Mixing
Which double cone splits a sphere into two equal volumes? [closed]
$begingroup$
Consider a double cone (like a past and future light cone) centered at the origin. Now imagine a sphere centered at the origin.
What angle of the slope of the double cone makes it so that the it splits the sphere into 3 pieces such that the volume(s) inside the double-cone is equal to the volume outside the double cone?
(This is easier in 2 dimensions where the answer is simply 45 degrees!)
euclidean-geometry
asked Jan 23 at 22:58
zoobyzooby
1,022716
$endgroup$
closed as off-topic by Eevee Trainer, John Douma, darij grinberg, metamorphy, amd Jan 24 at 5:59
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, John Douma, darij grinberg, metamorphy, amd
If this question can be reworded to fit the rules in the help center, please edit the question.
|
show 1 more comment
$begingroup$
Consider a double cone (like a past and future light cone) centered at the origin. Now imagine a sphere centered at the origin.
What angle of the slope of the double cone makes it so that the it splits the sphere into 3 pieces such that the volume(s) inside the double-cone is equal to the volume outside the double cone?
(This is easier in 2 dimensions where the answer is simply 45 degrees!)
euclidean-geometry
asked Jan 23 at 22:58
zoobyzooby
1,022716
$endgroup$
closed as off-topic by Eevee Trainer, John Douma, darij grinberg, metamorphy, amd Jan 24 at 5:59
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, John Douma, darij grinberg, metamorphy, amd
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
My intuition may be off here but why would the sphere be split into $3$ pieces (as opposed to $2$)?
$endgroup$
– pwerth
Jan 23 at 22:59
$begingroup$
Yes, counting the inside of the double cone as one piece so-to-speak.
$endgroup$
– zooby
Jan 23 at 23:06
$begingroup$
Have you tried setting up the integral and getting an expression that depends upon the angle?
$endgroup$
– John Douma
Jan 23 at 23:13
1
$begingroup$
All these answers for a question that evidences no work on the asker’s part!
$endgroup$
– amd
Jan 23 at 23:38
1
$begingroup$
You can easily look up formulas for the area of a cone and spherical cap. Express them in terms of the radius of the sphere and cone angle and solve for the angle.
$endgroup$
– amd
Jan 23 at 23:40
|
show 1 more comment
$begingroup$
Consider a double cone (like a past and future light cone) centered at the origin. Now imagine a sphere centered at the origin.
What angle of the slope of the double cone makes it so that the it splits the sphere into 3 pieces such that the volume(s) inside the double-cone is equal to the volume outside the double cone?
(This is easier in 2 dimensions where the answer is simply 45 degrees!)
euclidean-geometry
asked Jan 23 at 22:58
zoobyzooby
1,022716
$endgroup$
Consider a double cone (like a past and future light cone) centered at the origin. Now imagine a sphere centered at the origin.
What angle of the slope of the double cone makes it so that the it splits the sphere into 3 pieces such that the volume(s) inside the double-cone is equal to the volume outside the double cone?
(This is easier in 2 dimensions where the answer is simply 45 degrees!)
euclidean-geometry
euclidean-geometry
asked Jan 23 at 22:58
zoobyzooby
1,022716
asked Jan 23 at 22:58
zoobyzooby
1,022716
asked Jan 23 at 22:58
zoobyzooby
1,022716
asked Jan 23 at 22:58
zoobyzooby
1,022716
asked Jan 23 at 22:58
zoobyzooby
1,022716
1,022716
closed as off-topic by Eevee Trainer, John Douma, darij grinberg, metamorphy, amd Jan 24 at 5:59
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, John Douma, darij grinberg, metamorphy, amd
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by Eevee Trainer, John Douma, darij grinberg, metamorphy, amd Jan 24 at 5:59
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Eevee Trainer, John Douma, darij grinberg, metamorphy, amd
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
My intuition may be off here but why would the sphere be split into $3$ pieces (as opposed to $2$)?
