Line Integrals of Vector Fields, Homework Conundrum
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I am a student and I have a conflict with a given answer in the textbook. The question is the following:
Evaluate the line integral $int_C mathbf{F} cdot dmathbf{r}$ for the given vector field $mathbf{F}$ and the specified curve $C$.
$mathbf{F} = mathbf{a} times mathbf{r}$, where $mathbf{a}$ is a constant vector, $mathbf{r} = langle x, y, z rangle$, and $C$ is a straight line segment from $mathbf{r}_1$ to $mathbf{r}_2$.
Here is my solution:
$$int_C mathbf{F} cdot dmathbf{r} = int_C (mathbf{a} times mathbf{r}) cdot dmathbf{r} = 0$$
because the triple product of coplanar vectors vanishes.
However, the solution given is $mathbf{r}_2 cdot (mathbf{a} times mathbf{r}_1)$.
multivariable-calculus line-integrals
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add a comment |
$begingroup$
I am a student and I have a conflict with a given answer in the textbook. The question is the following:
Evaluate the line integral $int_C mathbf{F} cdot dmathbf{r}$ for the given vector field $mathbf{F}$ and the specified curve $C$.
$mathbf{F} = mathbf{a} times mathbf{r}$, where $mathbf{a}$ is a constant vector, $mathbf{r} = langle x, y, z rangle$, and $C$ is a straight line segment from $mathbf{r}_1$ to $mathbf{r}_2$.
Here is my solution:
$$int_C mathbf{F} cdot dmathbf{r} = int_C (mathbf{a} times mathbf{r}) cdot dmathbf{r} = 0$$
because the triple product of coplanar vectors vanishes.
However, the solution given is $mathbf{r}_2 cdot (mathbf{a} times mathbf{r}_1)$.
multivariable-calculus line-integrals
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1
$begingroup$
I don't see a triple product of coplanar vectors. How about writing $r=r_1+t(r_2-r_1)$ and actually doing the integration?
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– Lord Shark the Unknown
Dec 10 '18 at 6:54
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@LordSharktheUnknown So I suppose I've learned that $mathbf{r}$ and $dmathbf{r}$ are not necessarily parallel. Thank you for the comment. I have solved the problem.
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– Davis Rash
Dec 10 '18 at 7:44
$begingroup$
@LordSharktheUnknown Oh my God, I have just now realized how stupid I am. Of course $mathbf{r}$ and $dmathbf{r}$ are not parallel. Shame.
$endgroup$
– Davis Rash
Dec 10 '18 at 7:47
add a comment |
$begingroup$
I am a student and I have a conflict with a given answer in the textbook. The question is the following:
Evaluate the line integral $int_C mathbf{F} cdot dmathbf{r}$ for the given vector field $mathbf{F}$ and the specified curve $C$.
$mathbf{F} = mathbf{a} times mathbf{r}$, where $mathbf{a}$ is a constant vector, $mathbf{r} = langle x, y, z rangle$, and $C$ is a straight line segment from $mathbf{r}_1$ to $mathbf{r}_2$.
Here is my solution:
$$int_C mathbf{F} cdot dmathbf{r} = int_C (mathbf{a} times mathbf{r}) cdot dmathbf{r} = 0$$
because the triple product of coplanar vectors vanishes.
However, the solution given is $mathbf{r}_2 cdot (mathbf{a} times mathbf{r}_1)$.
multivariable-calculus line-integrals
$endgroup$
I am a student and I have a conflict with a given answer in the textbook. The question is the following:
Evaluate the line integral $int_C mathbf{F} cdot dmathbf{r}$ for the given vector field $mathbf{F}$ and the specified curve $C$.
$mathbf{F} = mathbf{a} times mathbf{r}$, where $mathbf{a}$ is a constant vector, $mathbf{r} = langle x, y, z rangle$, and $C$ is a straight line segment from $mathbf{r}_1$ to $mathbf{r}_2$.
Here is my solution:
$$int_C mathbf{F} cdot dmathbf{r} = int_C (mathbf{a} times mathbf{r}) cdot dmathbf{r} = 0$$
because the triple product of coplanar vectors vanishes.
However, the solution given is $mathbf{r}_2 cdot (mathbf{a} times mathbf{r}_1)$.
multivariable-calculus line-integrals
multivariable-calculus line-integrals
asked Dec 10 '18 at 6:51
Davis RashDavis Rash
357412
357412
1
$begingroup$
I don't see a triple product of coplanar vectors. How about writing $r=r_1+t(r_2-r_1)$ and actually doing the integration?
$endgroup$
– Lord Shark the Unknown
Dec 10 '18 at 6:54
$begingroup$
@LordSharktheUnknown So I suppose I've learned that $mathbf{r}$ and $dmathbf{r}$ are not necessarily parallel. Thank you for the comment. I have solved the problem.
$endgroup$
– Davis Rash
Dec 10 '18 at 7:44
$begingroup$
@LordSharktheUnknown Oh my God, I have just now realized how stupid I am. Of course $mathbf{r}$ and $dmathbf{r}$ are not parallel. Shame.
