < 8}
bracks{z > {x^{2} + y^{2} over 2}}dd^{3}vec{r}
\[5mm] & underline{mbox{Use Cylindrical Coordinates:}}
\
V & = int_{0}^{2pi}
int_{-infty}^{infty}int_{0}^{infty}bracks{rho^{2} + z^{2} < 8}
bracks{z > {rho^{2} over 2}}rho,ddrho,dd z,ddphi
\[5mm] & =
piint_{0}^{infty}int_{0}^{infty}bracks{rho + z^{2} < 8}
bracks{z > {rho over 2}},ddrho,dd z
\[5mm] & =
piint_{0}^{infty}int_{0}^{infty}
bracks{{1 over 2},rho < z < root{8 - rho}},dd z,ddrho
\[5mm] & =
piint_{0}^{infty}
bracks{{1 over 2},rho < root{8 - rho}}
int_{rho/2}^{root{8 - rho}},dd z,ddrho
\[5mm] & =
piint_{0}^{color{red}{4}}
pars{root{8 - rho} - {1 over 2},rho},ddrho =
bbx{{4 over 3}pars{8root{2} - 7},pi} approx 18.0692
end{align}
Note that
$ds{0 < rho/2 < root{8 - rho} implies rho < color{red}{4}}$.
answered Jan 27 at 21:42
Felix MarinFelix Marin
68.8k7109146
$endgroup$
add a comment |
$begingroup$
A geometric view of the problem will be much of help to solve it.
One is a sphere of radius $sqrt{8}$ centered at the origin.
The other is a paraboloid of revolution, given by the revolution of $z=x^2/2$
around the $z$ axis, thus with the vertex at the origin.
The volume between the two is given by revolution around the $z$ axis
of the 2D area delimited by a parabola and a circle.
I suppose you can compute that by "shells" or "washer" method.
answered Jan 27 at 19:11
G CabG Cab
20.4k31341
$endgroup$
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
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newcommand{ds}[1]{displaystyle{#1}}
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newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
V & stackrel{mrm{def.}}{equiv} iiint_{mathbb{R}^{large 3}}bracks{x^{2} + y^{2} + z^{2} < 8}
bracks{z > {x^{2} + y^{2} over 2}}dd^{3}vec{r}
\[5mm] & underline{mbox{Use Cylindrical Coordinates:}}
\
V & = int_{0}^{2pi}
int_{-infty}^{infty}int_{0}^{infty}bracks{rho^{2} + z^{2} < 8}
bracks{z > {rho^{2} over 2}}rho,ddrho,dd z,ddphi
\[5mm] & =
piint_{0}^{infty}int_{0}^{infty}bracks{rho + z^{2} < 8}
bracks{z > {rho over 2}},ddrho,dd z
\[5mm] & =
piint_{0}^{infty}int_{0}^{infty}
bracks{{1 over 2},rho < z < root{8 - rho}},dd z,ddrho
\[5mm] & =
piint_{0}^{infty}
bracks{{1 over 2},rho < root{8 - rho}}
int_{rho/2}^{root{8 - rho}},dd z,ddrho
\[5mm] & =
piint_{0}^{color{red}{4}}
pars{root{8 - rho} - {1 over 2},rho},ddrho =
bbx{{4 over 3}pars{8root{2} - 7},pi} approx 18.0692
end{align}
Note that
$ds{0 < rho/2 < root{8 - rho} implies rho < color{red}{4}}$.
answered Jan 27 at 21:42
Felix MarinFelix Marin
68.8k7109146
$endgroup$
add a comment |
$begingroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
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newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
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newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
V & stackrel{mrm{def.}}{equiv} iiint_{mathbb{R}^{large 3}}bracks{x^{2} + y^{2} + z^{2} < 8}
bracks{z > {x^{2} + y^{2} over 2}}dd^{3}vec{r}
\[5mm] & underline{mbox{Use Cylindrical Coordinates:}}
\
V & = int_{0}^{2pi}
int_{-infty}^{infty}int_{0}^{infty}bracks{rho^{2} + z^{2} < 8}
bracks{z > {rho^{2} over 2}}rho,ddrho,dd z,ddphi
\[5mm] & =
piint_{0}^{infty}int_{0}^{infty}bracks{rho + z^{2} < 8}
bracks{z > {rho over 2}},ddrho,dd z
\[5mm] & =
piint_{0}^{infty}int_{0}^{infty}
bracks{{1 over 2},rho < z < root{8 - rho}},dd z,ddrho
\[5mm] & =
piint_{0}^{infty}
bracks{{1 over 2},rho < root{8 - rho}}
int_{rho/2}^{root{8 - rho}},dd z,ddrho
\[5mm] & =
piint_{0}^{color{red}{4}}
pars{root{8 - rho} - {1 over 2},rho},ddrho =
bbx{{4 over 3}pars{8root{2} - 7},pi} approx 18.0692
end{align}
Note that
$ds{0 < rho/2 < root{8 - rho} implies rho < color{red}{4}}$.
