Why doesn't integrating over a sphere with $phi$ between $0$ and $2pi$ and $theta$ between $0$ and $pi$ work?
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One way of integrating over a sphere with $p = 1$ is by integrating over $p$ from $0$ to $1$, $phi$ from $0$ to $pi$, and $theta$ from $0$ to $2 pi$
Here is a link that can graph parametric surfaces in spherical coordinates: http://www.math.uri.edu/~bkaskosz/flashmo/sphplot/
If you enter in what is above, then you will find that the graph is indeed, a sphere. Furthermore, if you integrate $p^2sin (phi)$ over these ranges, you will get $frac {4pi}{3}$ which is the expected value.
I noticed that $p$ from $0$ to $1$, $phi$ from $0$ to $2pi$, and $theta$ from $0$ to $pi$ also seems to create a sphere.
Note the subtle change: $phi$ is from $0$ to $2pi$ and $theta$ is from $0$ to $1pi$. If you plug this in to the grapher, you find that what you get resembles a sphere.
However, when you integrate $p^2sin (phi)$ over $p$ from $0$ to $1$, $phi$ from $0$ to $2pi$, and $theta$ from $0$ to $pi$, you get $0$. Why is that?
integration multivariable-calculus spherical-coordinates multiple-integral
edited Nov 23 at 9:01
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up vote
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One way of integrating over a sphere with $p = 1$ is by integrating over $p$ from $0$ to $1$, $phi$ from $0$ to $pi$, and $theta$ from $0$ to $2 pi$
Here is a link that can graph parametric surfaces in spherical coordinates: http://www.math.uri.edu/~bkaskosz/flashmo/sphplot/
If you enter in what is above, then you will find that the graph is indeed, a sphere. Furthermore, if you integrate $p^2sin (phi)$ over these ranges, you will get $frac {4pi}{3}$ which is the expected value.
I noticed that $p$ from $0$ to $1$, $phi$ from $0$ to $2pi$, and $theta$ from $0$ to $pi$ also seems to create a sphere.
Note the subtle change: $phi$ is from $0$ to $2pi$ and $theta$ is from $0$ to $1pi$. If you plug this in to the grapher, you find that what you get resembles a sphere.
However, when you integrate $p^2sin (phi)$ over $p$ from $0$ to $1$, $phi$ from $0$ to $2pi$, and $theta$ from $0$ to $pi$, you get $0$. Why is that?
integration multivariable-calculus spherical-coordinates multiple-integral
edited Nov 23 at 9:01
idea
2,02531024
asked Nov 23 at 7:29
Typical Highschooler
1117
add a comment |
up vote
2
down vote
favorite
up vote
2
down vote
favorite
One way of integrating over a sphere with $p = 1$ is by integrating over $p$ from $0$ to $1$, $phi$ from $0$ to $pi$, and $theta$ from $0$ to $2 pi$
Here is a link that can graph parametric surfaces in spherical coordinates: http://www.math.uri.edu/~bkaskosz/flashmo/sphplot/
If you enter in what is above, then you will find that the graph is indeed, a sphere. Furthermore, if you integrate $p^2sin (phi)$ over these ranges, you will get $frac {4pi}{3}$ which is the expected value.
I noticed that $p$ from $0$ to $1$, $phi$ from $0$ to $2pi$, and $theta$ from $0$ to $pi$ also seems to create a sphere.
Note the subtle change: $phi$ is from $0$ to $2pi$ and $theta$ is from $0$ to $1pi$. If you plug this in to the grapher, you find that what you get resembles a sphere.
