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Line integral on triangle with given vertices
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Let $F$ be a vector field such that $$vecF=langle x^2,y^2,zrangle$$ Integration over the line segments which form the triangle with vertices $(0,0,0)$,$(0,2,0)$,$(0,0,2)$ can be achieved by parametrizing each segment and then evaluating $$sum_i=1^3int_C_ivecF(r_i^(x)(t),r_i^(y)(t),r_i^(z)(t))cdot vecr^prime_i(t),dt$$ with $vecr_i(t)$ being the parametrization of the $i$-th segment $C_i$.
However, taking into account that $F$ is a vector field with potential function $$f(x,y,z)=frac2x^3+2y^3-3z^26$$ and thus conservative, according to the generalised fundumental theorem of line integrals, since all the components $F_1(x,y,z),F_2(x,y,z),F_3(x,y,z)$ of $vecF$ are continuous in $mathbbR^3$, it follows that $$oint_CvecF,dvecr=fbig(r^(x)(0),r^(y)(0),r^(z)(0)big)-fbig(r^(x)(0),r^(y)(0),r^(z)(0)big)=0$$ given that $r(t)$ is a parametrization of $C$ (the whole triangle), which is a closed loop (same starting and endpoint). Is this true?
integration multivariable-calculus vector-analysis
asked Aug 3 at 21:00
Andrewtz98
1397
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up vote
0
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Let $F$ be a vector field such that $$vecF=langle x^2,y^2,zrangle$$ Integration over the line segments which form the triangle with vertices $(0,0,0)$,$(0,2,0)$,$(0,0,2)$ can be achieved by parametrizing each segment and then evaluating $$sum_i=1^3int_C_ivecF(r_i^(x)(t),r_i^(y)(t),r_i^(z)(t))cdot vecr^prime_i(t),dt$$ with $vecr_i(t)$ being the parametrization of the $i$-th segment $C_i$.
However, taking into account that $F$ is a vector field with potential function $$f(x,y,z)=frac2x^3+2y^3-3z^26$$ and thus conservative, according to the generalised fundumental theorem of line integrals, since all the components $F_1(x,y,z),F_2(x,y,z),F_3(x,y,z)$ of $vecF$ are continuous in $mathbbR^3$, it follows that $$oint_CvecF,dvecr=fbig(r^(x)(0),r^(y)(0),r^(z)(0)big)-fbig(r^(x)(0),r^(y)(0),r^(z)(0)big)=0$$ given that $r(t)$ is a parametrization of $C$ (the whole triangle), which is a closed loop (same starting and endpoint). Is this true?
integration multivariable-calculus vector-analysis
asked Aug 3 at 21:00
Andrewtz98
1397
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $F$ be a vector field such that $$vecF=langle x^2,y^2,zrangle$$ Integration over the line segments which form the triangle with vertices $(0,0,0)$,$(0,2,0)$,$(0,0,2)$ can be achieved by parametrizing each segment and then evaluating $$sum_i=1^3int_C_ivecF(r_i^(x)(t),r_i^(y)(t),r_i^(z)(t))cdot vecr^prime_i(t),dt$$ with $vecr_i(t)$ being the parametrization of the $i$-th segment $C_i$.
However, taking into account that $F$ is a vector field with potential function $$f(x,y,z)=frac2x^3+2y^3-3z^26$$ and thus conservative, according to the generalised fundumental theorem of line integrals, since all the components $F_1(x,y,z),F_2(x,y,z),F_3(x,y,z)$ of $vecF$ are continuous in $mathbbR^3$, it follows that $$oint_CvecF,dvecr=fbig(r^(x)(0),r^(y)(0),r^(z)(0)big)-fbig(r^(x)(0),r^(y)(0),r^(z)(0)big)=0$$ given that $r(t)$ is a parametrization of $C$ (the whole triangle), which is a closed loop (same starting and endpoint). Is this true?
integration multivariable-calculus vector-analysis
asked Aug 3 at 21:00
Andrewtz98
1397
Let $F$ be a vector field such that $$vecF=langle x^2,y^2,zrangle$$ Integration over the line segments which form the triangle with vertices $(0,0,0)$,$(0,2,0)$,$(0,0,2)$ can be achieved by parametrizing each segment and then evaluating $$sum_i=1^3int_C_ivecF(r_i^(x)(t),r_i^(y)(t),r_i^(z)(t))cdot vecr^prime_i(t),dt$$ with $vecr_i(t)$ being the parametrization of the $i$-th segment $C_i$.
