Using Stokes' Theorem to evaluate an integral around a triangular path












3












$begingroup$


Problem:



Use Stokes’ theorem to evaluate the integral
$I = intlimits_C textbf{F} centerdot textbf{ds}$
when $textbf{F}$ is the vector field
$textbf{F} = 3zxtextbf{i} + 3xytextbf{j} + yztextbf{k}$
and C is the path consisting of the three edges
of the triangle $∆ABC$ shown in



enter image description here



formed by the portion of the plane
$x + y + z = 1$
in the first octant of 3-space, oriented as
shown.



Attempt at Solution:



I know that by Stokes' theorem, $intlimits_C textbf{F} centerdot textbf{ds} = iintlimits_S (nabla times textbf{F}) centerdot textbf{dS}$, where $S$ is the surface closed by $C$. So, evaluating $nabla times textbf{F}$, I found it to be equal to $ztextbf{i} + 3xtextbf{j} + 3ytextbf{k}$. Substituting $z = 1 - x - y$, $nabla times textbf{F} = (1 - x - y)textbf{i} + 3xtextbf{j} + 3ytextbf{k}$.



Next, to find $textbf{dS}$, I took the gradient of $x + y + z = 1$, which turned out to be equal to $<1, 1, 1>$. Thus, $(nabla times textbf{F}) centerdot textbf{dS} = 1 + 2x + 2y$.



For the limits of integration, I used the projection of the surface onto the $xy$-plane, which produced the image of a triangle bounded by $x = 0$, $y = 0$, and $y = -x$. Accordingly, I chose the limits of integration as follows: $int_0^1 int_0^{-x} (1 + 2x + 2y)dydx$. After performing integration, I found the result to be $-frac{5}{6}$. However, this answer was incorrect. Where did I go wrong?






integration







share|cite|improve this question









$endgroup$












  • $begingroup$
    was it off by a factor of $sqrt{3}$?
    $endgroup$
    – user66081
    Aug 13 '14 at 22:16












  • $begingroup$
    Unfortunately, I am not sure. The possible answers (multiple choice) are $I = $ 1, $frac{1}{2}$, $frac{5}{6}$, $frac{2}{3}$, or $frac{7}{6}$. I attempted $frac{5}{6}$ thinking I had misplaced a negative sign in my calculations, but I was incorrect.
    $endgroup$
    – Swamp G
    Aug 13 '14 at 22:22








  • 1




    $begingroup$
    it is $y = 1-x$
    $endgroup$
    – user66081
    Aug 13 '14 at 22:29










  • $begingroup$
    Why is it not off by a factor of root $3$? You are supposed to dot the curl with a unit normal and $<1,1,1>$ is not unit
    $endgroup$
    – Sorfosh
    Aug 19 '18 at 3:02










  • $begingroup$
    @Sorfosh But then you will have to calculate the scalar $dS$, which, with the parametrization $(x, y, 1 - x - y)$, will give $dS = sqrt 3 ,dx dy$.
    $endgroup$
    – Maxim
    Sep 24 '18 at 21:24


















3












$begingroup$


Problem:



Use Stokes’ theorem to evaluate the integral
$I = intlimits_C textbf{F} centerdot textbf{ds}$
when $textbf{F}$ is the vector field
$textbf{F} = 3zxtextbf{i} + 3xytextbf{j} + yztextbf{k}$
and C is the path consisting of the three edges
of the triangle $∆ABC$ shown in



enter image description here



formed by the portion of the plane
$x + y + z = 1$
in the first octant of 3-space, oriented as
shown.



Attempt at Solution:



I know that by Stokes' theorem, $intlimits_C textbf{F} centerdot textbf{ds} = iintlimits_S (nabla times textbf{F}) centerdot textbf{dS}$, where $S$ is the surface closed by $C$. So, evaluating $nabla times textbf{F}$, I found it to be equal to $ztextbf{i} + 3xtextbf{j} + 3ytextbf{k}$. Substituting $z = 1 - x - y$, $nabla times textbf{F} = (1 - x - y)textbf{i} + 3xtextbf{j} + 3ytextbf{k}$.