However, when you integrate $p^2sin (phi)$ over $p$ from $0$ to $1$, $phi$ from $0$ to $2pi$, and $theta$ from $0$ to $pi$, you get $0$. Why is that?
integration multivariable-calculus spherical-coordinates multiple-integral
edited Nov 23 at 9:01
idea
2,02531024
asked Nov 23 at 7:29
Typical Highschooler
1117
One way of integrating over a sphere with $p = 1$ is by integrating over $p$ from $0$ to $1$, $phi$ from $0$ to $pi$, and $theta$ from $0$ to $2 pi$
Here is a link that can graph parametric surfaces in spherical coordinates: http://www.math.uri.edu/~bkaskosz/flashmo/sphplot/
If you enter in what is above, then you will find that the graph is indeed, a sphere. Furthermore, if you integrate $p^2sin (phi)$ over these ranges, you will get $frac {4pi}{3}$ which is the expected value.
I noticed that $p$ from $0$ to $1$, $phi$ from $0$ to $2pi$, and $theta$ from $0$ to $pi$ also seems to create a sphere.
Note the subtle change: $phi$ is from $0$ to $2pi$ and $theta$ is from $0$ to $1pi$. If you plug this in to the grapher, you find that what you get resembles a sphere.
However, when you integrate $p^2sin (phi)$ over $p$ from $0$ to $1$, $phi$ from $0$ to $2pi$, and $theta$ from $0$ to $pi$, you get $0$. Why is that?
integration multivariable-calculus spherical-coordinates multiple-integral
integration multivariable-calculus spherical-coordinates multiple-integral
edited Nov 23 at 9:01
idea
2,02531024
asked Nov 23 at 7:29
Typical Highschooler
1117
edited Nov 23 at 9:01
idea
2,02531024
asked Nov 23 at 7:29
Typical Highschooler
1117
edited Nov 23 at 9:01
idea
2,02531024
edited Nov 23 at 9:01
idea
2,02531024
edited Nov 23 at 9:01
idea
2,02531024
2,02531024
asked Nov 23 at 7:29
Typical Highschooler
1117
asked Nov 23 at 7:29
Typical Highschooler
1117
asked Nov 23 at 7:29
Typical Highschooler
1117
1117
add a comment |
add a comment |
2 Answers
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oldest
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up vote
1
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When you change to spherical coordinates (with the convention that you seem to be using), the absolute value of the Jacobian determinant is $|rho^2 sin phi|$, and this does only simplify to $rho^2 sinphi$ if $sinphi ge 0$, which is the case in the interval $0 le phi le pi$, but not in the interval $pi < phi < 2 pi$. But if you instead integrate $|rho^2 sin phi|$ over the ranges that you suggest, then you indeed get $4pi/3$.
answered Nov 23 at 8:36
Hans Lundmark
34.8k564111
add a comment |
up vote
1
down vote
The key fact is that the sphere $$x^2+y^2+z^2=R^2$$ is parametrized by
- $x=Rcos theta sin phi$
- $y=Rsin theta sin phi$
- $z=Rcos phi$
with
- $theta in [0,2pi)$
- $phi in[0,pi]$
and at each point $P(x,y,z)$ on the sphere corresponds an unique pair for the parameters $(theta, phi)$.
Of course we can also use different range for the parametrization as for example
- $theta in [-pi,pi)$
- $phi in[0,pi]$
(credits http://mathworld.wolfram.com/SphericalCoordinates.html)
answered Nov 23 at 7:34
gimusi
92.3k84495
I don't think this is the issue here. As the OP has noted, the unconventional ranges $theta in [0,pi)$ and $phi in [0,2 pi)$ also give a sphere.
– Hans Lundmark
Nov 23 at 8:38
@HansLundmark Yes of course, maybe I misunderstood the doubt ofthe OP. I agree we can introduce (infinitely) many different parametrization for the sphere by suitable choices of the range for $theta$ and $phi$.
