Find the integral of the vector field












2












$begingroup$



Let the vector field
$$vec{F}left(vec{x}right)=begin{pmatrix}{x_1^2+2x_3}\ x_1x_2\
x_3^2-2x_1end{pmatrix}$$ Compute the integral $int _Cvec{F}left(vec{x}right)dvec{x}$ from the origin to the point
$P(1/2/3)$ if $C$ is a straight line from the origin to $P$.




So in the book they only give us answers, but not how to get the answers. I calculated the integral and got



$$int _0^112t^2-8t$$ which equals to zero, however in the book the answers is $9frac{2}{3}$. I think the book is wrong because I just don't see how we can get that answer. Or am I wrong?






calculus integration multivariable-calculus definite-integrals vectors







share|cite|improve this question











$endgroup$












  • $begingroup$
    it is $9frac23=6$ or there is some notation issue there?
    $endgroup$
    – Masacroso
    Dec 5 '18 at 1:55










  • $begingroup$
    It just says $9frac{2}{:3}:$. I dont think it equals to 6.@Masacroso
    $endgroup$
    – Naochi
    Dec 5 '18 at 2:01












  • $begingroup$
    I could also show how I got it via Photo, I think it would take me too long to type it all out :(
    $endgroup$
    – Naochi
    Dec 5 '18 at 2:05










  • $begingroup$
    I had a mistake in my previous calculations (I wrote $z^3$ instead of $z^2$), but it seems that it is correct now
    $endgroup$
    – Masacroso
    Dec 5 '18 at 2:19


















2












$begingroup$



Let the vector field
$$vec{F}left(vec{x}right)=begin{pmatrix}{x_1^2+2x_3}\ x_1x_2\
x_3^2-2x_1end{pmatrix}$$ Compute the integral $int _Cvec{F}left(vec{x}right)dvec{x}$ from the origin to the point
$P(1/2/3)$ if $C$ is a straight line from the origin to $P$.




So in the book they only give us answers, but not how to get the answers. I calculated the integral and got



$$int _0^112t^2-8t$$ which equals to zero, however in the book the answers is $9frac{2}{3}$. I think the book is wrong because I just don't see how we can get that answer. Or am I wrong?






calculus integration multivariable-calculus definite-integrals vectors







share|cite|improve this question











$endgroup$












  • $begingroup$
    it is $9frac23=6$ or there is some notation issue there?
    $endgroup$
    – Masacroso
    Dec 5 '18 at 1:55










  • $begingroup$
    It just says $9frac{2}{:3}:$. I dont think it equals to 6.@Masacroso
    $endgroup$
    – Naochi
    Dec 5 '18 at 2:01












  • $begingroup$
    I could also show how I got it via Photo, I think it would take me too long to type it all out :(
    $endgroup$
    – Naochi
    Dec 5 '18 at 2:05










  • $begingroup$
    I had a mistake in my previous calculations (I wrote $z^3$ instead of $z^2$), but it seems that it is correct now
    $endgroup$
    – Masacroso
    Dec 5 '18 at 2:19
















2












2








2





$begingroup$



Let the vector field
$$vec{F}left(vec{x}right)=begin{pmatrix}{x_1^2+2x_3}\ x_1x_2\
x_3^2-2x_1end{pmatrix}$$ Compute the integral $int _Cvec{F}left(vec{x}right)dvec{x}$ from the origin to the point
$P(1/2/3)$ if $C$ is a straight line from the origin to $P$.




So in the book they only give us answers, but not how to get the answers. I calculated the integral and got



$$int _0^112t^2-8t$$ which equals to zero, however in the book the answers is $9frac{2}{3}$. I think the book is wrong because I just don't see how we can get that answer. Or am I wrong?






calculus integration multivariable-calculus definite-integrals vectors







share|cite|improve this question











$endgroup$





Let the vector field
$$vec{F}left(vec{x}right)=begin{pmatrix}{x_1^2+2x_3}\ x_1x_2\
x_3^2-2x_1end{pmatrix}$$ Compute the integral $int _Cvec{F}left(vec{x}right)dvec{x}$ from the origin to the point
$P(1/2/3)$ if $C$ is a straight line from the origin to $P$.




