When to include $r$ when converting to polar coordinates?
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When evaluating integrals, if you convert to polar/cylindrical coordinates, I know you have to include $r$ ($r, dr , dtheta$).
However, when you parametrize first (for line integrals, or surface integrals) do you still include $r$? For example, I'll parametrize $x = r costheta$, $y = r sintheta$, $z = z$. Plug in as $F(r(r,theta))$, and do the cross product of the partial derivatives. Is the extra $r$ already included in this process?
calculus multivariable-calculus
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Masacroso
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When evaluating integrals, if you convert to polar/cylindrical coordinates, I know you have to include $r$ ($r, dr , dtheta$).
However, when you parametrize first (for line integrals, or surface integrals) do you still include $r$? For example, I'll parametrize $x = r costheta$, $y = r sintheta$, $z = z$. Plug in as $F(r(r,theta))$, and do the cross product of the partial derivatives. Is the extra $r$ already included in this process?
calculus multivariable-calculus
edited yesterday
Masacroso
12.2k41746
asked yesterday
LtLame
63
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LtLame is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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put an example please, there are a lot of different kinds of integrals
– Masacroso
yesterday
add a comment |
up vote
1
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up vote
1
down vote
favorite
When evaluating integrals, if you convert to polar/cylindrical coordinates, I know you have to include $r$ ($r, dr , dtheta$).
However, when you parametrize first (for line integrals, or surface integrals) do you still include $r$? For example, I'll parametrize $x = r costheta$, $y = r sintheta$, $z = z$. Plug in as $F(r(r,theta))$, and do the cross product of the partial derivatives. Is the extra $r$ already included in this process?
calculus multivariable-calculus
edited yesterday
Masacroso
12.2k41746
asked yesterday
LtLame
63
New contributor
LtLame is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
When evaluating integrals, if you convert to polar/cylindrical coordinates, I know you have to include $r$ ($r, dr , dtheta$).
However, when you parametrize first (for line integrals, or surface integrals) do you still include $r$? For example, I'll parametrize $x = r costheta$, $y = r sintheta$, $z = z$. Plug in as $F(r(r,theta))$, and do the cross product of the partial derivatives. Is the extra $r$ already included in this process?
calculus multivariable-calculus
calculus multivariable-calculus
edited yesterday
Masacroso
12.2k41746
asked yesterday
LtLame
63
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edited yesterday
Masacroso
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asked yesterday
LtLame
63
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edited yesterday
Masacroso
12.2k41746
edited yesterday
Masacroso
12.2k41746
edited yesterday
Masacroso
12.2k41746
12.2k41746
asked yesterday
LtLame
63
New contributor
LtLame is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked yesterday
LtLame
63
asked yesterday
LtLame
63
63
New contributor
LtLame is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
LtLame is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
LtLame is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
put an example please, there are a lot of different kinds of integrals
– Masacroso
yesterday
add a comment |
put an example please, there are a lot of different kinds of integrals
– Masacroso
yesterday
put an example please, there are a lot of different kinds of integrals
– Masacroso
yesterday
put an example please, there are a lot of different kinds of integrals
– Masacroso
yesterday
add a comment |
1 Answer
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In the conversion $dxdy=rdrdtheta$, the factor of $r$ is a Jacobian determinant. This generalises the result $du=u'dx$ in a single-integral substitution. You always need to include Jacobians; the real question is what the Jacobian should be for a particular problem. In general, a switch between two sets of variables $u_i,,v_j$ has $d^n u = |det J|d^n v$ with $J_{ij}:=frac{partial u_i}{partial v_j}$. I recommend proving $dxdy=rdrdtheta$ as an exercise.
answered yesterday
J.G.
18.4k21932
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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
In the conversion $dxdy=rdrdtheta$, the factor of $r$ is a Jacobian determinant. This generalises the result $du=u'dx$ in a single-integral substitution. You always need to include Jacobians; the real question is what the Jacobian should be for a particular problem. In general, a switch between two sets of variables $u_i,,v_j$ has $d^n u = |det J|d^n v$ with $J_{ij}:=frac{partial u_i}{partial v_j}$. I recommend proving $dxdy=rdrdtheta$ as an exercise.
answered yesterday
J.G.
18.4k21932
add a comment |
up vote
0
down vote
In the conversion $dxdy=rdrdtheta$, the factor of $r$ is a Jacobian determinant. This generalises the result $du=u'dx$ in a single-integral substitution. You always need to include Jacobians; the real question is what the Jacobian should be for a particular problem. In general, a switch between two sets of variables $u_i,,v_j$ has $d^n u = |det J|d^n v$ with $J_{ij}:=frac{partial u_i}{partial v_j}$. I recommend proving $dxdy=rdrdtheta$ as an exercise.
answered yesterday
J.G.
18.4k21932
add a comment |
up vote
0
down vote
up vote
0
down vote
In the conversion $dxdy=rdrdtheta$, the factor of $r$ is a Jacobian determinant. This generalises the result $du=u'dx$ in a single-integral substitution. You always need to include Jacobians; the real question is what the Jacobian should be for a particular problem. In general, a switch between two sets of variables $u_i,,v_j$ has $d^n u = |det J|d^n v$ with $J_{ij}:=frac{partial u_i}{partial v_j}$. I recommend proving $dxdy=rdrdtheta$ as an exercise.
answered yesterday
J.G.
18.4k21932
In the conversion $dxdy=rdrdtheta$, the factor of $r$ is a Jacobian determinant. This generalises the result $du=u'dx$ in a single-integral substitution. You always need to include Jacobians; the real question is what the Jacobian should be for a particular problem. In general, a switch between two sets of variables $u_i,,v_j$ has $d^n u = |det J|d^n v$ with $J_{ij}:=frac{partial u_i}{partial v_j}$. I recommend proving $dxdy=rdrdtheta$ as an exercise.
answered yesterday
J.G.
18.4k21932
answered yesterday
J.G.
18.4k21932
answered yesterday
J.G.
18.4k21932
answered yesterday
J.G.
18.4k21932
18.4k21932
add a comment |
add a comment |
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put an example please, there are a lot of different kinds of integrals
– Masacroso
yesterday