Find the volume of a rotationally symmetric 3D body
$begingroup$
let $S$ be the set of points $(x,y,z)$ satisfying $ x^2 + y^2 + z^2 = 1 $ and $ 0leq z leq frac{1}{sqrt{2}} $
let $D$ be the $ 3 D $ body obtained by taking the union of all segments connecting $(0,0,0)$ to the points of $S$
Find the volume of the body.
Here is my trial :
Let
$g(t) = (0,0,0) + t(x,y,z)$ such that t $ in [0,1]$
by the giving $ S $ in the question we get that :
$ D =$ {$0 leq x^2 + y^2 + z^2 leq 1$}
so we want to do triple integral on $ D$ to find the volume :
$$
int_0^{2pi}
int_{pi/4}^{pi/2}
int_0^1 r^2 sin(phi) { dr} d{phi } d{theta}
$$
the thing is when i move to spherical coordinates i get different answer than the basic coordinates .
Spherical Answer : $ frac{sqrt{2}pi}{3} $
x,y,z coordinates Answer $ frac{5pi}{6sqrt{2}} $
integration multivariable-calculus jacobian
edited Dec 24 '18 at 17:23
gt6989b
36.1k22557
asked Dec 24 '18 at 16:27
Mather Mather
4088
$endgroup$
|
show 2 more comments
$begingroup$
let $S$ be the set of points $(x,y,z)$ satisfying $ x^2 + y^2 + z^2 = 1 $ and $ 0leq z leq frac{1}{sqrt{2}} $
let $D$ be the $ 3 D $ body obtained by taking the union of all segments connecting $(0,0,0)$ to the points of $S$
Find the volume of the body.
Here is my trial :
Let
$g(t) = (0,0,0) + t(x,y,z)$ such that t $ in [0,1]$
by the giving $ S $ in the question we get that :
$ D =$ {$0 leq x^2 + y^2 + z^2 leq 1$}
so we want to do triple integral on $ D$ to find the volume :
$$
int_0^{2pi}
int_{pi/4}^{pi/2}
int_0^1 r^2 sin(phi) { dr} d{phi } d{theta}
$$
the thing is when i move to spherical coordinates i get different answer than the basic coordinates .
Spherical Answer : $ frac{sqrt{2}pi}{3} $
x,y,z coordinates Answer $ frac{5pi}{6sqrt{2}} $
integration multivariable-calculus jacobian
edited Dec 24 '18 at 17:23
gt6989b
36.1k22557
asked Dec 24 '18 at 16:27
Mather Mather
4088
$endgroup$
$begingroup$
it would help if you would write in the actual integrals you are computing and the calculations. looks like an error in arithmetic likely
$endgroup$
– gt6989b
Dec 24 '18 at 16:32
$begingroup$
i have updated the integral
$endgroup$
– Mather
Dec 24 '18 at 16:49
$begingroup$
Are you sure that is the answer you obtain through the spherical integral?
$endgroup$
– Shubham Johri
Dec 24 '18 at 16:50
$begingroup$
i am not sure of the answer i tried to calculate the question in different ways but i failed to calculate in the spherical coordinates , and according to the answer down there the one in the x,y,z is right but the one in the spherical is wrong
$endgroup$
– Mather
Dec 24 '18 at 16:51
$begingroup$
@Mather The answer obtained through spherical co-ordinates is correct
$endgroup$
– Shubham Johri
Dec 24 '18 at 17:12
|
show 2 more comments
$begingroup$
let $S$ be the set of points $(x,y,z)$ satisfying $ x^2 + y^2 + z^2 = 1 $ and $ 0leq z leq frac{1}{sqrt{2}} $
let $D$ be the $ 3 D $ body obtained by taking the union of all segments connecting $(0,0,0)$ to the points of $S$
Find the volume of the body.
Here is my trial :
Let
$g(t) = (0,0,0) + t(x,y,z)$ such that t $ in [0,1]$
by the giving $ S $ in the question we get that :
$ D =$ {$0 leq x^2 + y^2 + z^2 leq 1$}
so we want to do triple integral on $ D$ to find the volume :
$$
int_0^{2pi}
int_{pi/4}^{pi/2}
int_0^1 r^2 sin(phi) { dr} d{phi } d{theta}
$$
the thing is when i move to spherical coordinates i get different answer than the basic coordinates .
