Mixing
Showing that the surface area of a zone of a sphere that lies between two parallel planes is $2pi Rh$
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Show that the surface area of a zone of a sphere that lies
between two parallel planes is $2pi Rh$,
Where $R$ is the radius of the sphere and $h$ is the distance between the
planes.
If you are wondering what is interesting about this ?
The fact that the surface area depends only on distance between the planes, and not where they cut the sphere.
I am looking to understand a calculus based solution.
calculus integration
edited Feb 7 at 1:03
Matthew Conroy
10.2k32836
asked Oct 13 '14 at 6:05
Fusion2
34418
bumped to the homepage by Community♦ 14 hours ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
add a comment |
up vote
1
down vote
favorite
Show that the surface area of a zone of a sphere that lies
between two parallel planes is $2pi Rh$,
Where $R$ is the radius of the sphere and $h$ is the distance between the
planes.
If you are wondering what is interesting about this ?
The fact that the surface area depends only on distance between the planes, and not where they cut the sphere.
I am looking to understand a calculus based solution.
calculus integration
edited Feb 7 at 1:03
Matthew Conroy
10.2k32836
asked Oct 13 '14 at 6:05
Fusion2
34418
bumped to the homepage by Community♦ 14 hours ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Check mathworld.wolfram.com/Zone.html
– Macavity
Oct 13 '14 at 6:13
i looked at it, its complicated, is there an easy way to do it, may be by using parameters
– Fusion2
Oct 13 '14 at 6:29
2
@KittenButcher The presented solution is using "parameters" and uses some basic calculus and integration. You should instead try to work through the solution and ask a question about a particular part of the presented proof.
– AlexR
Oct 13 '14 at 6:32
You can use calculus to find the surface area of a spherical cap and then subtract.
– Michael Burr
Aug 24 '17 at 11:43
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Show that the surface area of a zone of a sphere that lies
between two parallel planes is $2pi Rh$,
Where $R$ is the radius of the sphere and $h$ is the distance between the
planes.
If you are wondering what is interesting about this ?
The fact that the surface area depends only on distance between the planes, and not where they cut the sphere.
I am looking to understand a calculus based solution.
calculus integration
edited Feb 7 at 1:03
Matthew Conroy
10.2k32836
asked Oct 13 '14 at 6:05
Fusion2
34418
Show that the surface area of a zone of a sphere that lies
between two parallel planes is $2pi Rh$,
Where $R$ is the radius of the sphere and $h$ is the distance between the
planes.
If you are wondering what is interesting about this ?
The fact that the surface area depends only on distance between the planes, and not where they cut the sphere.
I am looking to understand a calculus based solution.
calculus integration
calculus integration
edited Feb 7 at 1:03
Matthew Conroy
10.2k32836
asked Oct 13 '14 at 6:05
Fusion2
34418
edited Feb 7 at 1:03
Matthew Conroy
10.2k32836
asked Oct 13 '14 at 6:05
Fusion2
34418
edited Feb 7 at 1:03
Matthew Conroy
10.2k32836
edited Feb 7 at 1:03
Matthew Conroy
10.2k32836
edited Feb 7 at 1:03
Matthew Conroy
10.2k32836
10.2k32836
asked Oct 13 '14 at 6:05
Fusion2
34418
asked Oct 13 '14 at 6:05
Fusion2
34418
asked Oct 13 '14 at 6:05
Fusion2
34418
34418
bumped to the homepage by Community♦ 14 hours ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
bumped to the homepage by Community♦ 14 hours ago
This question has answers that may be good or bad; the system has marked it active so that they can be reviewed.
Check mathworld.wolfram.com/Zone.html
– Macavity
Oct 13 '14 at 6:13
i looked at it, its complicated, is there an easy way to do it, may be by using parameters
– Fusion2
Oct 13 '14 at 6:29
2
@KittenButcher The presented solution is using "parameters" and uses some basic calculus and integration. You should instead try to work through the solution and ask a question about a particular part of the presented proof.
– AlexR
Oct 13 '14 at 6:32
You can use calculus to find the surface area of a spherical cap and then subtract.
– Michael Burr
Aug 24 '17 at 11:43
add a comment |
Check mathworld.wolfram.com/Zone.html
– Macavity
Oct 13 '14 at 6:13
i looked at it, its complicated, is there an easy way to do it, may be by using parameters
– Fusion2
Oct 13 '14 at 6:29
2
@KittenButcher The presented solution is using "parameters" and uses some basic calculus and integration. You should instead try to work through the solution and ask a question about a particular part of the presented proof.
– AlexR
Oct 13 '14 at 6:32
You can use calculus to find the surface area of a spherical cap and then subtract.
