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Jacobian of just one variable - Azimuthal average of this function
The Next CEO of Stack OverflowSolving defined integralVolume of a cube in spherical polarsLine integrals - Surface areaSurface Integral do I use a jacobian?Order of integration in spherical coordinatesAnalytical Solution to IntegralHow to prove that the integral below in independent of $theta$ and $phi$?How to calculate the integral $int_a^infty int_0^2 pirho, mathrm e^i k rho cos(phi-theta) sinphi, mathrm d rho, mathrm dphi$?Trouble Evaluating Triple Integral: $iiint sqrt3sqrtx^2+z^2 dV$How to evaluate $int_pi/2^pi sqrt1 - frac12cos^2 x + sin x sin 2x ;mathrmdx$
$begingroup$
Ideally I would like to get an analytical expression for the azimuthal average of a certain function:
$$ f(x,y) = mathrmMaxleft (0, , 1-fracx^2R_x^2 - fracy^2R_y^2 right )^3/2 $$
for which I can write my azimuthal average can be written:
$$ F_az(r) = frac12piint_0 ^2pi mathrmMaxleft (0, , 1-fracrho^2cos^2phiR_x^2 - fracrho^2sin^2phiR_y^2 right )^3/2 mathrmdphi$$
Now. I want to change to an elliptical coordinate system $u = rho cosphi/R_x$, $v = rho sinphi/R_x$...
If the integral were over $phi$ and $rho$, I could do just calculate the Jacobian.
But my integral is in one variable only! How do I compute it?
I tried symbolic integration on Mathematica but I would not stop calculating... so I got nothing from there.
Would anyone know if there is an analytical solution? How would I start computing it?
integration
asked Mar 27 at 17:25
SuperCiociaSuperCiocia
295213
$endgroup$
add a comment |
$begingroup$
Ideally I would like to get an analytical expression for the azimuthal average of a certain function:
$$ f(x,y) = mathrmMaxleft (0, , 1-fracx^2R_x^2 - fracy^2R_y^2 right )^3/2 $$
for which I can write my azimuthal average can be written:
$$ F_az(r) = frac12piint_0 ^2pi mathrmMaxleft (0, , 1-fracrho^2cos^2phiR_x^2 - fracrho^2sin^2phiR_y^2 right )^3/2 mathrmdphi$$
Now. I want to change to an elliptical coordinate system $u = rho cosphi/R_x$, $v = rho sinphi/R_x$...
If the integral were over $phi$ and $rho$, I could do just calculate the Jacobian.
But my integral is in one variable only! How do I compute it?
I tried symbolic integration on Mathematica but I would not stop calculating... so I got nothing from there.
Would anyone know if there is an analytical solution? How would I start computing it?
integration
asked Mar 27 at 17:25
SuperCiociaSuperCiocia
295213
$endgroup$
$begingroup$
what is the relationship between the integrand and $phi$?
$endgroup$
– gt6989b
Mar 27 at 17:27
$begingroup$
usual cartesian and plane polars, $phi = arctan(y/x)$
$endgroup$
– SuperCiocia
Mar 27 at 17:28
$begingroup$
I modified the question to add more details about wat I have tried
$endgroup$
– SuperCiocia
Mar 27 at 18:16
add a comment |
$begingroup$
Ideally I would like to get an analytical expression for the azimuthal average of a certain function:
$$ f(x,y) = mathrmMaxleft (0, , 1-fracx^2R_x^2 - fracy^2R_y^2 right )^3/2 $$
for which I can write my azimuthal average can be written:
$$ F_az(r) = frac12piint_0 ^2pi mathrmMaxleft (0, , 1-fracrho^2cos^2phiR_x^2 - fracrho^2sin^2phiR_y^2 right )^3/2 mathrmdphi$$
Now. I want to change to an elliptical coordinate system $u = rho cosphi/R_x$, $v = rho sinphi/R_x$...
If the integral were over $phi$ and $rho$, I could do just calculate the Jacobian.
But my integral is in one variable only! How do I compute it?
I tried symbolic integration on Mathematica but I would not stop calculating... so I got nothing from there.
Would anyone know if there is an analytical solution? How would I start computing it?
integration
asked Mar 27 at 17:25
SuperCiociaSuperCiocia
295213
$endgroup$
Ideally I would like to get an analytical expression for the azimuthal average of a certain function:
$$ f(x,y) = mathrmMaxleft (0, , 1-fracx^2R_x^2 - fracy^2R_y^2 right )^3/2 $$
for which I can write my azimuthal average can be written:
$$ F_az(r) = frac12piint_0 ^2pi mathrmMaxleft (0, , 1-fracrho^2cos^2phiR_x^2 - fracrho^2sin^2phiR_y^2 right )^3/2 mathrmdphi$$
Now. I want to change to an elliptical coordinate system $u = rho cosphi/R_x$, $v = rho sinphi/R_x$...