Spherical Answer : $ frac{sqrt{2}pi}{3} $
x,y,z coordinates Answer $ frac{5pi}{6sqrt{2}} $
integration multivariable-calculus jacobian
edited Dec 24 '18 at 17:23
gt6989b
36.1k22557
asked Dec 24 '18 at 16:27
Mather Mather
4088
$endgroup$
let $S$ be the set of points $(x,y,z)$ satisfying $ x^2 + y^2 + z^2 = 1 $ and $ 0leq z leq frac{1}{sqrt{2}} $
let $D$ be the $ 3 D $ body obtained by taking the union of all segments connecting $(0,0,0)$ to the points of $S$
Find the volume of the body.
Here is my trial :
Let
$g(t) = (0,0,0) + t(x,y,z)$ such that t $ in [0,1]$
by the giving $ S $ in the question we get that :
$ D =$ {$0 leq x^2 + y^2 + z^2 leq 1$}
so we want to do triple integral on $ D$ to find the volume :
$$
int_0^{2pi}
int_{pi/4}^{pi/2}
int_0^1 r^2 sin(phi) { dr} d{phi } d{theta}
$$
the thing is when i move to spherical coordinates i get different answer than the basic coordinates .
Spherical Answer : $ frac{sqrt{2}pi}{3} $
x,y,z coordinates Answer $ frac{5pi}{6sqrt{2}} $
integration multivariable-calculus jacobian
integration multivariable-calculus jacobian
edited Dec 24 '18 at 17:23
gt6989b
36.1k22557
asked Dec 24 '18 at 16:27
Mather Mather
4088
edited Dec 24 '18 at 17:23
gt6989b
36.1k22557
asked Dec 24 '18 at 16:27
Mather Mather
4088
edited Dec 24 '18 at 17:23
gt6989b
36.1k22557
edited Dec 24 '18 at 17:23
gt6989b
36.1k22557
edited Dec 24 '18 at 17:23
gt6989b
36.1k22557
36.1k22557
asked Dec 24 '18 at 16:27
Mather Mather
4088
asked Dec 24 '18 at 16:27
Mather Mather
4088
asked Dec 24 '18 at 16:27
Mather Mather
4088
4088
$begingroup$
it would help if you would write in the actual integrals you are computing and the calculations. looks like an error in arithmetic likely
$endgroup$
– gt6989b
Dec 24 '18 at 16:32
$begingroup$
i have updated the integral
$endgroup$
– Mather
Dec 24 '18 at 16:49
$begingroup$
Are you sure that is the answer you obtain through the spherical integral?
$endgroup$
– Shubham Johri
Dec 24 '18 at 16:50
$begingroup$
i am not sure of the answer i tried to calculate the question in different ways but i failed to calculate in the spherical coordinates , and according to the answer down there the one in the x,y,z is right but the one in the spherical is wrong
$endgroup$
– Mather
Dec 24 '18 at 16:51
$begingroup$
@Mather The answer obtained through spherical co-ordinates is correct
$endgroup$
– Shubham Johri
Dec 24 '18 at 17:12
|
show 2 more comments
$begingroup$
it would help if you would write in the actual integrals you are computing and the calculations. looks like an error in arithmetic likely
$endgroup$
– gt6989b
Dec 24 '18 at 16:32
$begingroup$
i have updated the integral
$endgroup$
– Mather
Dec 24 '18 at 16:49
$begingroup$
Are you sure that is the answer you obtain through the spherical integral?