So in the book they only give us answers, but not how to get the answers. I calculated the integral and got



$$int _0^112t^2-8t$$ which equals to zero, however in the book the answers is $9frac{2}{3}$. I think the book is wrong because I just don't see how we can get that answer. Or am I wrong?






calculus integration multivariable-calculus definite-integrals vectors




calculus integration multivariable-calculus definite-integrals vectors






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 5 '18 at 2:09









Masacroso

13.1k41746




13.1k41746










asked Dec 5 '18 at 1:39









NaochiNaochi

626




626












  • $begingroup$
    it is $9frac23=6$ or there is some notation issue there?
    $endgroup$
    – Masacroso
    Dec 5 '18 at 1:55










  • $begingroup$
    It just says $9frac{2}{:3}:$. I dont think it equals to 6.@Masacroso
    $endgroup$
    – Naochi
    Dec 5 '18 at 2:01












  • $begingroup$
    I could also show how I got it via Photo, I think it would take me too long to type it all out :(
    $endgroup$
    – Naochi
    Dec 5 '18 at 2:05










  • $begingroup$
    I had a mistake in my previous calculations (I wrote $z^3$ instead of $z^2$), but it seems that it is correct now
    $endgroup$
    – Masacroso
    Dec 5 '18 at 2:19




















  • $begingroup$
    it is $9frac23=6$ or there is some notation issue there?
    $endgroup$
    – Masacroso
    Dec 5 '18 at 1:55










  • $begingroup$
    It just says $9frac{2}{:3}:$. I dont think it equals to 6.@Masacroso
    $endgroup$
    – Naochi
    Dec 5 '18 at 2:01












  • $begingroup$
    I could also show how I got it via Photo, I think it would take me too long to type it all out :(
    $endgroup$
    – Naochi
    Dec 5 '18 at 2:05










  • $begingroup$
    I had a mistake in my previous calculations (I wrote $z^3$ instead of $z^2$), but it seems that it is correct now
    $endgroup$
    – Masacroso
    Dec 5 '18 at 2:19


















$begingroup$
it is $9frac23=6$ or there is some notation issue there?
$endgroup$
– Masacroso
Dec 5 '18 at 1:55




$begingroup$
it is $9frac23=6$ or there is some notation issue there?
$endgroup$
– Masacroso
Dec 5 '18 at 1:55












$begingroup$
It just says $9frac{2}{:3}:$. I dont think it equals to 6.@Masacroso
$endgroup$
– Naochi
Dec 5 '18 at 2:01






$begingroup$
It just says $9frac{2}{:3}:$. I dont think it equals to 6.@Masacroso
$endgroup$
– Naochi
Dec 5 '18 at 2:01














$begingroup$
I could also show how I got it via Photo, I think it would take me too long to type it all out :(
$endgroup$
– Naochi
Dec 5 '18 at 2:05




$begingroup$
I could also show how I got it via Photo, I think it would take me too long to type it all out :(
$endgroup$
– Naochi
Dec 5 '18 at 2:05












$begingroup$
I had a mistake in my previous calculations (I wrote $z^3$ instead of $z^2$), but it seems that it is correct now
$endgroup$
– Masacroso
Dec 5 '18 at 2:19






$begingroup$
I had a mistake in my previous calculations (I wrote $z^3$ instead of $z^2$), but it seems that it is correct now
$endgroup$
– Masacroso
Dec 5 '18 at 2:19












1 Answer
1






active

oldest

votes


















3












$begingroup$

You have that $F(x,y,z)=(x^2+2z,xy,z^2-2x)$, and you want to evaluate



$$int_C F(x,y,z)cdot (dx,dy,dz)$$



where $C$ is the straight line parametrized by $gamma:[0,1]toBbb R^3,,tmapsto t P$ for $P:=(1,2,3)$. Hence



$$int_C F(x,y,z)cdot (dx,dy,dz)=int_0^1 (Fcirc gamma)(t)cdot gamma'(t), dt\
=int_0^1 (t^2+6t), dt+2int_0^12t^2, dt+3int_0^1 (9t^2-2t), dt\
=frac13+3+2frac23+3left(3-1right)=frac53+9$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Why is $[0,1]$?
    $endgroup$
    – manooooh
    Dec 5 '18 at 2:16






