Line Integral Harmonization












1














Is there a connection between line integrals over scalar fields and line integrals over vector fields? For example, do the pair $f(x, y)$ and $F(x, y)$ which stand in a potential function and gradient relationship have the same line integral for any specified curve, when $F(x, y)$ is drawn in the $xy$-plane of the $3$-space where $f(x, y)$ lives? I feel like I'm missing some geometric connection that justifies calling both by the name "line integral."






calculus geometry multivariable-calculus definite-integrals line-integrals







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    Is there a connection between line integrals over scalar fields and line integrals over vector fields? For example, do the pair $f(x, y)$ and $F(x, y)$ which stand in a potential function and gradient relationship have the same line integral for any specified curve, when $F(x, y)$ is drawn in the $xy$-plane of the $3$-space where $f(x, y)$ lives? I feel like I'm missing some geometric connection that justifies calling both by the name "line integral."






    calculus geometry multivariable-calculus definite-integrals line-integrals







    share|cite|improve this question

























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      1







      Is there a connection between line integrals over scalar fields and line integrals over vector fields? For example, do the pair $f(x, y)$ and $F(x, y)$ which stand in a potential function and gradient relationship have the same line integral for any specified curve, when $F(x, y)$ is drawn in the $xy$-plane of the $3$-space where $f(x, y)$ lives? I feel like I'm missing some geometric connection that justifies calling both by the name "line integral."






      calculus geometry multivariable-calculus definite-integrals line-integrals







      share|cite|improve this question













      Is there a connection between line integrals over scalar fields and line integrals over vector fields? For example, do the pair $f(x, y)$ and $F(x, y)$ which stand in a potential function and gradient relationship have the same line integral for any specified curve, when $F(x, y)$ is drawn in the $xy$-plane of the $3$-space where $f(x, y)$ lives? I feel like I'm missing some geometric connection that justifies calling both by the name "line integral."






      calculus geometry multivariable-calculus definite-integrals line-integrals




      calculus geometry multivariable-calculus definite-integrals line-integrals






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      asked 2 days ago









      user10478

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          Line integrals over scalar and vector fields have different physical interpretations. A line integral over a scalar field is like finding the area between the graph of the curve and the xy plane. A line integral over a vector field is interpreted as work.






          share|cite|improve this answer








          New contributor




          Tim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.


















          • same comment as above
            – G Cab
            2 days ago



















          0














          The scalar field $f(x,y)$ is a function $f:mathbb{R}^2 to mathbb{R}$ and can be thinked as the ''height'' of a surface over the $xy$ plane.
          Given a line $gamma(t)=(x(t),y(t))$ parametrized by $tin[a,b]$, the line integral of $f$ on this line is defined as
          $$
          int_gamma f(x,y)ds=int_a^b f(x(t),y(t)sqrt{left(frac{dx}{dt}right)^2+left(frac{dy}{dt}right)^2} dt
          $$
          (note that here the $ds$ is a length along the path $gamma$)



          This is essentially an extension to a ''curved coordinate'' of the usual integral of a real function of a real variable and represents the area between the surface $z=f(x,y)$ and the line $gamma$, as you can see intuitively in the nice animated figure in the wikipedia.



          The integral of a vector field $vec F(x,y)$ over $gamma$ is an entirely different thing.
          In this case $vec F(x,y)$ cannot be wieved as a surface, and the intuitive meaning of the integral is not an area but the work done by the force $vec F$ on the path $gamma$.



          In this case the integral is defined as
          $$
          int_gamma vec Fcdot dvec r = int_gamma left(F_x dx+F_ydyright)=int_a^b left(F_x(x(t),y(t))frac{dx}{dt}+F_y(x(t),y(t))frac{dy}{dt} right)dt
          $$
          (note that here $dvec r$ is a vector, and $F_x dx+F_ydy$ is a differential 1-form)



          If $vec F(x,y)$ is the gradient of the scalar field $f(x,y)$ than this means that $F_x dx+F_ydy$ is an exact form and we have:



          $$
          int_gamma vec Fcdot dvec r=f(b)-f(a)
          $$
          that is essentially a generalization of the fundamental theorem of calculus.






          share|cite|improve this answer





















          • not only work ! it could be for instance the resultant force on a wire subject to distributed load (pressures)
            – G Cab
            2 days ago











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          2 Answers
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          2 Answers
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          active

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          0














          Line integrals over scalar and vector fields have different physical interpretations. A line integral over a scalar field is like finding the area between the graph of the curve and the xy plane. A line integral over a vector field is interpreted as work.






          share|cite|improve this answer








          New contributor




          Tim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.


















