How to prove that $text{area}( S_r(x) cap B_R(0)) leq text{area}(S_R(0))$?
$begingroup$
Consider the following surface in $mathbb{R^3}$:
$$Sigma = S_r(x) cap B_R(0)$$
Where $S_r(x)$ is a sphere of radius $r$ centered at $x$ and $B_R(0)$ is an open ball of radius $R$ centered at the origin.
I'd like to show that:
$$text{area}( S_r(x) cap B_R(0)) leq text{area}(S_R(0))$$
I tried to evaluate the following surface integral:
$$int_{Sigma} dS$$
But it's not clear to me how the surface $Sigma$ can be parameterized.
I assumed that $x = (p,0,0)$ and then use spherical coordinates. The expression I got is not very helpful.
Is there another easier way to prove that inequality?
calculus measure-theory multivariable-calculus multiple-integral
asked Dec 3 '18 at 12:45
SantosSantos
621514
$endgroup$
add a comment |
$begingroup$
Consider the following surface in $mathbb{R^3}$:
$$Sigma = S_r(x) cap B_R(0)$$
Where $S_r(x)$ is a sphere of radius $r$ centered at $x$ and $B_R(0)$ is an open ball of radius $R$ centered at the origin.
I'd like to show that:
$$text{area}( S_r(x) cap B_R(0)) leq text{area}(S_R(0))$$
I tried to evaluate the following surface integral:
$$int_{Sigma} dS$$
But it's not clear to me how the surface $Sigma$ can be parameterized.
I assumed that $x = (p,0,0)$ and then use spherical coordinates. The expression I got is not very helpful.
Is there another easier way to prove that inequality?
calculus measure-theory multivariable-calculus multiple-integral
asked Dec 3 '18 at 12:45
SantosSantos
621514
$endgroup$
1
$begingroup$
Are you imagining a relationship between $r$ and $R$? It seems to me this makes most sense as a question when $r$ is much bigger than $R$. (in particular, if $S_r(x)$ is contained in $B_R$ it is trivial)
$endgroup$
– T_M
Dec 3 '18 at 20:21
add a comment |
$begingroup$
Consider the following surface in $mathbb{R^3}$:
$$Sigma = S_r(x) cap B_R(0)$$
Where $S_r(x)$ is a sphere of radius $r$ centered at $x$ and $B_R(0)$ is an open ball of radius $R$ centered at the origin.
I'd like to show that:
$$text{area}( S_r(x) cap B_R(0)) leq text{area}(S_R(0))$$
I tried to evaluate the following surface integral:
$$int_{Sigma} dS$$
But it's not clear to me how the surface $Sigma$ can be parameterized.
I assumed that $x = (p,0,0)$ and then use spherical coordinates. The expression I got is not very helpful.
Is there another easier way to prove that inequality?
calculus measure-theory multivariable-calculus multiple-integral
asked Dec 3 '18 at 12:45
SantosSantos
621514
$endgroup$
Consider the following surface in $mathbb{R^3}$:
$$Sigma = S_r(x) cap B_R(0)$$
Where $S_r(x)$ is a sphere of radius $r$ centered at $x$ and $B_R(0)$ is an open ball of radius $R$ centered at the origin.
I'd like to show that:
$$text{area}( S_r(x) cap B_R(0)) leq text{area}(S_R(0))$$
I tried to evaluate the following surface integral:
$$int_{Sigma} dS$$
But it's not clear to me how the surface $Sigma$ can be parameterized.
I assumed that $x = (p,0,0)$ and then use spherical coordinates. The expression I got is not very helpful.
Is there another easier way to prove that inequality?
calculus measure-theory multivariable-calculus multiple-integral
calculus measure-theory multivariable-calculus multiple-integral
asked Dec 3 '18 at 12:45
SantosSantos
621514
asked Dec 3 '18 at 12:45
SantosSantos
621514
edited Dec 3 '18 at 18:40
Santos
asked Dec 3 '18 at 12:45
SantosSantos
621514
asked Dec 3 '18 at 12:45
SantosSantos
621514
asked Dec 3 '18 at 12:45
SantosSantos
621514
621514
1
$begingroup$
Are you imagining a relationship between $r$ and $R$? It seems to me this makes most sense as a question when $r$ is much bigger than $R$. (in particular, if $S_r(x)$ is contained in $B_R$ it is trivial)
$endgroup$
– T_M
Dec 3 '18 at 20:21
add a comment |
1
$begingroup$
Are you imagining a relationship between $r$ and $R$? It seems to me this makes most sense as a question when $r$ is much bigger than $R$. (in particular, if $S_r(x)$ is contained in $B_R$ it is trivial)
$endgroup$
– T_M
Dec 3 '18 at 20:21
1
1
$begingroup$
Are you imagining a relationship between $r$ and $R$? It seems to me this makes most sense as a question when $r$ is much bigger than $R$. (in particular, if $S_r(x)$ is contained in $B_R$ it is trivial)
$endgroup$
– T_M
Dec 3 '18 at 20:21
$begingroup$
Are you imagining a relationship between $r$ and $R$? It seems to me this makes most sense as a question when $r$ is much bigger than $R$. (in particular, if $S_r(x)$ is contained in $B_R$ it is trivial)
$endgroup$
– T_M
Dec 3 '18 at 20:21
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Here's the 2D version which you should be able to extend to 3D without too much difficulty. As mentioned in the comments you should assume $r>R$ because the other case is "trivial" (i.e., if $r<R$ then you can translate $S_r(x)$ into the interior of $B_R(0)$).