  • 1




    $begingroup$
    @manooooh you can parametrize as you want. I choose parametrize using the interval $[0,1]$ because the calculus are easier
    $endgroup$
    – Masacroso
    Dec 5 '18 at 2:18










  • $begingroup$
    Oh I see now, so I did make a mistake then? Because I did it all together and did not split them up. $int _0^1begin{pmatrix}t^2-6t\ 2t^2\ 9t^2-2tend{pmatrix}cdot begin{pmatrix}1\ 2\ 3end{pmatrix}dt$ edit:oml I see my mistake now, I didn't multiply them. xD
    $endgroup$
    – Naochi
    Dec 5 '18 at 2:20












  • $begingroup$
    Ok but then your answer would be $32/3$? I guess I must recheck because I got $int _0^127t^2-6t+4t^2+t^2-6t=14/3$
    $endgroup$
    – Naochi
    Dec 5 '18 at 2:24






  • 1




    $begingroup$
    @Naochi yes, $frac{32}3=frac53+9$, this is the result. I checked my answer using Wolfram Mathematica (assuming the data of the problem is correct)
    $endgroup$
    – Masacroso
    Dec 5 '18 at 2:30













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1 Answer
1






active

oldest

votes








1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









3












$begingroup$

You have that $F(x,y,z)=(x^2+2z,xy,z^2-2x)$, and you want to evaluate



$$int_C F(x,y,z)cdot (dx,dy,dz)$$



where $C$ is the straight line parametrized by $gamma:[0,1]toBbb R^3,,tmapsto t P$ for $P:=(1,2,3)$. Hence



$$int_C F(x,y,z)cdot (dx,dy,dz)=int_0^1 (Fcirc gamma)(t)cdot gamma'(t), dt\
=int_0^1 (t^2+6t), dt+2int_0^12t^2, dt+3int_0^1 (9t^2-2t), dt\
=frac13+3+2frac23+3left(3-1right)=frac53+9$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Why is $[0,1]$?
    $endgroup$
    – manooooh
    Dec 5 '18 at 2:16






  • 1




    $begingroup$
    @manooooh you can parametrize as you want. I choose parametrize using the interval $[0,1]$ because the calculus are easier
    $endgroup$
    – Masacroso
    Dec 5 '18 at 2:18










  • $begingroup$
    Oh I see now, so I did make a mistake then? Because I did it all together and did not split them up. $int _0^1begin{pmatrix}t^2-6t\ 2t^2\ 9t^2-2tend{pmatrix}cdot begin{pmatrix}1\ 2\ 3end{pmatrix}dt$ edit:oml I see my mistake now, I didn't multiply them. xD
    $endgroup$
    – Naochi
    Dec 5 '18 at 2:20












  • $begingroup$
    Ok but then your answer would be $32/3$? I guess I must recheck because I got $int _0^127t^2-6t+4t^2+t^2-6t=14/3$
    $endgroup$
    – Naochi
    Dec 5 '18 at 2:24






  • 1




    $begingroup$
    @Naochi yes, $frac{32}3=frac53+9$, this is the result. I checked my answer using Wolfram Mathematica (assuming the data of the problem is correct)
    $endgroup$
    – Masacroso
    Dec 5 '18 at 2:30


















3












$begingroup$

You have that $F(x,y,z)=(x^2+2z,xy,z^2-2x)$, and you want to evaluate



$$int_C F(x,y,z)cdot (dx,dy,dz)$$



where $C$ is the straight line parametrized by $gamma:[0,1]toBbb R^3,,tmapsto t P$ for $P:=(1,2,3)$. Hence



$$int_C F(x,y,z)cdot (dx,dy,dz)=int_0^1 (Fcirc gamma)(t)cdot gamma'(t), dt\
=int_0^1 (t^2+6t), dt+2int_0^12t^2, dt+3int_0^1 (9t^2-2t), dt\
=frac13+3+2frac23+3left(3-1right)=frac53+9$$






share|cite|improve this answer











$endgroup$













  • $begingroup$
    Why is $[0,1]$?
    $endgroup$
    – manooooh
    Dec 5 '18 at 2:16