Next, to find $textbf{dS}$, I took the gradient of $x + y + z = 1$, which turned out to be equal to $<1, 1, 1>$. Thus, $(nabla times textbf{F}) centerdot textbf{dS} = 1 + 2x + 2y$.



For the limits of integration, I used the projection of the surface onto the $xy$-plane, which produced the image of a triangle bounded by $x = 0$, $y = 0$, and $y = -x$. Accordingly, I chose the limits of integration as follows: $int_0^1 int_0^{-x} (1 + 2x + 2y)dydx$. After performing integration, I found the result to be $-frac{5}{6}$. However, this answer was incorrect. Where did I go wrong?






integration







share|cite|improve this question









$endgroup$












  • $begingroup$
    was it off by a factor of $sqrt{3}$?
    $endgroup$
    – user66081
    Aug 13 '14 at 22:16












  • $begingroup$
    Unfortunately, I am not sure. The possible answers (multiple choice) are $I = $ 1, $frac{1}{2}$, $frac{5}{6}$, $frac{2}{3}$, or $frac{7}{6}$. I attempted $frac{5}{6}$ thinking I had misplaced a negative sign in my calculations, but I was incorrect.
    $endgroup$
    – Swamp G
    Aug 13 '14 at 22:22








  • 1




    $begingroup$
    it is $y = 1-x$
    $endgroup$
    – user66081
    Aug 13 '14 at 22:29










  • $begingroup$
    Why is it not off by a factor of root $3$? You are supposed to dot the curl with a unit normal and $<1,1,1>$ is not unit
    $endgroup$
    – Sorfosh
    Aug 19 '18 at 3:02










  • $begingroup$
    @Sorfosh But then you will have to calculate the scalar $dS$, which, with the parametrization $(x, y, 1 - x - y)$, will give $dS = sqrt 3 ,dx dy$.
    $endgroup$
    – Maxim
    Sep 24 '18 at 21:24
















3












3








3


1



$begingroup$


Problem:



Use Stokes’ theorem to evaluate the integral
$I = intlimits_C textbf{F} centerdot textbf{ds}$
when $textbf{F}$ is the vector field
$textbf{F} = 3zxtextbf{i} + 3xytextbf{j} + yztextbf{k}$
and C is the path consisting of the three edges
of the triangle $∆ABC$ shown in



enter image description here



formed by the portion of the plane
$x + y + z = 1$
in the first octant of 3-space, oriented as
shown.



Attempt at Solution:



I know that by Stokes' theorem, $intlimits_C textbf{F} centerdot textbf{ds} = iintlimits_S (nabla times textbf{F}) centerdot textbf{dS}$, where $S$ is the surface closed by $C$. So, evaluating $nabla times textbf{F}$, I found it to be equal to $ztextbf{i} + 3xtextbf{j} + 3ytextbf{k}$. Substituting $z = 1 - x - y$, $nabla times textbf{F} = (1 - x - y)textbf{i} + 3xtextbf{j} + 3ytextbf{k}$.



Next, to find $textbf{dS}$, I took the gradient of $x + y + z = 1$, which turned out to be equal to $<1, 1, 1>$. Thus, $(nabla times textbf{F}) centerdot textbf{dS} = 1 + 2x + 2y$.



For the limits of integration, I used the projection of the surface onto the $xy$-plane, which produced the image of a triangle bounded by $x = 0$, $y = 0$, and $y = -x$. Accordingly, I chose the limits of integration as follows: $int_0^1 int_0^{-x} (1 + 2x + 2y)dydx$. After performing integration, I found the result to be $-frac{5}{6}$. However, this answer was incorrect. Where did I go wrong?






integration







share|cite|improve this question









$endgroup$




Problem:



Use Stokes’ theorem to evaluate the integral
$I = intlimits_C textbf{F} centerdot textbf{ds}$
when $textbf{F}$ is the vector field
$textbf{F} = 3zxtextbf{i} + 3xytextbf{j} + yztextbf{k}$
and C is the path consisting of the three edges
of the triangle $∆ABC$ shown in



enter image description here



formed by the portion of the plane
$x + y + z = 1$
in the first octant of 3-space, oriented as
shown.