– Michael Burr
Aug 24 '17 at 11:43
Check mathworld.wolfram.com/Zone.html
– Macavity
Oct 13 '14 at 6:13
Check mathworld.wolfram.com/Zone.html
– Macavity
Oct 13 '14 at 6:13
i looked at it, its complicated, is there an easy way to do it, may be by using parameters
– Fusion2
Oct 13 '14 at 6:29
i looked at it, its complicated, is there an easy way to do it, may be by using parameters
– Fusion2
Oct 13 '14 at 6:29
2
2
@KittenButcher The presented solution is using "parameters" and uses some basic calculus and integration. You should instead try to work through the solution and ask a question about a particular part of the presented proof.
– AlexR
Oct 13 '14 at 6:32
@KittenButcher The presented solution is using "parameters" and uses some basic calculus and integration. You should instead try to work through the solution and ask a question about a particular part of the presented proof.
– AlexR
Oct 13 '14 at 6:32
You can use calculus to find the surface area of a spherical cap and then subtract.
– Michael Burr
Aug 24 '17 at 11:43
You can use calculus to find the surface area of a spherical cap and then subtract.
– Michael Burr
Aug 24 '17 at 11:43
add a comment |
2 Answers
2
active
oldest
votes
up vote
0
down vote
A sphere is generated by rotating the right portion of a circle centered at the origin and with radius R about the y-axis, that is, by rotating the curve x=√R^2-y^2 about the y-axis, so the surface area of the required zone would be ∫ (from -c to c) 2π*√R^2-y^2*√R^2/R^2-y^2 dy, now we know that 2c=h( assume that c>0), we got h*∫ 2πR dy=2πRh
answered Aug 24 '17 at 10:03
shark
1
Please use MathJax.
– José Carlos Santos
Aug 24 '17 at 10:24
add a comment |
up vote
0
down vote
Consider an "infinitesimal latitude annulus" $A$ of geographical width $Deltatheta$, positioned at the geographical latitude $thetainleft]-{piover2},{piover2}right[$. Its area is given by
$${rm area}(A)=2pi Rcosthetacdot RDeltatheta .$$
Now a glance at the "infinitesimal right triangle" with hypotenuse $RDeltatheta$ reveals that the increment of the $z$-coordinate across $A$ amounts to $Delta z=costheta> RDeltatheta$. It follows that in fact
$$ {rm area}(A)=2pi R>Delta z .$$
You can put it this way: The projection of $S^2$ along horizontal rays onto the cylinder touching $S^2$ along the equator is area-preserving.
answered Aug 24 '17 at 11:39
Christian Blatter
170k7111324
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
A sphere is generated by rotating the right portion of a circle centered at the origin and with radius R about the y-axis, that is, by rotating the curve x=√R^2-y^2 about the y-axis, so the surface area of the required zone would be ∫ (from -c to c) 2π*√R^2-y^2*√R^2/R^2-y^2 dy, now we know that 2c=h( assume that c>0), we got h*∫ 2πR dy=2πRh
answered Aug 24 '17 at 10:03
shark
1
Please use MathJax.
– José Carlos Santos
Aug 24 '17 at 10:24
add a comment |
up vote
0
down vote
A sphere is generated by rotating the right portion of a circle centered at the origin and with radius R about the y-axis, that is, by rotating the curve x=√R^2-y^2 about the y-axis, so the surface area of the required zone would be ∫ (from -c to c) 2π*√R^2-y^2*√R^2/R^2-y^2 dy, now we know that 2c=h( assume that c>0), we got h*∫ 2πR dy=2πRh
answered Aug 24 '17 at 10:03
shark
1
Please use MathJax.
– José Carlos Santos
Aug 24 '17 at 10:24
add a comment |
up vote
0
down vote
up vote
0
down vote
A sphere is generated by rotating the right portion of a circle centered at the origin and with radius R about the y-axis, that is, by rotating the curve x=√R^2-y^2 about the y-axis, so the surface area of the required zone would be ∫ (from -c to c) 2π*√R^2-y^2*√R^2/R^2-y^2 dy, now we know that 2c=h( assume that c>0), we got h*∫ 2πR dy=2πRh
answered Aug 24 '17 at 10:03
shark
1
A sphere is generated by rotating the right portion of a circle centered at the origin and with radius R about the y-axis, that is, by rotating the curve x=√R^2-y^2 about the y-axis, so the surface area of the required zone would be ∫ (from -c to c) 2π*√R^2-y^2*√R^2/R^2-y^2 dy, now we know that 2c=h( assume that c>0), we got h*∫ 2πR dy=2πRh
answered Aug 24 '17 at 10:03
shark
1
answered Aug 24 '17 at 10:03
shark
1
answered Aug 24 '17 at 10:03
shark
1
answered Aug 24 '17 at 10:03
shark
1
1
Please use MathJax.
