Line Integral Harmonization
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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
asked Jan 3 at 17:17
user10478user10478
472211
$endgroup$
add a comment |
$begingroup$
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 Jan 3 at 17:17
user10478user10478
472211
$endgroup$
add a comment |
$begingroup$
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 Jan 3 at 17:17
user10478user10478
472211
$endgroup$
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 Jan 3 at 17:17
user10478user10478
472211
asked Jan 3 at 17:17
user10478user10478
472211
asked Jan 3 at 17:17
user10478user10478
472211
asked Jan 3 at 17:17
user10478user10478
472211
asked Jan 3 at 17:17
user10478user10478
472211
472211
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
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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 Jan 3 at 19:59
TimTim
1
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$begingroup$
same comment as above
$endgroup$
– G Cab
Jan 3 at 20:26
add a comment |
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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 Jan 3 at 20:11
Emilio NovatiEmilio Novati
52.2k43474
$endgroup$
$begingroup$
not only work ! it could be for instance the resultant force on a wire subject to distributed load (pressures)
$endgroup$
– G Cab
Jan 3 at 20:26
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
$begingroup$
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 Jan 3 at 19:59
TimTim
1
$endgroup$
$begingroup$
same comment as above
$endgroup$
– G Cab
Jan 3 at 20:26
add a comment |
$begingroup$
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 Jan 3 at 19:59
TimTim
1
$endgroup$
$begingroup$
same comment as above
$endgroup$
– G Cab
Jan 3 at 20:26
add a comment |
$begingroup$
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 Jan 3 at 19:59
TimTim
1
$endgroup$
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 Jan 3 at 19:59
TimTim
1
answered Jan 3 at 19:59
TimTim
1
answered Jan 3 at 19:59
TimTim
1
answered Jan 3 at 19:59
TimTim
1
1
$begingroup$
same comment as above
$endgroup$
– G Cab
Jan 3 at 20:26
add a comment |
$begingroup$
same comment as above
$endgroup$
– G Cab
Jan 3 at 20:26
$begingroup$
same comment as above
$endgroup$
– G Cab
Jan 3 at 20:26
$begingroup$
same comment as above
$endgroup$
– G Cab
Jan 3 at 20:26
add a comment |
$begingroup$
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 Jan 3 at 20:11
Emilio NovatiEmilio Novati
52.2k43474
$endgroup$
$begingroup$
not only work ! it could be for instance the resultant force on a wire subject to distributed load (pressures)
$endgroup$
– G Cab
Jan 3 at 20:26
add a comment |
$begingroup$
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 Jan 3 at 20:11
Emilio NovatiEmilio Novati
52.2k43474
$endgroup$
$begingroup$
not only work ! it could be for instance the resultant force on a wire subject to distributed load (pressures)
$endgroup$
– G Cab
Jan 3 at 20:26
add a comment |
$begingroup$
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 Jan 3 at 20:11
Emilio NovatiEmilio Novati
52.2k43474
$endgroup$
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 Jan 3 at 20:11
Emilio NovatiEmilio Novati
52.2k43474
answered Jan 3 at 20:11
Emilio NovatiEmilio Novati
52.2k43474
answered Jan 3 at 20:11
Emilio NovatiEmilio Novati
52.2k43474
answered Jan 3 at 20:11
Emilio NovatiEmilio Novati
52.2k43474
52.2k43474
$begingroup$
not only work ! it could be for instance the resultant force on a wire subject to distributed load (pressures)
$endgroup$
– G Cab
Jan 3 at 20:26
add a comment |
$begingroup$
not only work ! it could be for instance the resultant force on a wire subject to distributed load (pressures)
$endgroup$
– G Cab
Jan 3 at 20:26
$begingroup$
not only work ! it could be for instance the resultant force on a wire subject to distributed load (pressures)
$endgroup$
– G Cab
Jan 3 at 20:26
$begingroup$
not only work ! it could be for instance the resultant force on a wire subject to distributed load (pressures)
$endgroup$
– G Cab
Jan 3 at 20:26
add a comment |
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