Attempt at Solution:



I know that by Stokes' theorem, $intlimits_C textbf{F} centerdot textbf{ds} = iintlimits_S (nabla times textbf{F}) centerdot textbf{dS}$, where $S$ is the surface closed by $C$. So, evaluating $nabla times textbf{F}$, I found it to be equal to $ztextbf{i} + 3xtextbf{j} + 3ytextbf{k}$. Substituting $z = 1 - x - y$, $nabla times textbf{F} = (1 - x - y)textbf{i} + 3xtextbf{j} + 3ytextbf{k}$.



Next, to find $textbf{dS}$, I took the gradient of $x + y + z = 1$, which turned out to be equal to $<1, 1, 1>$. Thus, $(nabla times textbf{F}) centerdot textbf{dS} = 1 + 2x + 2y$.



For the limits of integration, I used the projection of the surface onto the $xy$-plane, which produced the image of a triangle bounded by $x = 0$, $y = 0$, and $y = -x$. Accordingly, I chose the limits of integration as follows: $int_0^1 int_0^{-x} (1 + 2x + 2y)dydx$. After performing integration, I found the result to be $-frac{5}{6}$. However, this answer was incorrect. Where did I go wrong?






integration




integration






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Aug 13 '14 at 22:08









Swamp GSwamp G

142310




142310












  • $begingroup$
    was it off by a factor of $sqrt{3}$?
    $endgroup$
    – user66081
    Aug 13 '14 at 22:16












  • $begingroup$
    Unfortunately, I am not sure. The possible answers (multiple choice) are $I = $ 1, $frac{1}{2}$, $frac{5}{6}$, $frac{2}{3}$, or $frac{7}{6}$. I attempted $frac{5}{6}$ thinking I had misplaced a negative sign in my calculations, but I was incorrect.
    $endgroup$
    – Swamp G
    Aug 13 '14 at 22:22








  • 1




    $begingroup$
    it is $y = 1-x$
    $endgroup$
    – user66081
    Aug 13 '14 at 22:29










  • $begingroup$
    Why is it not off by a factor of root $3$? You are supposed to dot the curl with a unit normal and $<1,1,1>$ is not unit
    $endgroup$
    – Sorfosh
    Aug 19 '18 at 3:02










  • $begingroup$
    @Sorfosh But then you will have to calculate the scalar $dS$, which, with the parametrization $(x, y, 1 - x - y)$, will give $dS = sqrt 3 ,dx dy$.
    $endgroup$
    – Maxim
    Sep 24 '18 at 21:24




















  • $begingroup$
    was it off by a factor of $sqrt{3}$?
    $endgroup$
    – user66081
    Aug 13 '14 at 22:16












  • $begingroup$
    Unfortunately, I am not sure. The possible answers (multiple choice) are $I = $ 1, $frac{1}{2}$, $frac{5}{6}$, $frac{2}{3}$, or $frac{7}{6}$. I attempted $frac{5}{6}$ thinking I had misplaced a negative sign in my calculations, but I was incorrect.
    $endgroup$
    – Swamp G
    Aug 13 '14 at 22:22








  • 1




    $begingroup$
    it is $y = 1-x$
    $endgroup$
    – user66081
    Aug 13 '14 at 22:29










  • $begingroup$
    Why is it not off by a factor of root $3$? You are supposed to dot the curl with a unit normal and $<1,1,1>$ is not unit
    $endgroup$
    – Sorfosh
    Aug 19 '18 at 3:02










  • $begingroup$
    @Sorfosh But then you will have to calculate the scalar $dS$, which, with the parametrization $(x, y, 1 - x - y)$, will give $dS = sqrt 3 ,dx dy$.
    $endgroup$
    – Maxim
    Sep 24 '18 at 21:24


