You can reorient the problem so that $S_r$ is centered at $(0,0)$:
To calculate the length of the arc inside $B_R$ we parametrize as:
$$
mathbf{r}(t) = langle r cos t, rsin trangle
$$
where $|t| leq alpha = arcsin(frac{R'}{r})$, where $R'$ is half the chord length (the dashed line). So
$$
text{length}= int_{-alpha}^{alpha} |mathbf{r}'(t)| ; dt = 2rarcsin(frac{R'}{r}) leq 2r arcsin(frac{R}{r})
$$
since $R'leq R$ and $arcsin$ is increasing. Lastly you can use convexity of $arcsin$ on $[0,1]$ to show that
$$
arcsin(frac{R}{r})leq frac{pi}{2}frac{R}{r}
$$
(convexity implies that $arcsin$ lies beneath the line segment joining $(0,0)$ to $(1,pi/2)$). So we end up with
$$
text{length} leq pi R leq 2pi R
$$
(twice as good of an inequality as we needed).
This should work in 3D as well; reorient so that $B_R$ is vertically above the origin, parametrize with spherical coordinates, use trigonometry as above to find bounds for the polar angle. One analogous step you should take is to bound the radius of the spherical cap the way we bounded the chord above.
answered Dec 4 '18 at 1:40
user25959user25959
1,573816
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
Here's the 2D version which you should be able to extend to 3D without too much difficulty. As mentioned in the comments you should assume $r>R$ because the other case is "trivial" (i.e., if $r<R$ then you can translate $S_r(x)$ into the interior of $B_R(0)$).
You can reorient the problem so that $S_r$ is centered at $(0,0)$:
To calculate the length of the arc inside $B_R$ we parametrize as:
$$
mathbf{r}(t) = langle r cos t, rsin trangle
$$
where $|t| leq alpha = arcsin(frac{R'}{r})$, where $R'$ is half the chord length (the dashed line). So
$$
text{length}= int_{-alpha}^{alpha} |mathbf{r}'(t)| ; dt = 2rarcsin(frac{R'}{r}) leq 2r arcsin(frac{R}{r})
$$
since $R'leq R$ and $arcsin$ is increasing. Lastly you can use convexity of $arcsin$ on $[0,1]$ to show that
$$
arcsin(frac{R}{r})leq frac{pi}{2}frac{R}{r}
$$
(convexity implies that $arcsin$ lies beneath the line segment joining $(0,0)$ to $(1,pi/2)$). So we end up with
$$
text{length} leq pi R leq 2pi R
$$
(twice as good of an inequality as we needed).
This should work in 3D as well; reorient so that $B_R$ is vertically above the origin, parametrize with spherical coordinates, use trigonometry as above to find bounds for the polar angle. One analogous step you should take is to bound the radius of the spherical cap the way we bounded the chord above.
answered Dec 4 '18 at 1:40
user25959user25959
1,573816
$endgroup$
add a comment |
$begingroup$
Here's the 2D version which you should be able to extend to 3D without too much difficulty. As mentioned in the comments you should assume $r>R$ because the other case is "trivial" (i.e., if $r<R$ then you can translate $S_r(x)$ into the interior of $B_R(0)$).
