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

13k41746




13k41746










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

13k41746




13k41746












  • $begingroup$
    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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Quarter-circle Tiles

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up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
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Mont Emei

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Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
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