– gimusi
Nov 23 at 8:45
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
up vote
1
down vote
accepted
When you change to spherical coordinates (with the convention that you seem to be using), the absolute value of the Jacobian determinant is $|rho^2 sin phi|$, and this does only simplify to $rho^2 sinphi$ if $sinphi ge 0$, which is the case in the interval $0 le phi le pi$, but not in the interval $pi < phi < 2 pi$. But if you instead integrate $|rho^2 sin phi|$ over the ranges that you suggest, then you indeed get $4pi/3$.
answered Nov 23 at 8:36
Hans Lundmark
34.8k564111
add a comment |
up vote
1
down vote
accepted
When you change to spherical coordinates (with the convention that you seem to be using), the absolute value of the Jacobian determinant is $|rho^2 sin phi|$, and this does only simplify to $rho^2 sinphi$ if $sinphi ge 0$, which is the case in the interval $0 le phi le pi$, but not in the interval $pi < phi < 2 pi$. But if you instead integrate $|rho^2 sin phi|$ over the ranges that you suggest, then you indeed get $4pi/3$.
answered Nov 23 at 8:36
Hans Lundmark
34.8k564111
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
When you change to spherical coordinates (with the convention that you seem to be using), the absolute value of the Jacobian determinant is $|rho^2 sin phi|$, and this does only simplify to $rho^2 sinphi$ if $sinphi ge 0$, which is the case in the interval $0 le phi le pi$, but not in the interval $pi < phi < 2 pi$. But if you instead integrate $|rho^2 sin phi|$ over the ranges that you suggest, then you indeed get $4pi/3$.
answered Nov 23 at 8:36
Hans Lundmark
34.8k564111
When you change to spherical coordinates (with the convention that you seem to be using), the absolute value of the Jacobian determinant is $|rho^2 sin phi|$, and this does only simplify to $rho^2 sinphi$ if $sinphi ge 0$, which is the case in the interval $0 le phi le pi$, but not in the interval $pi < phi < 2 pi$. But if you instead integrate $|rho^2 sin phi|$ over the ranges that you suggest, then you indeed get $4pi/3$.
answered Nov 23 at 8:36
Hans Lundmark
34.8k564111
answered Nov 23 at 8:36
Hans Lundmark
34.8k564111
answered Nov 23 at 8:36
Hans Lundmark
34.8k564111
answered Nov 23 at 8:36
Hans Lundmark
34.8k564111
34.8k564111
add a comment |
add a comment |
up vote
1
down vote
The key fact is that the sphere $$x^2+y^2+z^2=R^2$$ is parametrized by
- $x=Rcos theta sin phi$
- $y=Rsin theta sin phi$
- $z=Rcos phi$
with
- $theta in [0,2pi)$
- $phi in[0,pi]$
and at each point $P(x,y,z)$ on the sphere corresponds an unique pair for the parameters $(theta, phi)$.
Of course we can also use different range for the parametrization as for example
- $theta in [-pi,pi)$
- $phi in[0,pi]$
(credits http://mathworld.wolfram.com/SphericalCoordinates.html)
answered Nov 23 at 7:34
gimusi
92.3k84495
I don't think this is the issue here. As the OP has noted, the unconventional ranges $theta in [0,pi)$ and $phi in [0,2 pi)$ also give a sphere.
– Hans Lundmark
Nov 23 at 8:38
@HansLundmark Yes of course, maybe I misunderstood the doubt ofthe OP. I agree we can introduce (infinitely) many different parametrization for the sphere by suitable choices of the range for $theta$ and $phi$.
– gimusi
Nov 23 at 8:45
add a comment |
up vote
1
down vote
The key fact is that the sphere $$x^2+y^2+z^2=R^2$$ is parametrized by
- $x=Rcos theta sin phi$
- $y=Rsin theta sin phi$
- $z=Rcos phi$
with
- $theta in [0,2pi)$
- $phi in[0,pi]$
and at each point $P(x,y,z)$ on the sphere corresponds an unique pair for the parameters $(theta, phi)$.
Of course we can also use different range for the parametrization as for example
- $theta in [-pi,pi)$
- $phi in[0,pi]$
(credits http://mathworld.wolfram.com/SphericalCoordinates.html)
answered Nov 23 at 7:34
gimusi
92.3k84495
I don't think this is the issue here. As the OP has noted, the unconventional ranges $theta in [0,pi)$ and $phi in [0,2 pi)$ also give a sphere.
