Mixing
Line Integral Shift
$begingroup$
How does the integration identity $int_{a + c}^{b + c} f(x - c) dx = int_a^b f(x) dx$ extend to line integrals? Would a line integral of $f(x(t), y(t))$ shifted to $a + c leq t leq b + c$ become $f(x(t) - c, y(t) - c)$, $f(x(t - c), y(t - c))$, or something different? Does this depend on whether the line integral is with respect to $x$ or $y$ (scalar field), arclength (scalar field), or $vec r$ (vector field), and in each case, which aspects of the problem's geometry would be altered by the shift, and which would be invariant?
calculus integration geometry multivariable-calculus
asked Jan 7 at 20:53
user10478user10478
448211
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add a comment |
$begingroup$
How does the integration identity $int_{a + c}^{b + c} f(x - c) dx = int_a^b f(x) dx$ extend to line integrals? Would a line integral of $f(x(t), y(t))$ shifted to $a + c leq t leq b + c$ become $f(x(t) - c, y(t) - c)$, $f(x(t - c), y(t - c))$, or something different? Does this depend on whether the line integral is with respect to $x$ or $y$ (scalar field), arclength (scalar field), or $vec r$ (vector field), and in each case, which aspects of the problem's geometry would be altered by the shift, and which would be invariant?
calculus integration geometry multivariable-calculus
asked Jan 7 at 20:53
user10478user10478
448211
$endgroup$
add a comment |
$begingroup$
How does the integration identity $int_{a + c}^{b + c} f(x - c) dx = int_a^b f(x) dx$ extend to line integrals? Would a line integral of $f(x(t), y(t))$ shifted to $a + c leq t leq b + c$ become $f(x(t) - c, y(t) - c)$, $f(x(t - c), y(t - c))$, or something different? Does this depend on whether the line integral is with respect to $x$ or $y$ (scalar field), arclength (scalar field), or $vec r$ (vector field), and in each case, which aspects of the problem's geometry would be altered by the shift, and which would be invariant?
calculus integration geometry multivariable-calculus
asked Jan 7 at 20:53
user10478user10478
448211
$endgroup$
How does the integration identity $int_{a + c}^{b + c} f(x - c) dx = int_a^b f(x) dx$ extend to line integrals? Would a line integral of $f(x(t), y(t))$ shifted to $a + c leq t leq b + c$ become $f(x(t) - c, y(t) - c)$, $f(x(t - c), y(t - c))$, or something different? Does this depend on whether the line integral is with respect to $x$ or $y$ (scalar field), arclength (scalar field), or $vec r$ (vector field), and in each case, which aspects of the problem's geometry would be altered by the shift, and which would be invariant?
calculus integration geometry multivariable-calculus
calculus integration geometry multivariable-calculus
asked Jan 7 at 20:53
user10478user10478
448211
asked Jan 7 at 20:53
user10478user10478
448211
asked Jan 7 at 20:53
user10478user10478
448211
asked Jan 7 at 20:53
user10478user10478
448211
asked Jan 7 at 20:53
user10478user10478
448211
448211
add a comment |
add a comment |
1 Answer
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It should become $f(x(t-c),y(t-c))$. The line integral would be with respect to the variable that parameterises the line, namely $t$.
$$int_{tin[a,b]}f(x(t),y(t))mathrm dt=int_{tin[a+c,b+c]}f(x(t-c),y(t-c))mathrm dt$$
The reason why this works is because, as with the one dimensional example, changing the variables in this way keeps the path that we integrate along the same.
In the notation $int_{mathcal C}f(vec x),mathrm ds$, this choice of parameterisation is hidden, so no change of variables happen here.
answered Jan 7 at 21:45
John DoeJohn Doe
11.1k11238
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$begingroup$
It should become $f(x(t-c),y(t-c))$. The line integral would be with respect to the variable that parameterises the line, namely $t$.
$$int_{tin[a,b]}f(x(t),y(t))mathrm dt=int_{tin[a+c,b+c]}f(x(t-c),y(t-c))mathrm dt$$
The reason why this works is because, as with the one dimensional example, changing the variables in this way keeps the path that we integrate along the same.
In the notation $int_{mathcal C}f(vec x),mathrm ds$, this choice of parameterisation is hidden, so no change of variables happen here.
answered Jan 7 at 21:45
John DoeJohn Doe
11.1k11238
$endgroup$
add a comment |
$begingroup$
It should become $f(x(t-c),y(t-c))$. The line integral would be with respect to the variable that parameterises the line, namely $t$.
$$int_{tin[a,b]}f(x(t),y(t))mathrm dt=int_{tin[a+c,b+c]}f(x(t-c),y(t-c))mathrm dt$$
The reason why this works is because, as with the one dimensional example, changing the variables in this way keeps the path that we integrate along the same.
In the notation $int_{mathcal C}f(vec x),mathrm ds$, this choice of parameterisation is hidden, so no change of variables happen here.
answered Jan 7 at 21:45
John DoeJohn Doe
11.1k11238
$endgroup$
add a comment |
$begingroup$
It should become $f(x(t-c),y(t-c))$. The line integral would be with respect to the variable that parameterises the line, namely $t$.
$$int_{tin[a,b]}f(x(t),y(t))mathrm dt=int_{tin[a+c,b+c]}f(x(t-c),y(t-c))mathrm dt$$
The reason why this works is because, as with the one dimensional example, changing the variables in this way keeps the path that we integrate along the same.
In the notation $int_{mathcal C}f(vec x),mathrm ds$, this choice of parameterisation is hidden, so no change of variables happen here.
answered Jan 7 at 21:45
John DoeJohn Doe
11.1k11238
$endgroup$
It should become $f(x(t-c),y(t-c))$. The line integral would be with respect to the variable that parameterises the line, namely $t$.
$$int_{tin[a,b]}f(x(t),y(t))mathrm dt=int_{tin[a+c,b+c]}f(x(t-c),y(t-c))mathrm dt$$
The reason why this works is because, as with the one dimensional example, changing the variables in this way keeps the path that we integrate along the same.
In the notation $int_{mathcal C}f(vec x),mathrm ds$, this choice of parameterisation is hidden, so no change of variables happen here.
answered Jan 7 at 21:45
John DoeJohn Doe
11.1k11238
answered Jan 7 at 21:45
John DoeJohn Doe
11.1k11238
answered Jan 7 at 21:45
John DoeJohn Doe
11.1k11238
answered Jan 7 at 21:45
John DoeJohn Doe
11.1k11238
11.1k11238
add a comment |
add a comment |
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