How should I solve this triple integral?Finding the Limits of the Triple Integral (Spherical Coordinates)How to setup/evaluate a triple integral to show an interesting result in physics?Triple Integrals in Spherical Coordinates where (z-2)^2Triple integral in spherical / cylindrical coordinates - where's the error? Exercise checktriple integral over spherical region and of a shifted sphere integrandTriple Integral in spherical coordinatesCalculating triple integral over a region GTrying to evaluate this triple integral?Triple Integrals in Spherical Coordinates, problem with boundariesComputing Triple Integral Using Spherical Coordinates
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How should I solve this triple integral?
Finding the Limits of the Triple Integral (Spherical Coordinates)How to setup/evaluate a triple integral to show an interesting result in physics?Triple Integrals in Spherical Coordinates where (z-2)^2Triple integral in spherical / cylindrical coordinates - where's the error? Exercise checktriple integral over spherical region and of a shifted sphere integrandTriple Integral in spherical coordinatesCalculating triple integral over a region GTrying to evaluate this triple integral?Triple Integrals in Spherical Coordinates, problem with boundariesComputing Triple Integral Using Spherical Coordinates
$begingroup$
Hi everybody I have a triple integral I can't solve:
$$iiint sqrt x^2+y^2+z^2 ,dx ,dy ,dz $$
Which the region is between $z=sqrt x^2+y^2$ and $z=4$ .
The question says after using the spherical coordinates the answer is:
$$aLargeint _0^pi/4 frac(sin phi)dphicos^4phi $$
So I've used the spherical coordinates $0<r<4$ and $ 0<theta<2pi$ and $0<phi<fracpi2$. The things I don't get is first why is $phi$ varying between 0 and $fracpi4$ instead of $fracpi2$? And secondly where did that $cos^4phi$ came from?
By the way the question asks for the value of $a$.
integration spherical-coordinates multiple-integral
edited 2 days ago
Max
629317
asked 2 days ago
khoshrangkhoshrang
596
New contributor
khoshrang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
Hi everybody I have a triple integral I can't solve:
$$iiint sqrt x^2+y^2+z^2 ,dx ,dy ,dz $$
Which the region is between $z=sqrt x^2+y^2$ and $z=4$ .
The question says after using the spherical coordinates the answer is:
$$aLargeint _0^pi/4 frac(sin phi)dphicos^4phi $$
So I've used the spherical coordinates $0<r<4$ and $ 0<theta<2pi$ and $0<phi<fracpi2$. The things I don't get is first why is $phi$ varying between 0 and $fracpi4$ instead of $fracpi2$? And secondly where did that $cos^4phi$ came from?
By the way the question asks for the value of $a$.
integration spherical-coordinates multiple-integral
edited 2 days ago
Max
629317
asked 2 days ago
khoshrangkhoshrang
596
New contributor
khoshrang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
$begingroup$
EDIT: ok now i understand the ϕ<π/4 it's because if we put x=0 in z=√ x^2+y^2we will get z=+y and z = -y which explains the ϕ bounds
$endgroup$
– khoshrang
2 days ago
add a comment |
$begingroup$
Hi everybody I have a triple integral I can't solve:
$$iiint sqrt x^2+y^2+z^2 ,dx ,dy ,dz $$
Which the region is between $z=sqrt x^2+y^2$ and $z=4$ .
The question says after using the spherical coordinates the answer is:
$$aLargeint _0^pi/4 frac(sin phi)dphicos^4phi $$
So I've used the spherical coordinates $0<r<4$ and $ 0<theta<2pi$ and $0<phi<fracpi2$. The things I don't get is first why is $phi$ varying between 0 and $fracpi4$ instead of $fracpi2$? And secondly where did that $cos^4phi$ came from?
By the way the question asks for the value of $a$.
integration spherical-coordinates multiple-integral
edited 2 days ago
Max
629317
asked 2 days ago
khoshrangkhoshrang
596
New contributor
khoshrang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
Hi everybody I have a triple integral I can't solve:
$$iiint sqrt x^2+y^2+z^2 ,dx ,dy ,dz $$
Which the region is between $z=sqrt x^2+y^2$ and $z=4$ .
The question says after using the spherical coordinates the answer is:
$$aLargeint _0^pi/4 frac(sin phi)dphicos^4phi $$
So I've used the spherical coordinates $0<r<4$ and $ 0<theta<2pi$ and $0<phi<fracpi2$. The things I don't get is first why is $phi$ varying between 0 and $fracpi4$ instead of $fracpi2$? And secondly where did that $cos^4phi$ came from?