  • 1




    $begingroup$
    @manooooh you can parametrize as you want. I choose parametrize using the interval $[0,1]$ because the calculus are easier
    $endgroup$
    – Masacroso
    Dec 5 '18 at 2:18










  • $begingroup$
    Oh I see now, so I did make a mistake then? Because I did it all together and did not split them up. $int _0^1begin{pmatrix}t^2-6t\ 2t^2\ 9t^2-2tend{pmatrix}cdot begin{pmatrix}1\ 2\ 3end{pmatrix}dt$ edit:oml I see my mistake now, I didn't multiply them. xD
    $endgroup$
    – Naochi
    Dec 5 '18 at 2:20












  • $begingroup$
    Ok but then your answer would be $32/3$? I guess I must recheck because I got $int _0^127t^2-6t+4t^2+t^2-6t=14/3$
    $endgroup$
    – Naochi
    Dec 5 '18 at 2:24






  • 1




    $begingroup$
    @Naochi yes, $frac{32}3=frac53+9$, this is the result. I checked my answer using Wolfram Mathematica (assuming the data of the problem is correct)
    $endgroup$
    – Masacroso
    Dec 5 '18 at 2:30
















3












3








3





$begingroup$

You have that $F(x,y,z)=(x^2+2z,xy,z^2-2x)$, and you want to evaluate



$$int_C F(x,y,z)cdot (dx,dy,dz)$$



where $C$ is the straight line parametrized by $gamma:[0,1]toBbb R^3,,tmapsto t P$ for $P:=(1,2,3)$. Hence



$$int_C F(x,y,z)cdot (dx,dy,dz)=int_0^1 (Fcirc gamma)(t)cdot gamma'(t), dt\
=int_0^1 (t^2+6t), dt+2int_0^12t^2, dt+3int_0^1 (9t^2-2t), dt\
=frac13+3+2frac23+3left(3-1right)=frac53+9$$






share|cite|improve this answer











$endgroup$



You have that $F(x,y,z)=(x^2+2z,xy,z^2-2x)$, and you want to evaluate



$$int_C F(x,y,z)cdot (dx,dy,dz)$$



where $C$ is the straight line parametrized by $gamma:[0,1]toBbb R^3,,tmapsto t P$ for $P:=(1,2,3)$. Hence



$$int_C F(x,y,z)cdot (dx,dy,dz)=int_0^1 (Fcirc gamma)(t)cdot gamma'(t), dt\
=int_0^1 (t^2+6t), dt+2int_0^12t^2, dt+3int_0^1 (9t^2-2t), dt\
=frac13+3+2frac23+3left(3-1right)=frac53+9$$







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 5 '18 at 2:17

























answered Dec 5 '18 at 2:08









MasacrosoMasacroso

13.1k41746




13.1k41746












  • $begingroup$
    Why is $[0,1]$?
    $endgroup$
    – manooooh
    Dec 5 '18 at 2:16






  • 1




    $begingroup$
    @manooooh you can parametrize as you want. I choose parametrize using the interval $[0,1]$ because the calculus are easier
    $endgroup$
    – Masacroso
    Dec 5 '18 at 2:18










  • $begingroup$
    Oh I see now, so I did make a mistake then? Because I did it all together and did not split them up. $int _0^1begin{pmatrix}t^2-6t\ 2t^2\ 9t^2-2tend{pmatrix}cdot begin{pmatrix}1\ 2\ 3end{pmatrix}dt$ edit:oml I see my mistake now, I didn't multiply them. xD
    $endgroup$
    – Naochi
    Dec 5 '18 at 2:20












  • $begingroup$
    Ok but then your answer would be $32/3$? I guess I must recheck because I got $int _0^127t^2-6t+4t^2+t^2-6t=14/3$
    $endgroup$
    – Naochi
    Dec 5 '18 at 2:24






