Using Stokes' Theorem to evaluate an integral around a triangular path
$begingroup$
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
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
asked Aug 13 '14 at 22:08
Swamp GSwamp G
142310
$endgroup$
add a comment |
$begingroup$
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
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
asked Aug 13 '14 at 22:08
Swamp GSwamp G
142310
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
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
asked Aug 13 '14 at 22:08
Swamp GSwamp G
142310
$endgroup$
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
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
asked Aug 13 '14 at 22:08
Swamp GSwamp G
142310
asked Aug 13 '14 at 22:08
Swamp GSwamp G
142310
asked Aug 13 '14 at 22:08
Swamp GSwamp G
142310
asked Aug 13 '14 at 22:08
Swamp GSwamp G
142310
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
add a comment |
$begingroup$
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
add a comment |
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!
answered Aug 13 '14 at 22:37
Swamp GSwamp G
142310
$endgroup$
add a comment |
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1 Answer
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1 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!
answered Aug 13 '14 at 22:37
Swamp GSwamp G
142310
$endgroup$
add a comment |
$begingroup$
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!
answered Aug 13 '14 at 22:37
Swamp GSwamp G
142310
$endgroup$
add a comment |
$begingroup$
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!
answered Aug 13 '14 at 22:37
Swamp GSwamp G
142310
$endgroup$
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!
answered Aug 13 '14 at 22:37
Swamp GSwamp G
142310
answered Aug 13 '14 at 22:37
Swamp GSwamp G
142310
answered Aug 13 '14 at 22:37
Swamp GSwamp G
142310
answered Aug 13 '14 at 22:37
Swamp GSwamp G
142310
142310
add a comment |
add a comment |
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$begingroup$
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