          • same comment as above
            – G Cab
            2 days ago
















          0














          Line integrals over scalar and vector fields have different physical interpretations. A line integral over a scalar field is like finding the area between the graph of the curve and the xy plane. A line integral over a vector field is interpreted as work.






          share|cite|improve this answer








          New contributor




          Tim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.


















          • same comment as above
            – G Cab
            2 days ago














          0












          0








          0






          Line integrals over scalar and vector fields have different physical interpretations. A line integral over a scalar field is like finding the area between the graph of the curve and the xy plane. A line integral over a vector field is interpreted as work.






          share|cite|improve this answer








          New contributor




          Tim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.









          Line integrals over scalar and vector fields have different physical interpretations. A line integral over a scalar field is like finding the area between the graph of the curve and the xy plane. A line integral over a vector field is interpreted as work.







          share|cite|improve this answer








          New contributor




          Tim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.









          share|cite|improve this answer



          share|cite|improve this answer






          New contributor




          Tim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
          Check out our Code of Conduct.









          answered 2 days ago









          Tim

          1




          1




          New contributor




          Tim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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          New contributor





          Tim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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          Tim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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          • same comment as above
            – G Cab
            2 days ago


















          • same comment as above
            – G Cab
            2 days ago
















          same comment as above
          – G Cab
          2 days ago




          same comment as above
          – G Cab
          2 days ago











          0














          The scalar field $f(x,y)$ is a function $f:mathbb{R}^2 to mathbb{R}$ and can be thinked as the ''height'' of a surface over the $xy$ plane.
          Given a line $gamma(t)=(x(t),y(t))$ parametrized by $tin[a,b]$, the line integral of $f$ on this line is defined as
          $$
          int_gamma f(x,y)ds=int_a^b f(x(t),y(t)sqrt{left(frac{dx}{dt}right)^2+left(frac{dy}{dt}right)^2} dt
          $$
          (note that here the $ds$ is a length along the path $gamma$)



          This is essentially an extension to a ''curved coordinate'' of the usual integral of a real function of a real variable and represents the area between the surface $z=f(x,y)$ and the line $gamma$, as you can see intuitively in the nice animated figure in the wikipedia.



          The integral of a vector field $vec F(x,y)$ over $gamma$ is an entirely different thing.
          In this case $vec F(x,y)$ cannot be wieved as a surface, and the intuitive meaning of the integral is not an area but the work done by the force $vec F$ on the path $gamma$.



          In this case the integral is defined as
          $$
          int_gamma vec Fcdot dvec r = int_gamma left(F_x dx+F_ydyright)=int_a^b left(F_x(x(t),y(t))frac{dx}{dt}+F_y(x(t),y(t))frac{dy}{dt} right)dt
          $$
          (note that here $dvec r$ is a vector, and $F_x dx+F_ydy$ is a differential 1-form)



          If $vec F(x,y)$ is the gradient of the scalar field $f(x,y)$ than this means that $F_x dx+F_ydy$ is an exact form and we have:



          $$
          int_gamma vec Fcdot dvec r=f(b)-f(a)
          $$
          that is essentially a generalization of the fundamental theorem of calculus.






          share|cite|improve this answer





















          • not only work ! it could be for instance the resultant force on a wire subject to distributed load (pressures)
            – G Cab
            2 days ago
















          0














          The scalar field $f(x,y)$ is a function $f:mathbb{R}^2 to mathbb{R}$ and can be thinked as the ''height'' of a surface over the $xy$ plane.
          Given a line $gamma(t)=(x(t),y(t))$ parametrized by $tin[a,b]$, the line integral of $f$ on this line is defined as
          $$
          int_gamma f(x,y)ds=int_a^b f(x(t),y(t)sqrt{left(frac{dx}{dt}right)^2+left(frac{dy}{dt}right)^2} dt
          $$
          (note that here the $ds$ is a length along the path $gamma$)



          This is essentially an extension to a ''curved coordinate'' of the usual integral of a real function of a real variable and represents the area between the surface $z=f(x,y)$ and the line $gamma$, as you can see intuitively in the nice animated figure in the wikipedia.



          The integral of a vector field $vec F(x,y)$ over $gamma$ is an entirely different thing.
          In this case $vec F(x,y)$ cannot be wieved as a surface, and the intuitive meaning of the integral is not an area but the work done by the force $vec F$ on the path $gamma$.