$endgroup$
– pwerth
Jan 23 at 22:59
$begingroup$
Yes, counting the inside of the double cone as one piece so-to-speak.
$endgroup$
– zooby
Jan 23 at 23:06
$begingroup$
Have you tried setting up the integral and getting an expression that depends upon the angle?
$endgroup$
– John Douma
Jan 23 at 23:13
1
$begingroup$
All these answers for a question that evidences no work on the asker’s part!
$endgroup$
– amd
Jan 23 at 23:38
1
$begingroup$
You can easily look up formulas for the area of a cone and spherical cap. Express them in terms of the radius of the sphere and cone angle and solve for the angle.
$endgroup$
– amd
Jan 23 at 23:40
|
show 1 more comment
$begingroup$
My intuition may be off here but why would the sphere be split into $3$ pieces (as opposed to $2$)?
$endgroup$
– pwerth
Jan 23 at 22:59
$begingroup$
Yes, counting the inside of the double cone as one piece so-to-speak.
$endgroup$
– zooby
Jan 23 at 23:06
$begingroup$
Have you tried setting up the integral and getting an expression that depends upon the angle?
$endgroup$
– John Douma
Jan 23 at 23:13
1
$begingroup$
All these answers for a question that evidences no work on the asker’s part!
$endgroup$
– amd
Jan 23 at 23:38
1
$begingroup$
You can easily look up formulas for the area of a cone and spherical cap. Express them in terms of the radius of the sphere and cone angle and solve for the angle.
$endgroup$
– amd
Jan 23 at 23:40
$begingroup$
My intuition may be off here but why would the sphere be split into $3$ pieces (as opposed to $2$)?
$endgroup$
– pwerth
Jan 23 at 22:59
$begingroup$
My intuition may be off here but why would the sphere be split into $3$ pieces (as opposed to $2$)?
$endgroup$
– pwerth
Jan 23 at 22:59
$begingroup$
Yes, counting the inside of the double cone as one piece so-to-speak.
$endgroup$
– zooby
Jan 23 at 23:06
$begingroup$
Yes, counting the inside of the double cone as one piece so-to-speak.
$endgroup$
– zooby
Jan 23 at 23:06
$begingroup$
Have you tried setting up the integral and getting an expression that depends upon the angle?
$endgroup$
– John Douma
Jan 23 at 23:13
$begingroup$
Have you tried setting up the integral and getting an expression that depends upon the angle?
$endgroup$
– John Douma
Jan 23 at 23:13
1
1
$begingroup$
All these answers for a question that evidences no work on the asker’s part!
$endgroup$
– amd
Jan 23 at 23:38
$begingroup$
All these answers for a question that evidences no work on the asker’s part!
$endgroup$
– amd
Jan 23 at 23:38
1
1
$begingroup$
You can easily look up formulas for the area of a cone and spherical cap. Express them in terms of the radius of the sphere and cone angle and solve for the angle.
$endgroup$
– amd
Jan 23 at 23:40
$begingroup$
You can easily look up formulas for the area of a cone and spherical cap. Express them in terms of the radius of the sphere and cone angle and solve for the angle.
$endgroup$
– amd
Jan 23 at 23:40
|
show 1 more comment
3 Answers
3
active
oldest
votes
$begingroup$
Note that by symmetry, for the double cone to split the sphere into two pieces of equal volume, each hemisphere must also be split into two pieces of equal volume by each cone. So we can just work with the upper hemisphere. Since a sphere has volume $displaystylefrac{4pi r^{3}}{3}$, the volume a hemisphere is $displaystylefrac{2pi r^{3}}{3}$.
