Integral of Gaussian curvature over S
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Let f: $Bbb R^2 rightarrow Bbb R$ be a smooth function such that $f(x, y) = 0$ for all $(x, y)$ outside the unit disk, i.e., for all $(x, y)$ with $x^2 + y^2 geqq 1.$ Consider the surface $S$ in $Bbb R^3$ given by the graph of $f$ over the disk $x^2 + y^2 leqq 2.$ What can you say about the integral of the Gaussian curvature over S? Prove.
I assume Gauss-Bonnet is what I'd need to use here but I'm having trouble getting there with the information given - meaning I'm unsure of how to derive the values necessary for Gauss-Bonnet.
differential-geometry
asked Dec 4 at 19:45
jman63
33
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Let f: $Bbb R^2 rightarrow Bbb R$ be a smooth function such that $f(x, y) = 0$ for all $(x, y)$ outside the unit disk, i.e., for all $(x, y)$ with $x^2 + y^2 geqq 1.$ Consider the surface $S$ in $Bbb R^3$ given by the graph of $f$ over the disk $x^2 + y^2 leqq 2.$ What can you say about the integral of the Gaussian curvature over S? Prove.
I assume Gauss-Bonnet is what I'd need to use here but I'm having trouble getting there with the information given - meaning I'm unsure of how to derive the values necessary for Gauss-Bonnet.
differential-geometry
asked Dec 4 at 19:45
jman63
33
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let f: $Bbb R^2 rightarrow Bbb R$ be a smooth function such that $f(x, y) = 0$ for all $(x, y)$ outside the unit disk, i.e., for all $(x, y)$ with $x^2 + y^2 geqq 1.$ Consider the surface $S$ in $Bbb R^3$ given by the graph of $f$ over the disk $x^2 + y^2 leqq 2.$ What can you say about the integral of the Gaussian curvature over S? Prove.
I assume Gauss-Bonnet is what I'd need to use here but I'm having trouble getting there with the information given - meaning I'm unsure of how to derive the values necessary for Gauss-Bonnet.
differential-geometry
asked Dec 4 at 19:45
jman63
33
Let f: $Bbb R^2 rightarrow Bbb R$ be a smooth function such that $f(x, y) = 0$ for all $(x, y)$ outside the unit disk, i.e., for all $(x, y)$ with $x^2 + y^2 geqq 1.$ Consider the surface $S$ in $Bbb R^3$ given by the graph of $f$ over the disk $x^2 + y^2 leqq 2.$ What can you say about the integral of the Gaussian curvature over S? Prove.
I assume Gauss-Bonnet is what I'd need to use here but I'm having trouble getting there with the information given - meaning I'm unsure of how to derive the values necessary for Gauss-Bonnet.
differential-geometry
differential-geometry
asked Dec 4 at 19:45
jman63
33
asked Dec 4 at 19:45
jman63
33
asked Dec 4 at 19:45
jman63
33
asked Dec 4 at 19:45
jman63
33
asked Dec 4 at 19:45
jman63
33
33
add a comment |
add a comment |
1 Answer
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By the local Gauss-Bonnet theorem,
$$ int_{S} K , mathrm dA = 2pi -int_gamma k_g , mathrm ds, $$
where $gamma$ is the positively oriented circle $x^2+y^2=2$. Since $f(x,y)=0$ for $x^2+y^2geq 1$, it follows that $S$ is planar everywhere except in the unit circle. In particular, the geodesic curvature of $gamma$ can be easily computed by taking e.g. $(0,0,1)$ as the unit normal along $gamma$, and parametrizing $gamma$ accordingly.
Can you proceed?
answered Dec 4 at 19:52
MisterRiemann
5,7031624
I suppose you'd parametrize with cos, sin, and then just compute the integral once you've computed geodesic curvature?
– jman63
Dec 7 at 0:52
Yes! You could even compute the integral without parametrizing, but I usually mess up the signs if I do that.
– MisterRiemann
Dec 7 at 7:20
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
By the local Gauss-Bonnet theorem,
$$ int_{S} K , mathrm dA = 2pi -int_gamma k_g , mathrm ds, $$
where $gamma$ is the positively oriented circle $x^2+y^2=2$. Since $f(x,y)=0$ for $x^2+y^2geq 1$, it follows that $S$ is planar everywhere except in the unit circle. In particular, the geodesic curvature of $gamma$ can be easily computed by taking e.g. $(0,0,1)$ as the unit normal along $gamma$, and parametrizing $gamma$ accordingly.