          In this case the integral is defined as
          $$
          int_gamma vec Fcdot dvec r = int_gamma left(F_x dx+F_ydyright)=int_a^b left(F_x(x(t),y(t))frac{dx}{dt}+F_y(x(t),y(t))frac{dy}{dt} right)dt
          $$
          (note that here $dvec r$ is a vector, and $F_x dx+F_ydy$ is a differential 1-form)



          If $vec F(x,y)$ is the gradient of the scalar field $f(x,y)$ than this means that $F_x dx+F_ydy$ is an exact form and we have:



          $$
          int_gamma vec Fcdot dvec r=f(b)-f(a)
          $$
          that is essentially a generalization of the fundamental theorem of calculus.






          share|cite|improve this answer





















          • not only work ! it could be for instance the resultant force on a wire subject to distributed load (pressures)
            – G Cab
            2 days ago














          0












          0








          0






          The scalar field $f(x,y)$ is a function $f:mathbb{R}^2 to mathbb{R}$ and can be thinked as the ''height'' of a surface over the $xy$ plane.
          Given a line $gamma(t)=(x(t),y(t))$ parametrized by $tin[a,b]$, the line integral of $f$ on this line is defined as
          $$
          int_gamma f(x,y)ds=int_a^b f(x(t),y(t)sqrt{left(frac{dx}{dt}right)^2+left(frac{dy}{dt}right)^2} dt
          $$
          (note that here the $ds$ is a length along the path $gamma$)



          This is essentially an extension to a ''curved coordinate'' of the usual integral of a real function of a real variable and represents the area between the surface $z=f(x,y)$ and the line $gamma$, as you can see intuitively in the nice animated figure in the wikipedia.



          The integral of a vector field $vec F(x,y)$ over $gamma$ is an entirely different thing.
          In this case $vec F(x,y)$ cannot be wieved as a surface, and the intuitive meaning of the integral is not an area but the work done by the force $vec F$ on the path $gamma$.



          In this case the integral is defined as
          $$
          int_gamma vec Fcdot dvec r = int_gamma left(F_x dx+F_ydyright)=int_a^b left(F_x(x(t),y(t))frac{dx}{dt}+F_y(x(t),y(t))frac{dy}{dt} right)dt
          $$
          (note that here $dvec r$ is a vector, and $F_x dx+F_ydy$ is a differential 1-form)



          If $vec F(x,y)$ is the gradient of the scalar field $f(x,y)$ than this means that $F_x dx+F_ydy$ is an exact form and we have:



          $$
          int_gamma vec Fcdot dvec r=f(b)-f(a)
          $$
          that is essentially a generalization of the fundamental theorem of calculus.






          share|cite|improve this answer












          The scalar field $f(x,y)$ is a function $f:mathbb{R}^2 to mathbb{R}$ and can be thinked as the ''height'' of a surface over the $xy$ plane.
          Given a line $gamma(t)=(x(t),y(t))$ parametrized by $tin[a,b]$, the line integral of $f$ on this line is defined as
          $$
          int_gamma f(x,y)ds=int_a^b f(x(t),y(t)sqrt{left(frac{dx}{dt}right)^2+left(frac{dy}{dt}right)^2} dt
          $$
          (note that here the $ds$ is a length along the path $gamma$)



          This is essentially an extension to a ''curved coordinate'' of the usual integral of a real function of a real variable and represents the area between the surface $z=f(x,y)$ and the line $gamma$, as you can see intuitively in the nice animated figure in the wikipedia.



          The integral of a vector field $vec F(x,y)$ over $gamma$ is an entirely different thing.
          In this case $vec F(x,y)$ cannot be wieved as a surface, and the intuitive meaning of the integral is not an area but the work done by the force $vec F$ on the path $gamma$.



          In this case the integral is defined as
          $$
          int_gamma vec Fcdot dvec r = int_gamma left(F_x dx+F_ydyright)=int_a^b left(F_x(x(t),y(t))frac{dx}{dt}+F_y(x(t),y(t))frac{dy}{dt} right)dt
          $$
          (note that here $dvec r$ is a vector, and $F_x dx+F_ydy$ is a differential 1-form)



          If $vec F(x,y)$ is the gradient of the scalar field $f(x,y)$ than this means that $F_x dx+F_ydy$ is an exact form and we have:



          $$
          int_gamma vec Fcdot dvec r=f(b)-f(a)
          $$
          that is essentially a generalization of the fundamental theorem of calculus.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered 2 days ago









          Emilio Novati

          51.5k43472




          51.5k43472












          • not only work ! it could be for instance the resultant force on a wire subject to distributed load (pressures)
            – G Cab
            2 days ago


















          • not only work ! it could be for instance the resultant force on a wire subject to distributed load (pressures)
            – G Cab
            2 days ago
















          not only work ! it could be for instance the resultant force on a wire subject to distributed load (pressures)
          – G Cab
          2 days ago




          not only work ! it could be for instance the resultant force on a wire subject to distributed load (pressures)
          – G Cab
          2 days ago


















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