Using standard spherical coordinates $(theta,phi,rho)$, the equation of the sphere is $rho=r$ and the equation of the cone is $phi=phi_{0}$, where $phi_{0}in[0,pi/2]$ is the desired angle. We can then easily compute the volume of the sphere below the cone:
begin{align}
V &= int_{0}^{2pi}int_{0}^{phi_{0}}int_{0}^{r}rho^{2}sin{phi} drho dphi dtheta\
&=frac{r^{3}}{3}int_{0}^{2pi}left(-cos{phi}biggrvert^{phi_{0}}_{0}right) dtheta \
&= frac{2pi r^{3}}{3}(-cos{phi_{0}}+1)
end{align}
Now, we want this volume to be half of the volume of the hemisphere, so we should have
$$frac{2pi r^{3}}{3}(-cos{phi_{0}}+1)=frac{pi r^{3}}{3}$$
which yields
$$cos{phi_{0}}=frac{1}{2}$$
so the desired angle is
$$phi_{0}=cos^{-1}left(frac{1}{2}right)=frac{pi}{3}$$ I should mention that this angle $phi$ is measured from the top of the $z$-axis.
answered Jan 23 at 23:20
pwerthpwerth
3,265417
$endgroup$
$begingroup$
I get $phi=pi/3$
$endgroup$
– zooby
Jan 23 at 23:24
$begingroup$
@zooby Updated my answer to include that.
$endgroup$
– pwerth
Jan 23 at 23:27
$begingroup$
good calculus skills. I wonder if there's an intuitive explanation too... like that would make you think, "of course, the angle has to be 60 degrees."
$endgroup$
– zooby
Jan 24 at 5:48
add a comment |
$begingroup$
The volume of the portion of a cone inside a unit sphere equals the solid angle it subtends. You want this angle to be $dfracpi3$ steradians.
From Wolfram, the solid angle corresponding to the half aperture $theta$ is
$$pi(2(1-costheta)+sintheta).$$
You can solve for $theta$.
answered Jan 23 at 23:18
Yves DaoustYves Daoust
130k676229
$endgroup$
add a comment |
$begingroup$
For half volume you want: $phi = pi/3$ so:
answered Jan 23 at 23:29
David G. StorkDavid G. Stork
11.1k41432
$endgroup$
$begingroup$
I upvoted for a nice picture!
$endgroup$
– zooby
Jan 24 at 5:47
add a comment |
3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Note that by symmetry, for the double cone to split the sphere into two pieces of equal volume, each hemisphere must also be split into two pieces of equal volume by each cone. So we can just work with the upper hemisphere. Since a sphere has volume $displaystylefrac{4pi r^{3}}{3}$, the volume a hemisphere is $displaystylefrac{2pi r^{3}}{3}$.
Using standard spherical coordinates $(theta,phi,rho)$, the equation of the sphere is $rho=r$ and the equation of the cone is $phi=phi_{0}$, where $phi_{0}in[0,pi/2]$ is the desired angle. We can then easily compute the volume of the sphere below the cone:
begin{align}
V &= int_{0}^{2pi}int_{0}^{phi_{0}}int_{0}^{r}rho^{2}sin{phi} drho dphi dtheta\
&=frac{r^{3}}{3}int_{0}^{2pi}left(-cos{phi}biggrvert^{phi_{0}}_{0}right) dtheta \
&= frac{2pi r^{3}}{3}(-cos{phi_{0}}+1)
end{align}
Now, we want this volume to be half of the volume of the hemisphere, so we should have
$$frac{2pi r^{3}}{3}(-cos{phi_{0}}+1)=frac{pi r^{3}}{3}$$
which yields
$$cos{phi_{0}}=frac{1}{2}$$
so the desired angle is
$$phi_{0}=cos^{-1}left(frac{1}{2}right)=frac{pi}{3}$$ I should mention that this angle $phi$ is measured from the top of the $z$-axis.
answered Jan 23 at 23:20
pwerthpwerth
3,265417
$endgroup$
$begingroup$
I get $phi=pi/3$
$endgroup$
– zooby
Jan 23 at 23:24
$begingroup$
@zooby Updated my answer to include that.
$endgroup$
– pwerth
Jan 23 at 23:27
$begingroup$
good calculus skills. I wonder if there's an intuitive explanation too... like that would make you think, "of course, the angle has to be 60 degrees."