However, taking into account that $F$ is a vector field with potential function $$f(x,y,z)=frac2x^3+2y^3-3z^26$$ and thus conservative, according to the generalised fundumental theorem of line integrals, since all the components $F_1(x,y,z),F_2(x,y,z),F_3(x,y,z)$ of $vecF$ are continuous in $mathbbR^3$, it follows that $$oint_CvecF,dvecr=fbig(r^(x)(0),r^(y)(0),r^(z)(0)big)-fbig(r^(x)(0),r^(y)(0),r^(z)(0)big)=0$$ given that $r(t)$ is a parametrization of $C$ (the whole triangle), which is a closed loop (same starting and endpoint). Is this true?
integration multivariable-calculus vector-analysis
asked Aug 3 at 21:00
Andrewtz98
1397
asked Aug 3 at 21:00
Andrewtz98
1397
asked Aug 3 at 21:00
Andrewtz98
1397
asked Aug 3 at 21:00
Andrewtz98
1397
1397
add a comment |Â
add a comment |Â
2 Answers
2
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oldest
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1
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Yep, this is good, as there are no singularities in the potential function.
answered Aug 3 at 21:18
Isaac Browne
3,7112928
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up vote
1
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You have the correct conclusion, but didn’t really need to compute a scalar potential to reach it. $vec F$ is defined on all of $mathbb R^3$ and a simple calculation confirms that it is irrotational, so you can conclude that it is conservative.
answered Aug 3 at 23:28
amd
25.6k2943
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Yep, this is good, as there are no singularities in the potential function.
answered Aug 3 at 21:18
Isaac Browne
3,7112928
add a comment |Â
up vote
1
down vote
accepted
Yep, this is good, as there are no singularities in the potential function.
answered Aug 3 at 21:18
Isaac Browne
3,7112928
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Yep, this is good, as there are no singularities in the potential function.
answered Aug 3 at 21:18
Isaac Browne
3,7112928
Yep, this is good, as there are no singularities in the potential function.
answered Aug 3 at 21:18
Isaac Browne
3,7112928
answered Aug 3 at 21:18
Isaac Browne
3,7112928
answered Aug 3 at 21:18
Isaac Browne
3,7112928
answered Aug 3 at 21:18
Isaac Browne
3,7112928
3,7112928
add a comment |Â
add a comment |Â
up vote
1
down vote
You have the correct conclusion, but didn’t really need to compute a scalar potential to reach it. $vec F$ is defined on all of $mathbb R^3$ and a simple calculation confirms that it is irrotational, so you can conclude that it is conservative.
answered Aug 3 at 23:28
amd
25.6k2943
add a comment |Â
up vote
1
down vote
You have the correct conclusion, but didn’t really need to compute a scalar potential to reach it. $vec F$ is defined on all of $mathbb R^3$ and a simple calculation confirms that it is irrotational, so you can conclude that it is conservative.
answered Aug 3 at 23:28
amd
25.6k2943
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You have the correct conclusion, but didn’t really need to compute a scalar potential to reach it. $vec F$ is defined on all of $mathbb R^3$ and a simple calculation confirms that it is irrotational, so you can conclude that it is conservative.
answered Aug 3 at 23:28
amd
25.6k2943
You have the correct conclusion, but didn’t really need to compute a scalar potential to reach it. $vec F$ is defined on all of $mathbb R^3$ and a simple calculation confirms that it is irrotational, so you can conclude that it is conservative.
answered Aug 3 at 23:28
amd
25.6k2943
answered Aug 3 at 23:28
amd
25.6k2943
answered Aug 3 at 23:28
amd
25.6k2943
answered Aug 3 at 23:28
amd
25.6k2943
25.6k2943
add a comment |Â
add a comment |Â
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