By the way the question asks for the value of $a$.
integration spherical-coordinates multiple-integral
integration spherical-coordinates multiple-integral
edited 2 days ago
Max
629317
asked 2 days ago
khoshrangkhoshrang
596
New contributor
khoshrang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
Max
629317
asked 2 days ago
khoshrangkhoshrang
596
New contributor
khoshrang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 2 days ago
Max
629317
edited 2 days ago
Max
629317
edited 2 days ago
Max
629317
629317
asked 2 days ago
khoshrangkhoshrang
596
New contributor
khoshrang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 2 days ago
khoshrangkhoshrang
596
asked 2 days ago
khoshrangkhoshrang
596
596
New contributor
khoshrang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
khoshrang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
khoshrang is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$begingroup$
EDIT: ok now i understand the ϕ<π/4 it's because if we put x=0 in z=√ x^2+y^2we will get z=+y and z = -y which explains the ϕ bounds
$endgroup$
– khoshrang
2 days ago
add a comment |
$begingroup$
EDIT: ok now i understand the ϕ<π/4 it's because if we put x=0 in z=√ x^2+y^2we will get z=+y and z = -y which explains the ϕ bounds
$endgroup$
– khoshrang
2 days ago
$begingroup$
EDIT: ok now i understand the ϕ<π/4 it's because if we put x=0 in z=√ x^2+y^2we will get z=+y and z = -y which explains the ϕ bounds
$endgroup$
– khoshrang
2 days ago
$begingroup$
EDIT: ok now i understand the ϕ<π/4 it's because if we put x=0 in z=√ x^2+y^2we will get z=+y and z = -y which explains the ϕ bounds
$endgroup$
– khoshrang
2 days ago
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The plane $z=4$ has equation $r=4sec phi.$ So your limits for $r$ are $0leq r leq 4sec phi.$
answered 2 days ago
B. GoddardB. Goddard
19.6k21442
$endgroup$
1
$begingroup$
wow i can't believe how much time i wasted for this only because i was trying with the cylinder "r" concept . thank you
$endgroup$
– khoshrang
2 days ago
add a comment |
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1 Answer
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1 Answer
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active
oldest
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oldest
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active
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votes
$begingroup$
The plane $z=4$ has equation $r=4sec phi.$ So your limits for $r$ are $0leq r leq 4sec phi.$
answered 2 days ago
B. GoddardB. Goddard
19.6k21442
$endgroup$
1
$begingroup$
wow i can't believe how much time i wasted for this only because i was trying with the cylinder "r" concept . thank you
$endgroup$
– khoshrang
2 days ago
add a comment |
$begingroup$
The plane $z=4$ has equation $r=4sec phi.$ So your limits for $r$ are $0leq r leq 4sec phi.$
answered 2 days ago
B. GoddardB. Goddard
19.6k21442
$endgroup$
1
$begingroup$
wow i can't believe how much time i wasted for this only because i was trying with the cylinder "r" concept . thank you
$endgroup$
– khoshrang
2 days ago
add a comment |
$begingroup$
The plane $z=4$ has equation $r=4sec phi.$ So your limits for $r$ are $0leq r leq 4sec phi.$
answered 2 days ago
B. GoddardB. Goddard
19.6k21442
$endgroup$
The plane $z=4$ has equation $r=4sec phi.$ So your limits for $r$ are $0leq r leq 4sec phi.$
answered 2 days ago
B. GoddardB. Goddard
19.6k21442
answered 2 days ago
B. GoddardB. Goddard
19.6k21442
answered 2 days ago
B. GoddardB. Goddard
19.6k21442
answered 2 days ago
B. GoddardB. Goddard
19.6k21442
19.6k21442
1
$begingroup$
wow i can't believe how much time i wasted for this only because i was trying with the cylinder "r" concept . thank you
$endgroup$
– khoshrang
2 days ago
add a comment |
1
$begingroup$
wow i can't believe how much time i wasted for this only because i was trying with the cylinder "r" concept . thank you
$endgroup$
– khoshrang
2 days ago
1
1
$begingroup$
wow i can't believe how much time i wasted for this only because i was trying with the cylinder "r" concept . thank you
$endgroup$
– khoshrang
2 days ago
$begingroup$
wow i can't believe how much time i wasted for this only because i was trying with the cylinder "r" concept . thank you
$endgroup$
– khoshrang
2 days ago
add a comment |
khoshrang is a new contributor. Be nice, and check out our Code of Conduct.
khoshrang is a new contributor. Be nice, and check out our Code of Conduct.
khoshrang is a new contributor. Be nice, and check out our Code of Conduct.
khoshrang is a new contributor. Be nice, and check out our Code of Conduct.
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$begingroup$
EDIT: ok now i understand the ϕ<π/4 it's because if we put x=0 in z=√ x^2+y^2we will get z=+y and z = -y which explains the ϕ bounds
$endgroup$
– khoshrang
2 days ago