$endgroup$
– Shubham Johri
Dec 24 '18 at 16:50
$begingroup$
i am not sure of the answer i tried to calculate the question in different ways but i failed to calculate in the spherical coordinates , and according to the answer down there the one in the x,y,z is right but the one in the spherical is wrong
$endgroup$
– Mather
Dec 24 '18 at 16:51
$begingroup$
@Mather The answer obtained through spherical co-ordinates is correct
$endgroup$
– Shubham Johri
Dec 24 '18 at 17:12
$begingroup$
it would help if you would write in the actual integrals you are computing and the calculations. looks like an error in arithmetic likely
$endgroup$
– gt6989b
Dec 24 '18 at 16:32
$begingroup$
it would help if you would write in the actual integrals you are computing and the calculations. looks like an error in arithmetic likely
$endgroup$
– gt6989b
Dec 24 '18 at 16:32
$begingroup$
i have updated the integral
$endgroup$
– Mather
Dec 24 '18 at 16:49
$begingroup$
i have updated the integral
$endgroup$
– Mather
Dec 24 '18 at 16:49
$begingroup$
Are you sure that is the answer you obtain through the spherical integral?
$endgroup$
– Shubham Johri
Dec 24 '18 at 16:50
$begingroup$
Are you sure that is the answer you obtain through the spherical integral?
$endgroup$
– Shubham Johri
Dec 24 '18 at 16:50
$begingroup$
i am not sure of the answer i tried to calculate the question in different ways but i failed to calculate in the spherical coordinates , and according to the answer down there the one in the x,y,z is right but the one in the spherical is wrong
$endgroup$
– Mather
Dec 24 '18 at 16:51
$begingroup$
i am not sure of the answer i tried to calculate the question in different ways but i failed to calculate in the spherical coordinates , and according to the answer down there the one in the x,y,z is right but the one in the spherical is wrong
$endgroup$
– Mather
Dec 24 '18 at 16:51
$begingroup$
@Mather The answer obtained through spherical co-ordinates is correct
$endgroup$
– Shubham Johri
Dec 24 '18 at 17:12
$begingroup$
@Mather The answer obtained through spherical co-ordinates is correct
$endgroup$
– Shubham Johri
Dec 24 '18 at 17:12
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
The integral in cartesian co-ordinates looks like this: $I=I_1-I_2$
$displaystyle I_1=int_{-1}^1int_{-sqrt{1-y^2}}^{sqrt{1-y^2}}sqrt{1-x^2-y^2} dx dy=2pi/3\I_2=displaystyleint_{-frac1{sqrt2}}^{frac1{sqrt2}}int_{-sqrt{frac12-y^2}}^{sqrt{frac12-y^2}}sqrt{1-x^2-y^2}-sqrt{x^2+y^2} dx dy$
$I_1$ is just the volume of the hemsiphere. $I_2$ gives the volume of the inverted cone with spherical base that needs to be subtracted from $I_1$.
The answer obtained is the same as the one produced by the spherical integral: $sqrt2pi/3$.
answered Dec 24 '18 at 17:21
Shubham JohriShubham Johri
5,683918
$endgroup$
$begingroup$
i sliced the domain into circules as a function of z like the one up there
$endgroup$
– Mather
Dec 24 '18 at 17:22
$begingroup$
I'm adding more details, just wait a sec
$endgroup$
– Shubham Johri
Dec 24 '18 at 17:23
$begingroup$
The domain is a collection of annular discs parallel to the $z=0$ plane. The answer above has assumed it to be a collection of complete discs
$endgroup$
– Shubham Johri
Dec 24 '18 at 17:32
$begingroup$
but i dont understand why i cant slice the body into discs with radius $ 1-z^2 $
$endgroup$
– Mather
Dec 24 '18 at 17:44
$begingroup$
Do you know a good 3D surface plotter?
$endgroup$
– Shubham Johri
Dec 24 '18 at 17:46
|
show 3 more comments
$begingroup$
$$intlimits_{z=0}^{1/sqrt{2}} pi left( (1 - z^2) - z^2 right) dz = intlimits_{z=0}^{1/sqrt{2}} pi (1 - 2 z^2) dz = frac{sqrt{2} pi }{3}$$
Annular disk of inner radius $z$ and outer radius $sqrt{1 - z^2}$ and thickness $dz$ (in blue).
answered Dec 24 '18 at 16:33
David G. StorkDavid G. Stork
12.2k41836
$endgroup$
$begingroup$
yes i got this answer on basic coordinates but on spherical i got diffrent one
$endgroup$
– Mather
Dec 24 '18 at 16:34
$begingroup$
Show your full spherical integral.