answered Jan 27 at 21:42
Felix MarinFelix Marin
68.8k7109146
$endgroup$
add a comment |
$begingroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
newcommand{braces}[1]{leftlbrace,{#1},rightrbrace}
newcommand{bracks}[1]{leftlbrack,{#1},rightrbrack}
newcommand{dd}{mathrm{d}}
newcommand{ds}[1]{displaystyle{#1}}
newcommand{expo}[1]{,mathrm{e}^{#1},}
newcommand{ic}{mathrm{i}}
newcommand{mc}[1]{mathcal{#1}}
newcommand{mrm}[1]{mathrm{#1}}
newcommand{pars}[1]{left(,{#1},right)}
newcommand{partiald}[3]{frac{partial^{#1} #2}{partial #3^{#1}}}
newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
V & stackrel{mrm{def.}}{equiv} iiint_{mathbb{R}^{large 3}}bracks{x^{2} + y^{2} + z^{2} < 8}
bracks{z > {x^{2} + y^{2} over 2}}dd^{3}vec{r}
\[5mm] & underline{mbox{Use Cylindrical Coordinates:}}
\
V & = int_{0}^{2pi}
int_{-infty}^{infty}int_{0}^{infty}bracks{rho^{2} + z^{2} < 8}
bracks{z > {rho^{2} over 2}}rho,ddrho,dd z,ddphi
\[5mm] & =
piint_{0}^{infty}int_{0}^{infty}bracks{rho + z^{2} < 8}
bracks{z > {rho over 2}},ddrho,dd z
\[5mm] & =
piint_{0}^{infty}int_{0}^{infty}
bracks{{1 over 2},rho < z < root{8 - rho}},dd z,ddrho
\[5mm] & =
piint_{0}^{infty}
bracks{{1 over 2},rho < root{8 - rho}}
int_{rho/2}^{root{8 - rho}},dd z,ddrho
\[5mm] & =
piint_{0}^{color{red}{4}}
pars{root{8 - rho} - {1 over 2},rho},ddrho =
bbx{{4 over 3}pars{8root{2} - 7},pi} approx 18.0692
end{align}
Note that
$ds{0 < rho/2 < root{8 - rho} implies rho < color{red}{4}}$.
answered Jan 27 at 21:42
Felix MarinFelix Marin
68.8k7109146
$endgroup$
$newcommand{bbx}[1]{,bbox[15px,border:1px groove navy]{displaystyle{#1}},}
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newcommand{root}[2]{,sqrt[#1]{,{#2},},}
newcommand{totald}[3]{frac{mathrm{d}^{#1} #2}{mathrm{d} #3^{#1}}}
newcommand{verts}[1]{leftvert,{#1},rightvert}$
begin{align}
V & stackrel{mrm{def.}}{equiv} iiint_{mathbb{R}^{large 3}}bracks{x^{2} + y^{2} + z^{2} < 8}
bracks{z > {x^{2} + y^{2} over 2}}dd^{3}vec{r}
\[5mm] & underline{mbox{Use Cylindrical Coordinates:}}
\
V & = int_{0}^{2pi}
int_{-infty}^{infty}int_{0}^{infty}bracks{rho^{2} + z^{2} < 8}
bracks{z > {rho^{2} over 2}}rho,ddrho,dd z,ddphi
\[5mm] & =
piint_{0}^{infty}int_{0}^{infty}bracks{rho + z^{2} < 8}
bracks{z > {rho over 2}},ddrho,dd z
\[5mm] & =
piint_{0}^{infty}int_{0}^{infty}
bracks{{1 over 2},rho < z < root{8 - rho}},dd z,ddrho
\[5mm] & =
piint_{0}^{infty}
bracks{{1 over 2},rho < root{8 - rho}}
int_{rho/2}^{root{8 - rho}},dd z,ddrho
\[5mm] & =
piint_{0}^{color{red}{4}}
pars{root{8 - rho} - {1 over 2},rho},ddrho =
bbx{{4 over 3}pars{8root{2} - 7},pi} approx 18.0692
end{align}
Note that
$ds{0 < rho/2 < root{8 - rho} implies rho < color{red}{4}}$.