$endgroup$
– Davis Rash
Dec 10 '18 at 7:47
add a comment |
1
$begingroup$
I don't see a triple product of coplanar vectors. How about writing $r=r_1+t(r_2-r_1)$ and actually doing the integration?
$endgroup$
– Lord Shark the Unknown
Dec 10 '18 at 6:54
$begingroup$
@LordSharktheUnknown So I suppose I've learned that $mathbf{r}$ and $dmathbf{r}$ are not necessarily parallel. Thank you for the comment. I have solved the problem.
$endgroup$
– Davis Rash
Dec 10 '18 at 7:44
$begingroup$
@LordSharktheUnknown Oh my God, I have just now realized how stupid I am. Of course $mathbf{r}$ and $dmathbf{r}$ are not parallel. Shame.
$endgroup$
– Davis Rash
Dec 10 '18 at 7:47
1
1
$begingroup$
I don't see a triple product of coplanar vectors. How about writing $r=r_1+t(r_2-r_1)$ and actually doing the integration?
$endgroup$
– Lord Shark the Unknown
Dec 10 '18 at 6:54
$begingroup$
I don't see a triple product of coplanar vectors. How about writing $r=r_1+t(r_2-r_1)$ and actually doing the integration?
$endgroup$
– Lord Shark the Unknown
Dec 10 '18 at 6:54
$begingroup$
@LordSharktheUnknown So I suppose I've learned that $mathbf{r}$ and $dmathbf{r}$ are not necessarily parallel. Thank you for the comment. I have solved the problem.
$endgroup$
– Davis Rash
Dec 10 '18 at 7:44
$begingroup$
@LordSharktheUnknown So I suppose I've learned that $mathbf{r}$ and $dmathbf{r}$ are not necessarily parallel. Thank you for the comment. I have solved the problem.
$endgroup$
– Davis Rash
Dec 10 '18 at 7:44
$begingroup$
@LordSharktheUnknown Oh my God, I have just now realized how stupid I am. Of course $mathbf{r}$ and $dmathbf{r}$ are not parallel. Shame.
$endgroup$
– Davis Rash
Dec 10 '18 at 7:47
$begingroup$
@LordSharktheUnknown Oh my God, I have just now realized how stupid I am. Of course $mathbf{r}$ and $dmathbf{r}$ are not parallel. Shame.
$endgroup$
– Davis Rash
Dec 10 '18 at 7:47
add a comment |
1 Answer
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oldest
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$begingroup$
With the help of the comment by Lord Shark the Unknown, I let
begin{align}
mathbf{r} & = mathbf{r}_1 + t(mathbf{r}_2 - mathbf{r}_1) \
Longrightarrow quad dmathbf{r} & = (mathbf{r}_2 - mathbf{r}_1),dt.
end{align}
We now have
begin{align}
mathbf{a} times mathbf{r} & = mathbf{a} times mathbf{r}_1 + t(mathbf{r}_2 - mathbf{r}_1) \
& = mathbf{a} times mathbf{r}_1 + tmathbf{a} times (mathbf{r}_2 - mathbf{r}_1).
end{align}
And finally
begin{align}
int_C mathbf{F} cdot dmathbf{r} & = int_0^1 (mathbf{a} times mathbf{r}_1 + tmathbf{a} times (mathbf{r}_2 - mathbf{r}_1)) cdot (mathbf{r}_2 - mathbf{r}_1),dt \
& = int_0^1 (mathbf{a} times mathbf{r}_1) cdot (mathbf{r}_2 - mathbf{r}_1),dt \
& = int_0^1 (mathbf{a} times mathbf{r}_1) cdot mathbf{r}_2,dt = (mathbf{a} times mathbf{r}_1) cdot mathbf{r}_2.
end{align}
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add a comment |
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1 Answer
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$begingroup$
With the help of the comment by Lord Shark the Unknown, I let
begin{align}
mathbf{r} & = mathbf{r}_1 + t(mathbf{r}_2 - mathbf{r}_1) \
Longrightarrow quad dmathbf{r} & = (mathbf{r}_2 - mathbf{r}_1),dt.
end{align}
We now have
begin{align}
mathbf{a} times mathbf{r} & = mathbf{a} times mathbf{r}_1 + t(mathbf{r}_2 - mathbf{r}_1) \
& = mathbf{a} times mathbf{r}_1 + tmathbf{a} times (mathbf{r}_2 - mathbf{r}_1).
end{align}
And finally
begin{align}
int_C mathbf{F} cdot dmathbf{r} & = int_0^1 (mathbf{a} times mathbf{r}_1 + tmathbf{a} times (mathbf{r}_2 - mathbf{r}_1)) cdot (mathbf{r}_2 - mathbf{r}_1),dt \
& = int_0^1 (mathbf{a} times mathbf{r}_1) cdot (mathbf{r}_2 - mathbf{r}_1),dt \
& = int_0^1 (mathbf{a} times mathbf{r}_1) cdot mathbf{r}_2,dt = (mathbf{a} times mathbf{r}_1) cdot mathbf{r}_2.