$endgroup$
– David G. Stork
Dec 24 '18 at 16:34
$begingroup$
i updated it without the limits
$endgroup$
– Mather
Dec 24 '18 at 16:36
$begingroup$
You don't have the Jacobian.
$endgroup$
– David G. Stork
Dec 24 '18 at 16:36
$begingroup$
the jacobian is the integrand itself $ r^2sin(phi) $ isn't it ?
$endgroup$
– Mather
Dec 24 '18 at 16:38
|
show 2 more comments
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The integral in cartesian co-ordinates looks like this: $I=I_1-I_2$
$displaystyle I_1=int_{-1}^1int_{-sqrt{1-y^2}}^{sqrt{1-y^2}}sqrt{1-x^2-y^2} dx dy=2pi/3\I_2=displaystyleint_{-frac1{sqrt2}}^{frac1{sqrt2}}int_{-sqrt{frac12-y^2}}^{sqrt{frac12-y^2}}sqrt{1-x^2-y^2}-sqrt{x^2+y^2} dx dy$
$I_1$ is just the volume of the hemsiphere. $I_2$ gives the volume of the inverted cone with spherical base that needs to be subtracted from $I_1$.
The answer obtained is the same as the one produced by the spherical integral: $sqrt2pi/3$.
answered Dec 24 '18 at 17:21
Shubham JohriShubham Johri
5,683918
$endgroup$
$begingroup$
i sliced the domain into circules as a function of z like the one up there
$endgroup$
– Mather
Dec 24 '18 at 17:22
$begingroup$
I'm adding more details, just wait a sec
$endgroup$
– Shubham Johri
Dec 24 '18 at 17:23
$begingroup$
The domain is a collection of annular discs parallel to the $z=0$ plane. The answer above has assumed it to be a collection of complete discs
$endgroup$
– Shubham Johri
Dec 24 '18 at 17:32
$begingroup$
but i dont understand why i cant slice the body into discs with radius $ 1-z^2 $
$endgroup$
– Mather
Dec 24 '18 at 17:44
$begingroup$
Do you know a good 3D surface plotter?
$endgroup$
– Shubham Johri
Dec 24 '18 at 17:46
|
show 3 more comments
$begingroup$
The integral in cartesian co-ordinates looks like this: $I=I_1-I_2$
$displaystyle I_1=int_{-1}^1int_{-sqrt{1-y^2}}^{sqrt{1-y^2}}sqrt{1-x^2-y^2} dx dy=2pi/3\I_2=displaystyleint_{-frac1{sqrt2}}^{frac1{sqrt2}}int_{-sqrt{frac12-y^2}}^{sqrt{frac12-y^2}}sqrt{1-x^2-y^2}-sqrt{x^2+y^2} dx dy$
$I_1$ is just the volume of the hemsiphere. $I_2$ gives the volume of the inverted cone with spherical base that needs to be subtracted from $I_1$.
The answer obtained is the same as the one produced by the spherical integral: $sqrt2pi/3$.
answered Dec 24 '18 at 17:21
Shubham JohriShubham Johri
5,683918
$endgroup$
$begingroup$
i sliced the domain into circules as a function of z like the one up there
$endgroup$
– Mather
Dec 24 '18 at 17:22
$begingroup$
I'm adding more details, just wait a sec
$endgroup$
– Shubham Johri
Dec 24 '18 at 17:23
$begingroup$
The domain is a collection of annular discs parallel to the $z=0$ plane. The answer above has assumed it to be a collection of complete discs
$endgroup$
– Shubham Johri
Dec 24 '18 at 17:32
$begingroup$
but i dont understand why i cant slice the body into discs with radius $ 1-z^2 $
$endgroup$
– Mather
Dec 24 '18 at 17:44
$begingroup$
Do you know a good 3D surface plotter?