$endgroup$
– zooby
Jan 24 at 5:48
add a comment |
$begingroup$
Note that by symmetry, for the double cone to split the sphere into two pieces of equal volume, each hemisphere must also be split into two pieces of equal volume by each cone. So we can just work with the upper hemisphere. Since a sphere has volume $displaystylefrac{4pi r^{3}}{3}$, the volume a hemisphere is $displaystylefrac{2pi r^{3}}{3}$.
Using standard spherical coordinates $(theta,phi,rho)$, the equation of the sphere is $rho=r$ and the equation of the cone is $phi=phi_{0}$, where $phi_{0}in[0,pi/2]$ is the desired angle. We can then easily compute the volume of the sphere below the cone:
begin{align}
V &= int_{0}^{2pi}int_{0}^{phi_{0}}int_{0}^{r}rho^{2}sin{phi} drho dphi dtheta\
&=frac{r^{3}}{3}int_{0}^{2pi}left(-cos{phi}biggrvert^{phi_{0}}_{0}right) dtheta \
&= frac{2pi r^{3}}{3}(-cos{phi_{0}}+1)
end{align}
Now, we want this volume to be half of the volume of the hemisphere, so we should have
$$frac{2pi r^{3}}{3}(-cos{phi_{0}}+1)=frac{pi r^{3}}{3}$$
which yields
$$cos{phi_{0}}=frac{1}{2}$$
so the desired angle is
$$phi_{0}=cos^{-1}left(frac{1}{2}right)=frac{pi}{3}$$ I should mention that this angle $phi$ is measured from the top of the $z$-axis.
answered Jan 23 at 23:20
pwerthpwerth
3,265417
$endgroup$
$begingroup$
I get $phi=pi/3$
$endgroup$
– zooby
Jan 23 at 23:24
$begingroup$
@zooby Updated my answer to include that.
$endgroup$
– pwerth
Jan 23 at 23:27
$begingroup$
good calculus skills. I wonder if there's an intuitive explanation too... like that would make you think, "of course, the angle has to be 60 degrees."
$endgroup$
– zooby
Jan 24 at 5:48
add a comment |
$begingroup$
Note that by symmetry, for the double cone to split the sphere into two pieces of equal volume, each hemisphere must also be split into two pieces of equal volume by each cone. So we can just work with the upper hemisphere. Since a sphere has volume $displaystylefrac{4pi r^{3}}{3}$, the volume a hemisphere is $displaystylefrac{2pi r^{3}}{3}$.
Using standard spherical coordinates $(theta,phi,rho)$, the equation of the sphere is $rho=r$ and the equation of the cone is $phi=phi_{0}$, where $phi_{0}in[0,pi/2]$ is the desired angle. We can then easily compute the volume of the sphere below the cone:
begin{align}
V &= int_{0}^{2pi}int_{0}^{phi_{0}}int_{0}^{r}rho^{2}sin{phi} drho dphi dtheta\
&=frac{r^{3}}{3}int_{0}^{2pi}left(-cos{phi}biggrvert^{phi_{0}}_{0}right) dtheta \
&= frac{2pi r^{3}}{3}(-cos{phi_{0}}+1)
end{align}
Now, we want this volume to be half of the volume of the hemisphere, so we should have
$$frac{2pi r^{3}}{3}(-cos{phi_{0}}+1)=frac{pi r^{3}}{3}$$
which yields
$$cos{phi_{0}}=frac{1}{2}$$
so the desired angle is
$$phi_{0}=cos^{-1}left(frac{1}{2}right)=frac{pi}{3}$$ I should mention that this angle $phi$ is measured from the top of the $z$-axis.
answered Jan 23 at 23:20
pwerthpwerth
3,265417
$endgroup$
Note that by symmetry, for the double cone to split the sphere into two pieces of equal volume, each hemisphere must also be split into two pieces of equal volume by each cone. So we can just work with the upper hemisphere. Since a sphere has volume $displaystylefrac{4pi r^{3}}{3}$, the volume a hemisphere is $displaystylefrac{2pi r^{3}}{3}$.