Can you proceed?
answered Dec 4 at 19:52
MisterRiemann
5,7031624
I suppose you'd parametrize with cos, sin, and then just compute the integral once you've computed geodesic curvature?
– jman63
Dec 7 at 0:52
Yes! You could even compute the integral without parametrizing, but I usually mess up the signs if I do that.
– MisterRiemann
Dec 7 at 7:20
add a comment |
up vote
0
down vote
accepted
By the local Gauss-Bonnet theorem,
$$ int_{S} K , mathrm dA = 2pi -int_gamma k_g , mathrm ds, $$
where $gamma$ is the positively oriented circle $x^2+y^2=2$. Since $f(x,y)=0$ for $x^2+y^2geq 1$, it follows that $S$ is planar everywhere except in the unit circle. In particular, the geodesic curvature of $gamma$ can be easily computed by taking e.g. $(0,0,1)$ as the unit normal along $gamma$, and parametrizing $gamma$ accordingly.
Can you proceed?
answered Dec 4 at 19:52
MisterRiemann
5,7031624
I suppose you'd parametrize with cos, sin, and then just compute the integral once you've computed geodesic curvature?
– jman63
Dec 7 at 0:52
Yes! You could even compute the integral without parametrizing, but I usually mess up the signs if I do that.
– MisterRiemann
Dec 7 at 7:20
add a comment |
up vote
0
down vote
accepted
up vote
0
down vote
accepted
By the local Gauss-Bonnet theorem,
$$ int_{S} K , mathrm dA = 2pi -int_gamma k_g , mathrm ds, $$
where $gamma$ is the positively oriented circle $x^2+y^2=2$. Since $f(x,y)=0$ for $x^2+y^2geq 1$, it follows that $S$ is planar everywhere except in the unit circle. In particular, the geodesic curvature of $gamma$ can be easily computed by taking e.g. $(0,0,1)$ as the unit normal along $gamma$, and parametrizing $gamma$ accordingly.
Can you proceed?
answered Dec 4 at 19:52
MisterRiemann
5,7031624
By the local Gauss-Bonnet theorem,
$$ int_{S} K , mathrm dA = 2pi -int_gamma k_g , mathrm ds, $$
where $gamma$ is the positively oriented circle $x^2+y^2=2$. Since $f(x,y)=0$ for $x^2+y^2geq 1$, it follows that $S$ is planar everywhere except in the unit circle. In particular, the geodesic curvature of $gamma$ can be easily computed by taking e.g. $(0,0,1)$ as the unit normal along $gamma$, and parametrizing $gamma$ accordingly.
Can you proceed?
answered Dec 4 at 19:52
MisterRiemann
5,7031624
edited Dec 4 at 19:57
answered Dec 4 at 19:52
MisterRiemann
5,7031624
answered Dec 4 at 19:52
MisterRiemann
5,7031624
answered Dec 4 at 19:52
MisterRiemann
5,7031624
5,7031624
I suppose you'd parametrize with cos, sin, and then just compute the integral once you've computed geodesic curvature?
– jman63
Dec 7 at 0:52
Yes! You could even compute the integral without parametrizing, but I usually mess up the signs if I do that.
– MisterRiemann
Dec 7 at 7:20
add a comment |
I suppose you'd parametrize with cos, sin, and then just compute the integral once you've computed geodesic curvature?
– jman63
Dec 7 at 0:52
Yes! You could even compute the integral without parametrizing, but I usually mess up the signs if I do that.
– MisterRiemann
Dec 7 at 7:20
I suppose you'd parametrize with cos, sin, and then just compute the integral once you've computed geodesic curvature?
– jman63
Dec 7 at 0:52
I suppose you'd parametrize with cos, sin, and then just compute the integral once you've computed geodesic curvature?
– jman63
Dec 7 at 0:52
Yes! You could even compute the integral without parametrizing, but I usually mess up the signs if I do that.
– MisterRiemann
Dec 7 at 7:20
Yes! You could even compute the integral without parametrizing, but I usually mess up the signs if I do that.
– MisterRiemann
Dec 7 at 7:20
add a comment |
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