If the integral were over $phi$ and $rho$, I could do just calculate the Jacobian.
But my integral is in one variable only! How do I compute it?
I tried symbolic integration on Mathematica but I would not stop calculating... so I got nothing from there.
Would anyone know if there is an analytical solution? How would I start computing it?
integration
integration
asked Mar 27 at 17:25
SuperCiociaSuperCiocia
295213
asked Mar 27 at 17:25
SuperCiociaSuperCiocia
295213
edited Mar 27 at 18:15
SuperCiocia
asked Mar 27 at 17:25
SuperCiociaSuperCiocia
295213
asked Mar 27 at 17:25
SuperCiociaSuperCiocia
295213
asked Mar 27 at 17:25
SuperCiociaSuperCiocia
295213
295213
$begingroup$
what is the relationship between the integrand and $phi$?
$endgroup$
– gt6989b
Mar 27 at 17:27
$begingroup$
usual cartesian and plane polars, $phi = arctan(y/x)$
$endgroup$
– SuperCiocia
Mar 27 at 17:28
$begingroup$
I modified the question to add more details about wat I have tried
$endgroup$
– SuperCiocia
Mar 27 at 18:16
add a comment |
$begingroup$
what is the relationship between the integrand and $phi$?
$endgroup$
– gt6989b
Mar 27 at 17:27
$begingroup$
usual cartesian and plane polars, $phi = arctan(y/x)$
$endgroup$
– SuperCiocia
Mar 27 at 17:28
$begingroup$
I modified the question to add more details about wat I have tried
$endgroup$
– SuperCiocia
Mar 27 at 18:16
$begingroup$
what is the relationship between the integrand and $phi$?
$endgroup$
– gt6989b
Mar 27 at 17:27
$begingroup$
what is the relationship between the integrand and $phi$?
$endgroup$
– gt6989b
Mar 27 at 17:27
$begingroup$
usual cartesian and plane polars, $phi = arctan(y/x)$
$endgroup$
– SuperCiocia
Mar 27 at 17:28
$begingroup$
usual cartesian and plane polars, $phi = arctan(y/x)$
$endgroup$
– SuperCiocia
Mar 27 at 17:28
$begingroup$
I modified the question to add more details about wat I have tried
$endgroup$
– SuperCiocia
Mar 27 at 18:16
$begingroup$
I modified the question to add more details about wat I have tried
$endgroup$
– SuperCiocia
Mar 27 at 18:16
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
HINT
One stupid way
$$
frac12pi int_0^2pi f(phi)dphi
= frac12pi int_0^1 drho int_0^2pi f(phi)dphi
$$
answered Mar 27 at 18:20
gt6989bgt6989b
35.2k22557
$endgroup$
$begingroup$
Yeah but would that give an analytic solution for my $f$?
$endgroup$
– SuperCiocia
Mar 27 at 18:54
$begingroup$
@SuperCiocia you were looking to change coordinates -- this will give you a way to do so
$endgroup$
– gt6989b
Mar 27 at 19:14
$begingroup$
And I thank you for that. Mathematica still does not give me a symbolic integral. Do you know of a way to start solving my $f$? I.e. a change of variables that would make a known integral like Elliptic ones come out?
$endgroup$
– SuperCiocia
Mar 27 at 19:24
$begingroup$
@SuperCiocia unfortunately, not
$endgroup$
– gt6989b
Mar 27 at 19:49
add a comment |
Your Answer
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1 Answer
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oldest
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1 Answer
1
active
oldest
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active
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active
oldest
votes
$begingroup$
HINT
One stupid way
$$
frac12pi int_0^2pi f(phi)dphi
= frac12pi int_0^1 drho int_0^2pi f(phi)dphi
$$
answered Mar 27 at 18:20
gt6989bgt6989b
35.2k22557
$endgroup$
$begingroup$
Yeah but would that give an analytic solution for my $f$?
$endgroup$
– SuperCiocia
Mar 27 at 18:54
$begingroup$
@SuperCiocia you were looking to change coordinates -- this will give you a way to do so
$endgroup$
– gt6989b
Mar 27 at 19:14
$begingroup$
And I thank you for that. Mathematica still does not give me a symbolic integral. Do you know of a way to start solving my $f$? I.e. a change of variables that would make a known integral like Elliptic ones come out?
$endgroup$
– SuperCiocia
Mar 27 at 19:24
$begingroup$
@SuperCiocia unfortunately, not
$endgroup$
– gt6989b
Mar 27 at 19:49
add a comment |
$begingroup$
HINT
One stupid way
$$
frac12pi int_0^2pi f(phi)dphi
= frac12pi int_0^1 drho int_0^2pi f(phi)dphi
$$
answered Mar 27 at 18:20
gt6989bgt6989b
35.2k22557
$endgroup$
$begingroup$
Yeah but would that give an analytic solution for my $f$?