$begingroup$
was it off by a factor of $sqrt{3}$?
$endgroup$
– user66081
Aug 13 '14 at 22:16






$begingroup$
was it off by a factor of $sqrt{3}$?
$endgroup$
– user66081
Aug 13 '14 at 22:16














$begingroup$
Unfortunately, I am not sure. The possible answers (multiple choice) are $I = $ 1, $frac{1}{2}$, $frac{5}{6}$, $frac{2}{3}$, or $frac{7}{6}$. I attempted $frac{5}{6}$ thinking I had misplaced a negative sign in my calculations, but I was incorrect.
$endgroup$
– Swamp G
Aug 13 '14 at 22:22






$begingroup$
Unfortunately, I am not sure. The possible answers (multiple choice) are $I = $ 1, $frac{1}{2}$, $frac{5}{6}$, $frac{2}{3}$, or $frac{7}{6}$. I attempted $frac{5}{6}$ thinking I had misplaced a negative sign in my calculations, but I was incorrect.
$endgroup$
– Swamp G
Aug 13 '14 at 22:22






1




1




$begingroup$
it is $y = 1-x$
$endgroup$
– user66081
Aug 13 '14 at 22:29




$begingroup$
it is $y = 1-x$
$endgroup$
– user66081
Aug 13 '14 at 22:29












$begingroup$
Why is it not off by a factor of root $3$? You are supposed to dot the curl with a unit normal and $<1,1,1>$ is not unit
$endgroup$
– Sorfosh
Aug 19 '18 at 3:02




$begingroup$
Why is it not off by a factor of root $3$? You are supposed to dot the curl with a unit normal and $<1,1,1>$ is not unit
$endgroup$
– Sorfosh
Aug 19 '18 at 3:02












$begingroup$
@Sorfosh But then you will have to calculate the scalar $dS$, which, with the parametrization $(x, y, 1 - x - y)$, will give $dS = sqrt 3 ,dx dy$.
$endgroup$
– Maxim
Sep 24 '18 at 21:24






$begingroup$
@Sorfosh But then you will have to calculate the scalar $dS$, which, with the parametrization $(x, y, 1 - x - y)$, will give $dS = sqrt 3 ,dx dy$.
$endgroup$
– Maxim
Sep 24 '18 at 21:24












1 Answer
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As user66081 pointed out, I made the simple mistake of setting the upper $y$-boundary in the iterated integral to $-x$ instead of $1-x$. Correcting this, I found that $int_0^1int_0^{1-x} (1 + 2x + 2y)dydx = frac{7}{6}$. Thank you!






share|cite|improve this answer









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    $begingroup$

    As user66081 pointed out, I made the simple mistake of setting the upper $y$-boundary in the iterated integral to $-x$ instead of $1-x$. Correcting this, I found that $int_0^1int_0^{1-x} (1 + 2x + 2y)dydx = frac{7}{6}$. Thank you!






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      As user66081 pointed out, I made the simple mistake of setting the upper $y$-boundary in the iterated integral to $-x$ instead of $1-x$. Correcting this, I found that $int_0^1int_0^{1-x} (1 + 2x + 2y)dydx = frac{7}{6}$. Thank you!






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        As user66081 pointed out, I made the simple mistake of setting the upper $y$-boundary in the iterated integral to $-x$ instead of $1-x$. Correcting this, I found that $int_0^1int_0^{1-x} (1 + 2x + 2y)dydx = frac{7}{6}$. Thank you!






        share|cite|improve this answer









        $endgroup$



        As user66081 pointed out, I made the simple mistake of setting the upper $y$-boundary in the iterated integral to $-x$ instead of $1-x$. Correcting this, I found that $int_0^1int_0^{1-x} (1 + 2x + 2y)dydx = frac{7}{6}$. Thank you!







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Aug 13 '14 at 22:37









        Swamp GSwamp G

        142310




        142310






























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