Line Integrals of Vector Fields, Homework Conundrum












0












$begingroup$


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?
    $endgroup$
    – 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
















0












$begingroup$


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







share|cite|improve this question









$endgroup$








  • 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














0












0








0





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










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














  • 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










1 Answer
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0












$begingroup$

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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    1 Answer
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    1 Answer
    1






    active

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    active

    oldest

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    active

    oldest

    votes









    0












    $begingroup$

    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}






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      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}






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        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}






        share|cite|improve this answer









        $endgroup$



        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}







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 10 '18 at 7:43









        Davis RashDavis Rash

        357412




        357412






























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