You can reorient the problem so that $S_r$ is centered at $(0,0)$:
To calculate the length of the arc inside $B_R$ we parametrize as:
$$
mathbf{r}(t) = langle r cos t, rsin trangle
$$
where $|t| leq alpha = arcsin(frac{R'}{r})$, where $R'$ is half the chord length (the dashed line). So
$$
text{length}= int_{-alpha}^{alpha} |mathbf{r}'(t)| ; dt = 2rarcsin(frac{R'}{r}) leq 2r arcsin(frac{R}{r})
$$
since $R'leq R$ and $arcsin$ is increasing. Lastly you can use convexity of $arcsin$ on $[0,1]$ to show that
$$
arcsin(frac{R}{r})leq frac{pi}{2}frac{R}{r}
$$
(convexity implies that $arcsin$ lies beneath the line segment joining $(0,0)$ to $(1,pi/2)$). So we end up with
$$
text{length} leq pi R leq 2pi R
$$
(twice as good of an inequality as we needed).
This should work in 3D as well; reorient so that $B_R$ is vertically above the origin, parametrize with spherical coordinates, use trigonometry as above to find bounds for the polar angle. One analogous step you should take is to bound the radius of the spherical cap the way we bounded the chord above.
answered Dec 4 '18 at 1:40
user25959user25959
1,573816
$endgroup$
add a comment |
$begingroup$
Here's the 2D version which you should be able to extend to 3D without too much difficulty. As mentioned in the comments you should assume $r>R$ because the other case is "trivial" (i.e., if $r<R$ then you can translate $S_r(x)$ into the interior of $B_R(0)$).
You can reorient the problem so that $S_r$ is centered at $(0,0)$:
To calculate the length of the arc inside $B_R$ we parametrize as:
$$
mathbf{r}(t) = langle r cos t, rsin trangle
$$
where $|t| leq alpha = arcsin(frac{R'}{r})$, where $R'$ is half the chord length (the dashed line). So
$$
text{length}= int_{-alpha}^{alpha} |mathbf{r}'(t)| ; dt = 2rarcsin(frac{R'}{r}) leq 2r arcsin(frac{R}{r})
$$
since $R'leq R$ and $arcsin$ is increasing. Lastly you can use convexity of $arcsin$ on $[0,1]$ to show that
$$
arcsin(frac{R}{r})leq frac{pi}{2}frac{R}{r}
$$
(convexity implies that $arcsin$ lies beneath the line segment joining $(0,0)$ to $(1,pi/2)$). So we end up with
$$
text{length} leq pi R leq 2pi R
$$
(twice as good of an inequality as we needed).
This should work in 3D as well; reorient so that $B_R$ is vertically above the origin, parametrize with spherical coordinates, use trigonometry as above to find bounds for the polar angle. One analogous step you should take is to bound the radius of the spherical cap the way we bounded the chord above.
answered Dec 4 '18 at 1:40
user25959user25959
1,573816
$endgroup$
Here's the 2D version which you should be able to extend to 3D without too much difficulty. As mentioned in the comments you should assume $r>R$ because the other case is "trivial" (i.e., if $r<R$ then you can translate $S_r(x)$ into the interior of $B_R(0)$).
You can reorient the problem so that $S_r$ is centered at $(0,0)$:
To calculate the length of the arc inside $B_R$ we parametrize as:
$$
mathbf{r}(t) = langle r cos t, rsin trangle
$$
where $|t| leq alpha = arcsin(frac{R'}{r})$, where $R'$ is half the chord length (the dashed line). So
$$
text{length}= int_{-alpha}^{alpha} |mathbf{r}'(t)| ; dt = 2rarcsin(frac{R'}{r}) leq 2r arcsin(frac{R}{r})
$$
since $R'leq R$ and $arcsin$ is increasing. Lastly you can use convexity of $arcsin$ on $[0,1]$ to show that
$$
arcsin(frac{R}{r})leq frac{pi}{2}frac{R}{r}
$$
(convexity implies that $arcsin$ lies beneath the line segment joining $(0,0)$ to $(1,pi/2)$). So we end up with
$$
text{length} leq pi R leq 2pi R
$$
(twice as good of an inequality as we needed).
This should work in 3D as well; reorient so that $B_R$ is vertically above the origin, parametrize with spherical coordinates, use trigonometry as above to find bounds for the polar angle. One analogous step you should take is to bound the radius of the spherical cap the way we bounded the chord above.
answered Dec 4 '18 at 1:40
user25959user25959
1,573816
edited Dec 4 '18 at 1:51
answered Dec 4 '18 at 1:40
user25959user25959
1,573816
answered Dec 4 '18 at 1:40
user25959user25959
1,573816
answered Dec 4 '18 at 1:40
user25959user25959
1,573816
1,573816
add a comment |
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$begingroup$
Are you imagining a relationship between $r$ and $R$? It seems to me this makes most sense as a question when $r$ is much bigger than $R$. (in particular, if $S_r(x)$ is contained in $B_R$ it is trivial)
$endgroup$
– T_M
Dec 3 '18 at 20:21