$endgroup$
– Shubham Johri
Dec 24 '18 at 17:46
|
show 3 more comments
$begingroup$
The integral in cartesian co-ordinates looks like this: $I=I_1-I_2$
$displaystyle I_1=int_{-1}^1int_{-sqrt{1-y^2}}^{sqrt{1-y^2}}sqrt{1-x^2-y^2} dx dy=2pi/3\I_2=displaystyleint_{-frac1{sqrt2}}^{frac1{sqrt2}}int_{-sqrt{frac12-y^2}}^{sqrt{frac12-y^2}}sqrt{1-x^2-y^2}-sqrt{x^2+y^2} dx dy$
$I_1$ is just the volume of the hemsiphere. $I_2$ gives the volume of the inverted cone with spherical base that needs to be subtracted from $I_1$.
The answer obtained is the same as the one produced by the spherical integral: $sqrt2pi/3$.
answered Dec 24 '18 at 17:21
Shubham JohriShubham Johri
5,683918
$endgroup$
The integral in cartesian co-ordinates looks like this: $I=I_1-I_2$
$displaystyle I_1=int_{-1}^1int_{-sqrt{1-y^2}}^{sqrt{1-y^2}}sqrt{1-x^2-y^2} dx dy=2pi/3\I_2=displaystyleint_{-frac1{sqrt2}}^{frac1{sqrt2}}int_{-sqrt{frac12-y^2}}^{sqrt{frac12-y^2}}sqrt{1-x^2-y^2}-sqrt{x^2+y^2} dx dy$
$I_1$ is just the volume of the hemsiphere. $I_2$ gives the volume of the inverted cone with spherical base that needs to be subtracted from $I_1$.
The answer obtained is the same as the one produced by the spherical integral: $sqrt2pi/3$.
answered Dec 24 '18 at 17:21
Shubham JohriShubham Johri
5,683918
edited Dec 24 '18 at 18:41
answered Dec 24 '18 at 17:21
Shubham JohriShubham Johri
5,683918
answered Dec 24 '18 at 17:21
Shubham JohriShubham Johri
5,683918
answered Dec 24 '18 at 17:21
Shubham JohriShubham Johri
5,683918
5,683918
$begingroup$
i sliced the domain into circules as a function of z like the one up there
$endgroup$
– Mather
Dec 24 '18 at 17:22
$begingroup$
I'm adding more details, just wait a sec
$endgroup$
– Shubham Johri
Dec 24 '18 at 17:23
$begingroup$
The domain is a collection of annular discs parallel to the $z=0$ plane. The answer above has assumed it to be a collection of complete discs
$endgroup$
– Shubham Johri
Dec 24 '18 at 17:32
$begingroup$
but i dont understand why i cant slice the body into discs with radius $ 1-z^2 $
$endgroup$
– Mather
Dec 24 '18 at 17:44
$begingroup$
Do you know a good 3D surface plotter?
$endgroup$
– Shubham Johri
Dec 24 '18 at 17:46
|
show 3 more comments
$begingroup$
i sliced the domain into circules as a function of z like the one up there
$endgroup$
– Mather
Dec 24 '18 at 17:22
$begingroup$
I'm adding more details, just wait a sec
$endgroup$
– Shubham Johri
Dec 24 '18 at 17:23
$begingroup$
The domain is a collection of annular discs parallel to the $z=0$ plane. The answer above has assumed it to be a collection of complete discs
$endgroup$
– Shubham Johri
Dec 24 '18 at 17:32
$begingroup$
but i dont understand why i cant slice the body into discs with radius $ 1-z^2 $
$endgroup$
– Mather
Dec 24 '18 at 17:44
$begingroup$
Do you know a good 3D surface plotter?