answered Jan 27 at 21:42
Felix MarinFelix Marin
68.8k7109146
edited Jan 27 at 22:02
answered Jan 27 at 21:42
Felix MarinFelix Marin
68.8k7109146
answered Jan 27 at 21:42
Felix MarinFelix Marin
68.8k7109146
answered Jan 27 at 21:42
Felix MarinFelix Marin
68.8k7109146
68.8k7109146
add a comment |
add a comment |
$begingroup$
A geometric view of the problem will be much of help to solve it.
One is a sphere of radius $sqrt{8}$ centered at the origin.
The other is a paraboloid of revolution, given by the revolution of $z=x^2/2$
around the $z$ axis, thus with the vertex at the origin.
The volume between the two is given by revolution around the $z$ axis
of the 2D area delimited by a parabola and a circle.
I suppose you can compute that by "shells" or "washer" method.
answered Jan 27 at 19:11
G CabG Cab
20.4k31341
$endgroup$
add a comment |
$begingroup$
A geometric view of the problem will be much of help to solve it.
One is a sphere of radius $sqrt{8}$ centered at the origin.
The other is a paraboloid of revolution, given by the revolution of $z=x^2/2$
around the $z$ axis, thus with the vertex at the origin.
The volume between the two is given by revolution around the $z$ axis
of the 2D area delimited by a parabola and a circle.
I suppose you can compute that by "shells" or "washer" method.
answered Jan 27 at 19:11
G CabG Cab
20.4k31341
$endgroup$
add a comment |
$begingroup$
A geometric view of the problem will be much of help to solve it.
One is a sphere of radius $sqrt{8}$ centered at the origin.
The other is a paraboloid of revolution, given by the revolution of $z=x^2/2$
around the $z$ axis, thus with the vertex at the origin.
The volume between the two is given by revolution around the $z$ axis
of the 2D area delimited by a parabola and a circle.
I suppose you can compute that by "shells" or "washer" method.
answered Jan 27 at 19:11
G CabG Cab
20.4k31341
$endgroup$
A geometric view of the problem will be much of help to solve it.
One is a sphere of radius $sqrt{8}$ centered at the origin.
The other is a paraboloid of revolution, given by the revolution of $z=x^2/2$
around the $z$ axis, thus with the vertex at the origin.
The volume between the two is given by revolution around the $z$ axis
of the 2D area delimited by a parabola and a circle.
I suppose you can compute that by "shells" or "washer" method.
answered Jan 27 at 19:11
G CabG Cab
20.4k31341
answered Jan 27 at 19:11
G CabG Cab
20.4k31341
answered Jan 27 at 19:11
G CabG Cab
20.4k31341
answered Jan 27 at 19:11
G CabG Cab
20.4k31341
20.4k31341
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$begingroup$
There are two surfaces. Your approach was to find a consequence of both surface equations and then parametrize that, which gives a single surface. But that doesn't seem related to getting volume between the two starting surfaces.
$endgroup$
– coffeemath
Jan 27 at 12:31
$begingroup$
@coffeemath Why wouldn't it be related to getting the volume between the 2 of them? It really made sense... actually.. I think you're right, then what approach should I be using?
$endgroup$
– C. Cristi
Jan 27 at 12:34
$begingroup$
You have an almost identical exercise here: math.stackexchange.com/questions/3065012/…
$endgroup$
– Rafa Budría
Jan 27 at 12:37
$begingroup$
I tried to follow that example but still get the wrong answer: I will update in the question!
$endgroup$
– C. Cristi
Jan 27 at 12:47