Line Integral Harmonization
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
asked 2 days ago
user10478
405211
add a comment |
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
asked 2 days ago
user10478
405211
add a comment |
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
asked 2 days ago
user10478
405211
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
asked 2 days ago
user10478
405211
asked 2 days ago
user10478
405211
asked 2 days ago
user10478
405211
asked 2 days ago
user10478
405211
asked 2 days ago
user10478
405211
405211
add a comment |
add a comment |
2 Answers
2
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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.
answered 2 days ago
Tim
1
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
add a comment |
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.
answered 2 days ago
Emilio Novati
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
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
answered 2 days ago
Tim
1
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
add a comment |
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.
answered 2 days ago
Tim
1
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
add a comment |
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.
answered 2 days ago
Tim
1
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.
answered 2 days ago
Tim
1
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
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
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.
Check out our Code of Conduct.
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.
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
add a comment |
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
add a comment |
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.
answered 2 days ago
Emilio Novati
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
add a comment |
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.
answered 2 days ago
Emilio Novati
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
add a comment |
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.
answered 2 days ago
Emilio Novati
51.5k43472
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.
answered 2 days ago
Emilio Novati
51.5k43472
answered 2 days ago
Emilio Novati
51.5k43472
answered 2 days ago
Emilio Novati
51.5k43472
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
add a comment |
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
add a comment |
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