– José Carlos Santos
Aug 24 '17 at 10:24
add a comment |
Please use MathJax.
– José Carlos Santos
Aug 24 '17 at 10:24
Please use MathJax.
– José Carlos Santos
Aug 24 '17 at 10:24
Please use MathJax.
– José Carlos Santos
Aug 24 '17 at 10:24
add a comment |
up vote
0
down vote
Consider an "infinitesimal latitude annulus" $A$ of geographical width $Deltatheta$, positioned at the geographical latitude $thetainleft]-{piover2},{piover2}right[$. Its area is given by
$${rm area}(A)=2pi Rcosthetacdot RDeltatheta .$$
Now a glance at the "infinitesimal right triangle" with hypotenuse $RDeltatheta$ reveals that the increment of the $z$-coordinate across $A$ amounts to $Delta z=costheta> RDeltatheta$. It follows that in fact
$$ {rm area}(A)=2pi R>Delta z .$$
You can put it this way: The projection of $S^2$ along horizontal rays onto the cylinder touching $S^2$ along the equator is area-preserving.
answered Aug 24 '17 at 11:39
Christian Blatter
170k7111324
add a comment |
up vote
0
down vote
Consider an "infinitesimal latitude annulus" $A$ of geographical width $Deltatheta$, positioned at the geographical latitude $thetainleft]-{piover2},{piover2}right[$. Its area is given by
$${rm area}(A)=2pi Rcosthetacdot RDeltatheta .$$
Now a glance at the "infinitesimal right triangle" with hypotenuse $RDeltatheta$ reveals that the increment of the $z$-coordinate across $A$ amounts to $Delta z=costheta> RDeltatheta$. It follows that in fact
$$ {rm area}(A)=2pi R>Delta z .$$
You can put it this way: The projection of $S^2$ along horizontal rays onto the cylinder touching $S^2$ along the equator is area-preserving.
answered Aug 24 '17 at 11:39
Christian Blatter
170k7111324
add a comment |
up vote
0
down vote
up vote
0
down vote
Consider an "infinitesimal latitude annulus" $A$ of geographical width $Deltatheta$, positioned at the geographical latitude $thetainleft]-{piover2},{piover2}right[$. Its area is given by
$${rm area}(A)=2pi Rcosthetacdot RDeltatheta .$$
Now a glance at the "infinitesimal right triangle" with hypotenuse $RDeltatheta$ reveals that the increment of the $z$-coordinate across $A$ amounts to $Delta z=costheta> RDeltatheta$. It follows that in fact
$$ {rm area}(A)=2pi R>Delta z .$$
You can put it this way: The projection of $S^2$ along horizontal rays onto the cylinder touching $S^2$ along the equator is area-preserving.
answered Aug 24 '17 at 11:39
Christian Blatter
170k7111324
Consider an "infinitesimal latitude annulus" $A$ of geographical width $Deltatheta$, positioned at the geographical latitude $thetainleft]-{piover2},{piover2}right[$. Its area is given by
$${rm area}(A)=2pi Rcosthetacdot RDeltatheta .$$
Now a glance at the "infinitesimal right triangle" with hypotenuse $RDeltatheta$ reveals that the increment of the $z$-coordinate across $A$ amounts to $Delta z=costheta> RDeltatheta$. It follows that in fact
$$ {rm area}(A)=2pi R>Delta z .$$
You can put it this way: The projection of $S^2$ along horizontal rays onto the cylinder touching $S^2$ along the equator is area-preserving.
answered Aug 24 '17 at 11:39
Christian Blatter
170k7111324
answered Aug 24 '17 at 11:39
Christian Blatter
170k7111324
answered Aug 24 '17 at 11:39
Christian Blatter
170k7111324
answered Aug 24 '17 at 11:39
Christian Blatter
170k7111324
170k7111324
add a comment |
add a comment |
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Check mathworld.wolfram.com/Zone.html
– Macavity
Oct 13 '14 at 6:13
i looked at it, its complicated, is there an easy way to do it, may be by using parameters
– Fusion2
Oct 13 '14 at 6:29
2
@KittenButcher The presented solution is using "parameters" and uses some basic calculus and integration. You should instead try to work through the solution and ask a question about a particular part of the presented proof.
– AlexR
Oct 13 '14 at 6:32
You can use calculus to find the surface area of a spherical cap and then subtract.
– Michael Burr
Aug 24 '17 at 11:43