$endgroup$
– SuperCiocia
Mar 27 at 18:54
$begingroup$
@SuperCiocia you were looking to change coordinates -- this will give you a way to do so
$endgroup$
– gt6989b
Mar 27 at 19:14
$begingroup$
And I thank you for that. Mathematica still does not give me a symbolic integral. Do you know of a way to start solving my $f$? I.e. a change of variables that would make a known integral like Elliptic ones come out?
$endgroup$
– SuperCiocia
Mar 27 at 19:24
$begingroup$
@SuperCiocia unfortunately, not
$endgroup$
– gt6989b
Mar 27 at 19:49
add a comment |
$begingroup$
HINT
One stupid way
$$
frac12pi int_0^2pi f(phi)dphi
= frac12pi int_0^1 drho int_0^2pi f(phi)dphi
$$
answered Mar 27 at 18:20
gt6989bgt6989b
35.2k22557
$endgroup$
HINT
One stupid way
$$
frac12pi int_0^2pi f(phi)dphi
= frac12pi int_0^1 drho int_0^2pi f(phi)dphi
$$
answered Mar 27 at 18:20
gt6989bgt6989b
35.2k22557
answered Mar 27 at 18:20
gt6989bgt6989b
35.2k22557
answered Mar 27 at 18:20
gt6989bgt6989b
35.2k22557
answered Mar 27 at 18:20
gt6989bgt6989b
35.2k22557
35.2k22557
$begingroup$
Yeah but would that give an analytic solution for my $f$?
$endgroup$
– SuperCiocia
Mar 27 at 18:54
$begingroup$
@SuperCiocia you were looking to change coordinates -- this will give you a way to do so
$endgroup$
– gt6989b
Mar 27 at 19:14
$begingroup$
And I thank you for that. Mathematica still does not give me a symbolic integral. Do you know of a way to start solving my $f$? I.e. a change of variables that would make a known integral like Elliptic ones come out?
$endgroup$
– SuperCiocia
Mar 27 at 19:24
$begingroup$
@SuperCiocia unfortunately, not
$endgroup$
– gt6989b
Mar 27 at 19:49
add a comment |
$begingroup$
Yeah but would that give an analytic solution for my $f$?
$endgroup$
– SuperCiocia
Mar 27 at 18:54
$begingroup$
@SuperCiocia you were looking to change coordinates -- this will give you a way to do so
$endgroup$
– gt6989b
Mar 27 at 19:14
$begingroup$
And I thank you for that. Mathematica still does not give me a symbolic integral. Do you know of a way to start solving my $f$? I.e. a change of variables that would make a known integral like Elliptic ones come out?
$endgroup$
– SuperCiocia
Mar 27 at 19:24
$begingroup$
@SuperCiocia unfortunately, not
$endgroup$
– gt6989b
Mar 27 at 19:49
$begingroup$
Yeah but would that give an analytic solution for my $f$?
$endgroup$
– SuperCiocia
Mar 27 at 18:54
$begingroup$
Yeah but would that give an analytic solution for my $f$?
$endgroup$
– SuperCiocia
Mar 27 at 18:54
$begingroup$
@SuperCiocia you were looking to change coordinates -- this will give you a way to do so
$endgroup$
– gt6989b
Mar 27 at 19:14
$begingroup$
@SuperCiocia you were looking to change coordinates -- this will give you a way to do so
$endgroup$
– gt6989b
Mar 27 at 19:14
$begingroup$
And I thank you for that. Mathematica still does not give me a symbolic integral. Do you know of a way to start solving my $f$? I.e. a change of variables that would make a known integral like Elliptic ones come out?
$endgroup$
– SuperCiocia
Mar 27 at 19:24
$begingroup$
And I thank you for that. Mathematica still does not give me a symbolic integral. Do you know of a way to start solving my $f$? I.e. a change of variables that would make a known integral like Elliptic ones come out?
$endgroup$
– SuperCiocia
Mar 27 at 19:24
$begingroup$
@SuperCiocia unfortunately, not
$endgroup$
– gt6989b
Mar 27 at 19:49
$begingroup$
@SuperCiocia unfortunately, not
$endgroup$
– gt6989b
Mar 27 at 19:49
add a comment |
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$begingroup$
what is the relationship between the integrand and $phi$?
$endgroup$
– gt6989b
Mar 27 at 17:27
$begingroup$
usual cartesian and plane polars, $phi = arctan(y/x)$
$endgroup$
– SuperCiocia
Mar 27 at 17:28
$begingroup$
I modified the question to add more details about wat I have tried
$endgroup$
– SuperCiocia
Mar 27 at 18:16