  • 1




    $begingroup$
    @Naochi yes, $frac{32}3=frac53+9$, this is the result. I checked my answer using Wolfram Mathematica (assuming the data of the problem is correct)
    $endgroup$
    – Masacroso
    Dec 5 '18 at 2:30




















  • $begingroup$
    Why is $[0,1]$?
    $endgroup$
    – manooooh
    Dec 5 '18 at 2:16






  • 1




    $begingroup$
    @manooooh you can parametrize as you want. I choose parametrize using the interval $[0,1]$ because the calculus are easier
    $endgroup$
    – Masacroso
    Dec 5 '18 at 2:18










  • $begingroup$
    Oh I see now, so I did make a mistake then? Because I did it all together and did not split them up. $int _0^1begin{pmatrix}t^2-6t\ 2t^2\ 9t^2-2tend{pmatrix}cdot begin{pmatrix}1\ 2\ 3end{pmatrix}dt$ edit:oml I see my mistake now, I didn't multiply them. xD
    $endgroup$
    – Naochi
    Dec 5 '18 at 2:20












  • $begingroup$
    Ok but then your answer would be $32/3$? I guess I must recheck because I got $int _0^127t^2-6t+4t^2+t^2-6t=14/3$
    $endgroup$
    – Naochi
    Dec 5 '18 at 2:24






  • 1




    $begingroup$
    @Naochi yes, $frac{32}3=frac53+9$, this is the result. I checked my answer using Wolfram Mathematica (assuming the data of the problem is correct)
    $endgroup$
    – Masacroso
    Dec 5 '18 at 2:30


















$begingroup$
Why is $[0,1]$?
$endgroup$
– manooooh
Dec 5 '18 at 2:16




$begingroup$
Why is $[0,1]$?
$endgroup$
– manooooh
Dec 5 '18 at 2:16




1




1




$begingroup$
@manooooh you can parametrize as you want. I choose parametrize using the interval $[0,1]$ because the calculus are easier
$endgroup$
– Masacroso
Dec 5 '18 at 2:18




$begingroup$
@manooooh you can parametrize as you want. I choose parametrize using the interval $[0,1]$ because the calculus are easier
$endgroup$
– Masacroso
Dec 5 '18 at 2:18












$begingroup$
Oh I see now, so I did make a mistake then? Because I did it all together and did not split them up. $int _0^1begin{pmatrix}t^2-6t\ 2t^2\ 9t^2-2tend{pmatrix}cdot begin{pmatrix}1\ 2\ 3end{pmatrix}dt$ edit:oml I see my mistake now, I didn't multiply them. xD
$endgroup$
– Naochi
Dec 5 '18 at 2:20






$begingroup$
Oh I see now, so I did make a mistake then? Because I did it all together and did not split them up. $int _0^1begin{pmatrix}t^2-6t\ 2t^2\ 9t^2-2tend{pmatrix}cdot begin{pmatrix}1\ 2\ 3end{pmatrix}dt$ edit:oml I see my mistake now, I didn't multiply them. xD
$endgroup$
– Naochi
Dec 5 '18 at 2:20














$begingroup$
Ok but then your answer would be $32/3$? I guess I must recheck because I got $int _0^127t^2-6t+4t^2+t^2-6t=14/3$
$endgroup$
– Naochi
Dec 5 '18 at 2:24




$begingroup$
Ok but then your answer would be $32/3$? I guess I must recheck because I got $int _0^127t^2-6t+4t^2+t^2-6t=14/3$
$endgroup$
– Naochi
Dec 5 '18 at 2:24




1




1




$begingroup$
@Naochi yes, $frac{32}3=frac53+9$, this is the result. I checked my answer using Wolfram Mathematica (assuming the data of the problem is correct)
$endgroup$
– Masacroso
Dec 5 '18 at 2:30






$begingroup$
@Naochi yes, $frac{32}3=frac53+9$, this is the result. I checked my answer using Wolfram Mathematica (assuming the data of the problem is correct)
$endgroup$
– Masacroso
Dec 5 '18 at 2:30




















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Puebla de Zaragoza

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Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc
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Musa

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Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo
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