Using standard spherical coordinates $(theta,phi,rho)$, the equation of the sphere is $rho=r$ and the equation of the cone is $phi=phi_{0}$, where $phi_{0}in[0,pi/2]$ is the desired angle. We can then easily compute the volume of the sphere below the cone:
begin{align}
V &= int_{0}^{2pi}int_{0}^{phi_{0}}int_{0}^{r}rho^{2}sin{phi} drho dphi dtheta\
&=frac{r^{3}}{3}int_{0}^{2pi}left(-cos{phi}biggrvert^{phi_{0}}_{0}right) dtheta \
&= frac{2pi r^{3}}{3}(-cos{phi_{0}}+1)
end{align}
Now, we want this volume to be half of the volume of the hemisphere, so we should have
$$frac{2pi r^{3}}{3}(-cos{phi_{0}}+1)=frac{pi r^{3}}{3}$$
which yields
$$cos{phi_{0}}=frac{1}{2}$$
so the desired angle is
$$phi_{0}=cos^{-1}left(frac{1}{2}right)=frac{pi}{3}$$ I should mention that this angle $phi$ is measured from the top of the $z$-axis.
answered Jan 23 at 23:20
pwerthpwerth
3,265417
edited Jan 23 at 23:25
answered Jan 23 at 23:20
pwerthpwerth
3,265417
answered Jan 23 at 23:20
pwerthpwerth
3,265417
answered Jan 23 at 23:20
pwerthpwerth
3,265417
3,265417
$begingroup$
I get $phi=pi/3$
$endgroup$
– zooby
Jan 23 at 23:24
$begingroup$
@zooby Updated my answer to include that.
$endgroup$
– pwerth
Jan 23 at 23:27
$begingroup$
good calculus skills. I wonder if there's an intuitive explanation too... like that would make you think, "of course, the angle has to be 60 degrees."
$endgroup$
– zooby
Jan 24 at 5:48
add a comment |
$begingroup$
I get $phi=pi/3$
$endgroup$
– zooby
Jan 23 at 23:24
$begingroup$
@zooby Updated my answer to include that.
$endgroup$
– pwerth
Jan 23 at 23:27
$begingroup$
good calculus skills. I wonder if there's an intuitive explanation too... like that would make you think, "of course, the angle has to be 60 degrees."
$endgroup$
– zooby
Jan 24 at 5:48
$begingroup$
I get $phi=pi/3$
$endgroup$
– zooby
Jan 23 at 23:24
$begingroup$
I get $phi=pi/3$
$endgroup$
– zooby
Jan 23 at 23:24
$begingroup$
@zooby Updated my answer to include that.
$endgroup$
– pwerth
Jan 23 at 23:27
$begingroup$
@zooby Updated my answer to include that.
$endgroup$
– pwerth
Jan 23 at 23:27
$begingroup$
good calculus skills. I wonder if there's an intuitive explanation too... like that would make you think, "of course, the angle has to be 60 degrees."
$endgroup$
– zooby
Jan 24 at 5:48
$begingroup$
good calculus skills. I wonder if there's an intuitive explanation too... like that would make you think, "of course, the angle has to be 60 degrees."
$endgroup$
– zooby
Jan 24 at 5:48
add a comment |
$begingroup$
The volume of the portion of a cone inside a unit sphere equals the solid angle it subtends. You want this angle to be $dfracpi3$ steradians.
From Wolfram, the solid angle corresponding to the half aperture $theta$ is
$$pi(2(1-costheta)+sintheta).$$
You can solve for $theta$.
answered Jan 23 at 23:18
Yves DaoustYves Daoust
130k676229
$endgroup$
add a comment |
$begingroup$
The volume of the portion of a cone inside a unit sphere equals the solid angle it subtends. You want this angle to be $dfracpi3$ steradians.
