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
edited Nov 20 at 22:11
Masacroso
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asked Nov 20 at 22:03
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up vote
1
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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 Nov 20 at 22:11
Masacroso
12.8k41746
asked Nov 20 at 22:03
LtLame
63
put an example please, there are a lot of different kinds of integrals
– Masacroso
Nov 20 at 22:13
add a comment |
up vote
1
down vote
favorite
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 Nov 20 at 22:11
Masacroso
12.8k41746
asked Nov 20 at 22:03
LtLame
63
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 Nov 20 at 22:11
Masacroso
12.8k41746
asked Nov 20 at 22:03
LtLame
63
edited Nov 20 at 22:11
Masacroso
12.8k41746
asked Nov 20 at 22:03
LtLame
63
edited Nov 20 at 22:11
Masacroso
12.8k41746
edited Nov 20 at 22:11
Masacroso
12.8k41746
edited Nov 20 at 22:11
Masacroso
12.8k41746
12.8k41746
asked Nov 20 at 22:03
LtLame
63
asked Nov 20 at 22:03
LtLame
63
asked Nov 20 at 22:03
LtLame
63
63
put an example please, there are a lot of different kinds of integrals
– Masacroso
Nov 20 at 22:13
add a comment |
put an example please, there are a lot of different kinds of integrals
– Masacroso
Nov 20 at 22:13
put an example please, there are a lot of different kinds of integrals
– Masacroso
Nov 20 at 22:13
put an example please, there are a lot of different kinds of integrals
– Masacroso
Nov 20 at 22:13
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 Nov 20 at 22:18
J.G.
21.2k21933
add a comment |
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1 Answer
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1 Answer
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up vote
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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 Nov 20 at 22:18
J.G.
21.2k21933
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 Nov 20 at 22:18
J.G.
21.2k21933
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 Nov 20 at 22:18
J.G.
21.2k21933
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 Nov 20 at 22:18
J.G.
21.2k21933
answered Nov 20 at 22:18
J.G.
21.2k21933
answered Nov 20 at 22:18
J.G.
21.2k21933
answered Nov 20 at 22:18
J.G.
21.2k21933
21.2k21933
add a comment |
add a comment |
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put an example please, there are a lot of different kinds of integrals
– Masacroso
Nov 20 at 22:13