$endgroup$
– Shubham Johri
Dec 24 '18 at 17:46
$begingroup$
i sliced the domain into circules as a function of z like the one up there
$endgroup$
– Mather
Dec 24 '18 at 17:22
$begingroup$
i sliced the domain into circules as a function of z like the one up there
$endgroup$
– Mather
Dec 24 '18 at 17:22
$begingroup$
I'm adding more details, just wait a sec
$endgroup$
– Shubham Johri
Dec 24 '18 at 17:23
$begingroup$
I'm adding more details, just wait a sec
$endgroup$
– Shubham Johri
Dec 24 '18 at 17:23
$begingroup$
The domain is a collection of annular discs parallel to the $z=0$ plane. The answer above has assumed it to be a collection of complete discs
$endgroup$
– Shubham Johri
Dec 24 '18 at 17:32
$begingroup$
The domain is a collection of annular discs parallel to the $z=0$ plane. The answer above has assumed it to be a collection of complete discs
$endgroup$
– Shubham Johri
Dec 24 '18 at 17:32
$begingroup$
but i dont understand why i cant slice the body into discs with radius $ 1-z^2 $
$endgroup$
– Mather
Dec 24 '18 at 17:44
$begingroup$
but i dont understand why i cant slice the body into discs with radius $ 1-z^2 $
$endgroup$
– Mather
Dec 24 '18 at 17:44
$begingroup$
Do you know a good 3D surface plotter?
$endgroup$
– Shubham Johri
Dec 24 '18 at 17:46
$begingroup$
Do you know a good 3D surface plotter?
$endgroup$
– Shubham Johri
Dec 24 '18 at 17:46
|
show 3 more comments
$begingroup$
$$intlimits_{z=0}^{1/sqrt{2}} pi left( (1 - z^2) - z^2 right) dz = intlimits_{z=0}^{1/sqrt{2}} pi (1 - 2 z^2) dz = frac{sqrt{2} pi }{3}$$
Annular disk of inner radius $z$ and outer radius $sqrt{1 - z^2}$ and thickness $dz$ (in blue).
answered Dec 24 '18 at 16:33
David G. StorkDavid G. Stork
12.2k41836
$endgroup$
$begingroup$
yes i got this answer on basic coordinates but on spherical i got diffrent one
$endgroup$
– Mather
Dec 24 '18 at 16:34
$begingroup$
Show your full spherical integral.
$endgroup$
– David G. Stork
Dec 24 '18 at 16:34
$begingroup$
i updated it without the limits
$endgroup$
– Mather
Dec 24 '18 at 16:36
$begingroup$
You don't have the Jacobian.
$endgroup$
– David G. Stork
Dec 24 '18 at 16:36
$begingroup$
the jacobian is the integrand itself $ r^2sin(phi) $ isn't it ?
$endgroup$
– Mather
Dec 24 '18 at 16:38
|
show 2 more comments
$begingroup$
$$intlimits_{z=0}^{1/sqrt{2}} pi left( (1 - z^2) - z^2 right) dz = intlimits_{z=0}^{1/sqrt{2}} pi (1 - 2 z^2) dz = frac{sqrt{2} pi }{3}$$
Annular disk of inner radius $z$ and outer radius $sqrt{1 - z^2}$ and thickness $dz$ (in blue).
answered Dec 24 '18 at 16:33
David G. StorkDavid G. Stork
12.2k41836
$endgroup$
$begingroup$
yes i got this answer on basic coordinates but on spherical i got diffrent one
$endgroup$
– Mather
Dec 24 '18 at 16:34
$begingroup$
Show your full spherical integral.
$endgroup$
– David G. Stork
Dec 24 '18 at 16:34
$begingroup$
i updated it without the limits
$endgroup$
– Mather
Dec 24 '18 at 16:36
$begingroup$
You don't have the Jacobian.
$endgroup$
– David G. Stork
Dec 24 '18 at 16:36
$begingroup$
the jacobian is the integrand itself $ r^2sin(phi) $ isn't it ?
$endgroup$
– Mather
Dec 24 '18 at 16:38
|
show 2 more comments
$begingroup$
$$intlimits_{z=0}^{1/sqrt{2}} pi left( (1 - z^2) - z^2 right) dz = intlimits_{z=0}^{1/sqrt{2}} pi (1 - 2 z^2) dz = frac{sqrt{2} pi }{3}$$
Annular disk of inner radius $z$ and outer radius $sqrt{1 - z^2}$ and thickness $dz$ (in blue).
