When to include $r$ when converting to polar coordinates?












1














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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  • put an example please, there are a lot of different kinds of integrals
    – Masacroso
    Nov 20 '18 at 22:13
















1














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







share|cite|improve this question
























  • put an example please, there are a lot of different kinds of integrals
    – Masacroso
    Nov 20 '18 at 22:13














1












1








1







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







share|cite|improve this question















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






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edited Nov 20 '18 at 22:11









Masacroso

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asked Nov 20 '18 at 22:03









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  • put an example please, there are a lot of different kinds of integrals
    – Masacroso
    Nov 20 '18 at 22:13


















  • put an example please, there are a lot of different kinds of integrals
    – Masacroso
    Nov 20 '18 at 22:13
















put an example please, there are a lot of different kinds of integrals
– Masacroso
Nov 20 '18 at 22:13




put an example please, there are a lot of different kinds of integrals
– Masacroso
Nov 20 '18 at 22:13










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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.






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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.






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      0














      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.






      share|cite|improve this 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.






        share|cite|improve this answer












        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.







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        answered Nov 20 '18 at 22:18









        J.G.

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