end{align}
$endgroup$
add a comment |
$begingroup$
With the help of the comment by Lord Shark the Unknown, I let
begin{align}
mathbf{r} & = mathbf{r}_1 + t(mathbf{r}_2 - mathbf{r}_1) \
Longrightarrow quad dmathbf{r} & = (mathbf{r}_2 - mathbf{r}_1),dt.
end{align}
We now have
begin{align}
mathbf{a} times mathbf{r} & = mathbf{a} times mathbf{r}_1 + t(mathbf{r}_2 - mathbf{r}_1) \
& = mathbf{a} times mathbf{r}_1 + tmathbf{a} times (mathbf{r}_2 - mathbf{r}_1).
end{align}
And finally
begin{align}
int_C mathbf{F} cdot dmathbf{r} & = int_0^1 (mathbf{a} times mathbf{r}_1 + tmathbf{a} times (mathbf{r}_2 - mathbf{r}_1)) cdot (mathbf{r}_2 - mathbf{r}_1),dt \
& = int_0^1 (mathbf{a} times mathbf{r}_1) cdot (mathbf{r}_2 - mathbf{r}_1),dt \
& = int_0^1 (mathbf{a} times mathbf{r}_1) cdot mathbf{r}_2,dt = (mathbf{a} times mathbf{r}_1) cdot mathbf{r}_2.
end{align}
$endgroup$
add a comment |
$begingroup$
With the help of the comment by Lord Shark the Unknown, I let
begin{align}
mathbf{r} & = mathbf{r}_1 + t(mathbf{r}_2 - mathbf{r}_1) \
Longrightarrow quad dmathbf{r} & = (mathbf{r}_2 - mathbf{r}_1),dt.
end{align}
We now have
begin{align}
mathbf{a} times mathbf{r} & = mathbf{a} times mathbf{r}_1 + t(mathbf{r}_2 - mathbf{r}_1) \
& = mathbf{a} times mathbf{r}_1 + tmathbf{a} times (mathbf{r}_2 - mathbf{r}_1).
end{align}
And finally
begin{align}
int_C mathbf{F} cdot dmathbf{r} & = int_0^1 (mathbf{a} times mathbf{r}_1 + tmathbf{a} times (mathbf{r}_2 - mathbf{r}_1)) cdot (mathbf{r}_2 - mathbf{r}_1),dt \
& = int_0^1 (mathbf{a} times mathbf{r}_1) cdot (mathbf{r}_2 - mathbf{r}_1),dt \
& = int_0^1 (mathbf{a} times mathbf{r}_1) cdot mathbf{r}_2,dt = (mathbf{a} times mathbf{r}_1) cdot mathbf{r}_2.
end{align}
$endgroup$
With the help of the comment by Lord Shark the Unknown, I let
begin{align}
mathbf{r} & = mathbf{r}_1 + t(mathbf{r}_2 - mathbf{r}_1) \
Longrightarrow quad dmathbf{r} & = (mathbf{r}_2 - mathbf{r}_1),dt.
end{align}
We now have
begin{align}
mathbf{a} times mathbf{r} & = mathbf{a} times mathbf{r}_1 + t(mathbf{r}_2 - mathbf{r}_1) \
& = mathbf{a} times mathbf{r}_1 + tmathbf{a} times (mathbf{r}_2 - mathbf{r}_1).
end{align}
And finally
begin{align}
int_C mathbf{F} cdot dmathbf{r} & = int_0^1 (mathbf{a} times mathbf{r}_1 + tmathbf{a} times (mathbf{r}_2 - mathbf{r}_1)) cdot (mathbf{r}_2 - mathbf{r}_1),dt \
& = int_0^1 (mathbf{a} times mathbf{r}_1) cdot (mathbf{r}_2 - mathbf{r}_1),dt \
& = int_0^1 (mathbf{a} times mathbf{r}_1) cdot mathbf{r}_2,dt = (mathbf{a} times mathbf{r}_1) cdot mathbf{r}_2.
end{align}
answered Dec 10 '18 at 7:43
Davis RashDavis Rash
357412
357412
add a comment |
add a comment |
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$begingroup$
I don't see a triple product of coplanar vectors. How about writing $r=r_1+t(r_2-r_1)$ and actually doing the integration?
$endgroup$
– Lord Shark the Unknown
Dec 10 '18 at 6:54
$begingroup$
@LordSharktheUnknown So I suppose I've learned that $mathbf{r}$ and $dmathbf{r}$ are not necessarily parallel. Thank you for the comment. I have solved the problem.
$endgroup$
– Davis Rash
Dec 10 '18 at 7:44
$begingroup$
@LordSharktheUnknown Oh my God, I have just now realized how stupid I am. Of course $mathbf{r}$ and $dmathbf{r}$ are not parallel. Shame.
$endgroup$
– Davis Rash
Dec 10 '18 at 7:47