answered Dec 24 '18 at 16:33
David G. StorkDavid G. Stork
12.2k41836
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$$intlimits_{z=0}^{1/sqrt{2}} pi left( (1 - z^2) - z^2 right) dz = intlimits_{z=0}^{1/sqrt{2}} pi (1 - 2 z^2) dz = frac{sqrt{2} pi }{3}$$
Annular disk of inner radius $z$ and outer radius $sqrt{1 - z^2}$ and thickness $dz$ (in blue).
answered Dec 24 '18 at 16:33
David G. StorkDavid G. Stork
12.2k41836
edited Dec 24 '18 at 19:10
answered Dec 24 '18 at 16:33
David G. StorkDavid G. Stork
12.2k41836
answered Dec 24 '18 at 16:33
David G. StorkDavid G. Stork
12.2k41836
answered Dec 24 '18 at 16:33
David G. StorkDavid G. Stork
12.2k41836
12.2k41836
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yes i got this answer on basic coordinates but on spherical i got diffrent one
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– Mather
Dec 24 '18 at 16:34
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Show your full spherical integral.
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– David G. Stork
Dec 24 '18 at 16:34
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i updated it without the limits
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– Mather
Dec 24 '18 at 16:36
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You don't have the Jacobian.
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– David G. Stork
Dec 24 '18 at 16:36
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the jacobian is the integrand itself $ r^2sin(phi) $ isn't it ?
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– Mather
Dec 24 '18 at 16:38
|
show 2 more comments
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yes i got this answer on basic coordinates but on spherical i got diffrent one
$endgroup$
– Mather
Dec 24 '18 at 16:34
$begingroup$
Show your full spherical integral.
$endgroup$
– David G. Stork
Dec 24 '18 at 16:34
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i updated it without the limits
$endgroup$
– Mather
Dec 24 '18 at 16:36
$begingroup$
You don't have the Jacobian.
$endgroup$
– David G. Stork
Dec 24 '18 at 16:36
$begingroup$
the jacobian is the integrand itself $ r^2sin(phi) $ isn't it ?
$endgroup$
– Mather
Dec 24 '18 at 16:38
$begingroup$
yes i got this answer on basic coordinates but on spherical i got diffrent one
$endgroup$
– Mather
Dec 24 '18 at 16:34
$begingroup$
yes i got this answer on basic coordinates but on spherical i got diffrent one
$endgroup$
– Mather
Dec 24 '18 at 16:34
$begingroup$
Show your full spherical integral.
$endgroup$
– David G. Stork
Dec 24 '18 at 16:34
$begingroup$
Show your full spherical integral.
$endgroup$
– David G. Stork
Dec 24 '18 at 16:34
$begingroup$
i updated it without the limits
$endgroup$
– Mather
Dec 24 '18 at 16:36
$begingroup$
i updated it without the limits
$endgroup$
– Mather
Dec 24 '18 at 16:36
$begingroup$
You don't have the Jacobian.
$endgroup$
– David G. Stork
Dec 24 '18 at 16:36
$begingroup$
You don't have the Jacobian.
$endgroup$
– David G. Stork
Dec 24 '18 at 16:36
$begingroup$
the jacobian is the integrand itself $ r^2sin(phi) $ isn't it ?
$endgroup$
– Mather
Dec 24 '18 at 16:38
$begingroup$
the jacobian is the integrand itself $ r^2sin(phi) $ isn't it ?
$endgroup$
– Mather
Dec 24 '18 at 16:38
|
show 2 more comments
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it would help if you would write in the actual integrals you are computing and the calculations. looks like an error in arithmetic likely
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– gt6989b
Dec 24 '18 at 16:32
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i have updated the integral
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– Mather
Dec 24 '18 at 16:49
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Are you sure that is the answer you obtain through the spherical integral?
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– Shubham Johri
Dec 24 '18 at 16:50
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i am not sure of the answer i tried to calculate the question in different ways but i failed to calculate in the spherical coordinates , and according to the answer down there the one in the x,y,z is right but the one in the spherical is wrong
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– Mather
Dec 24 '18 at 16:51
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@Mather The answer obtained through spherical co-ordinates is correct
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– Shubham Johri
Dec 24 '18 at 17:12