– Hans Lundmark
Nov 23 at 8:38
@HansLundmark Yes of course, maybe I misunderstood the doubt ofthe OP. I agree we can introduce (infinitely) many different parametrization for the sphere by suitable choices of the range for $theta$ and $phi$.
– gimusi
Nov 23 at 8:45
add a comment |
up vote
1
down vote
up vote
1
down vote
The key fact is that the sphere $$x^2+y^2+z^2=R^2$$ is parametrized by
- $x=Rcos theta sin phi$
- $y=Rsin theta sin phi$
- $z=Rcos phi$
with
- $theta in [0,2pi)$
- $phi in[0,pi]$
and at each point $P(x,y,z)$ on the sphere corresponds an unique pair for the parameters $(theta, phi)$.
Of course we can also use different range for the parametrization as for example
- $theta in [-pi,pi)$
- $phi in[0,pi]$
(credits http://mathworld.wolfram.com/SphericalCoordinates.html)
answered Nov 23 at 7:34
gimusi
92.3k84495
The key fact is that the sphere $$x^2+y^2+z^2=R^2$$ is parametrized by
- $x=Rcos theta sin phi$
- $y=Rsin theta sin phi$
- $z=Rcos phi$
with
- $theta in [0,2pi)$
- $phi in[0,pi]$
and at each point $P(x,y,z)$ on the sphere corresponds an unique pair for the parameters $(theta, phi)$.
Of course we can also use different range for the parametrization as for example
- $theta in [-pi,pi)$
- $phi in[0,pi]$
(credits http://mathworld.wolfram.com/SphericalCoordinates.html)
answered Nov 23 at 7:34
gimusi
92.3k84495
edited Nov 23 at 8:47
answered Nov 23 at 7:34
gimusi
92.3k84495
answered Nov 23 at 7:34
gimusi
92.3k84495
answered Nov 23 at 7:34
gimusi
92.3k84495
92.3k84495
I don't think this is the issue here. As the OP has noted, the unconventional ranges $theta in [0,pi)$ and $phi in [0,2 pi)$ also give a sphere.
– Hans Lundmark
Nov 23 at 8:38
@HansLundmark Yes of course, maybe I misunderstood the doubt ofthe OP. I agree we can introduce (infinitely) many different parametrization for the sphere by suitable choices of the range for $theta$ and $phi$.
– gimusi
Nov 23 at 8:45
add a comment |
I don't think this is the issue here. As the OP has noted, the unconventional ranges $theta in [0,pi)$ and $phi in [0,2 pi)$ also give a sphere.
– Hans Lundmark
Nov 23 at 8:38
@HansLundmark Yes of course, maybe I misunderstood the doubt ofthe OP. I agree we can introduce (infinitely) many different parametrization for the sphere by suitable choices of the range for $theta$ and $phi$.
– gimusi
Nov 23 at 8:45
I don't think this is the issue here. As the OP has noted, the unconventional ranges $theta in [0,pi)$ and $phi in [0,2 pi)$ also give a sphere.
– Hans Lundmark
Nov 23 at 8:38
I don't think this is the issue here. As the OP has noted, the unconventional ranges $theta in [0,pi)$ and $phi in [0,2 pi)$ also give a sphere.
– Hans Lundmark
Nov 23 at 8:38
@HansLundmark Yes of course, maybe I misunderstood the doubt ofthe OP. I agree we can introduce (infinitely) many different parametrization for the sphere by suitable choices of the range for $theta$ and $phi$.
– gimusi
Nov 23 at 8:45
@HansLundmark Yes of course, maybe I misunderstood the doubt ofthe OP. I agree we can introduce (infinitely) many different parametrization for the sphere by suitable choices of the range for $theta$ and $phi$.
– gimusi
Nov 23 at 8:45
add a comment |
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