From Wolfram, the solid angle corresponding to the half aperture $theta$ is
$$pi(2(1-costheta)+sintheta).$$
You can solve for $theta$.
answered Jan 23 at 23:18
Yves DaoustYves Daoust
130k676229
$endgroup$
add a comment |
$begingroup$
The volume of the portion of a cone inside a unit sphere equals the solid angle it subtends. You want this angle to be $dfracpi3$ steradians.
From Wolfram, the solid angle corresponding to the half aperture $theta$ is
$$pi(2(1-costheta)+sintheta).$$
You can solve for $theta$.
answered Jan 23 at 23:18
Yves DaoustYves Daoust
130k676229
$endgroup$
The volume of the portion of a cone inside a unit sphere equals the solid angle it subtends. You want this angle to be $dfracpi3$ steradians.
From Wolfram, the solid angle corresponding to the half aperture $theta$ is
$$pi(2(1-costheta)+sintheta).$$
You can solve for $theta$.
answered Jan 23 at 23:18
Yves DaoustYves Daoust
130k676229
answered Jan 23 at 23:18
Yves DaoustYves Daoust
130k676229
answered Jan 23 at 23:18
Yves DaoustYves Daoust
130k676229
answered Jan 23 at 23:18
Yves DaoustYves Daoust
130k676229
130k676229
add a comment |
add a comment |
$begingroup$
For half volume you want: $phi = pi/3$ so:
answered Jan 23 at 23:29
David G. StorkDavid G. Stork
11.1k41432
$endgroup$
$begingroup$
I upvoted for a nice picture!
$endgroup$
– zooby
Jan 24 at 5:47
add a comment |
$begingroup$
For half volume you want: $phi = pi/3$ so:
answered Jan 23 at 23:29
David G. StorkDavid G. Stork
11.1k41432
$endgroup$
$begingroup$
I upvoted for a nice picture!
$endgroup$
– zooby
Jan 24 at 5:47
add a comment |
$begingroup$
For half volume you want: $phi = pi/3$ so:
answered Jan 23 at 23:29
David G. StorkDavid G. Stork
11.1k41432
$endgroup$
For half volume you want: $phi = pi/3$ so:
answered Jan 23 at 23:29
David G. StorkDavid G. Stork
11.1k41432
answered Jan 23 at 23:29
David G. StorkDavid G. Stork
11.1k41432
answered Jan 23 at 23:29
David G. StorkDavid G. Stork
11.1k41432
answered Jan 23 at 23:29
David G. StorkDavid G. Stork
11.1k41432
11.1k41432
$begingroup$
I upvoted for a nice picture!
$endgroup$
– zooby
Jan 24 at 5:47
add a comment |
$begingroup$
I upvoted for a nice picture!
$endgroup$
– zooby
Jan 24 at 5:47
$begingroup$
I upvoted for a nice picture!
$endgroup$
– zooby
Jan 24 at 5:47
$begingroup$
I upvoted for a nice picture!
$endgroup$
– zooby
Jan 24 at 5:47
add a comment |
$begingroup$
My intuition may be off here but why would the sphere be split into $3$ pieces (as opposed to $2$)?
$endgroup$
– pwerth
Jan 23 at 22:59
$begingroup$
Yes, counting the inside of the double cone as one piece so-to-speak.
$endgroup$
– zooby
Jan 23 at 23:06
$begingroup$
Have you tried setting up the integral and getting an expression that depends upon the angle?
$endgroup$
– John Douma
Jan 23 at 23:13
1
$begingroup$
All these answers for a question that evidences no work on the asker’s part!
$endgroup$
– amd
Jan 23 at 23:38
1
$begingroup$
You can easily look up formulas for the area of a cone and spherical cap. Express them in terms of the radius of the sphere and cone angle and solve for the angle.
$endgroup$
– amd
Jan 23 at 23:40