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Given rotor and curve find circulation of a vector field
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Given the curve $C$ of equation $$vec X=(3cos t,3sin t,6cos t),qquad0leq tleq2pi$$ oriented according imposes this parameterization, find the circulation of $vec f$ along $C$ if $vec finmathcal C^1$ in $mathbb R^3$ and $$textrot vec f(x,y,z)=(0,2x-y,z),$$ where $textrot$ means curl.
First I found the curve: $$begincasesx=3cos t\y=3sin t\z=6cos tendcasesequivbegincasesx^2+y^2=9\z=6dfrac x3endcasesequivbegincasesx^2+y^2=9\z-2x=0,endcases$$ so the normal becomes $$vec N=(-2,0,1).$$ Now I have to use the Stokes theorem, that is
$$begineqnarray*
&&ointlimits_Cvec f;text dvec s&(1)\\
&=&iintlimits_SBig.nablatimesvec fBig&(2)\\
&=&iintlimits_P_xy(0,2x-y,underbrace z_2x)cdot(-2,0,1);text dxtext dy&(3)\\
&=&2iintlimits_P_xyx;text dxtext dy&(4)\\
&undersettextUsing polar coordinates=&2int_0^2pitext dthetaint_0^3rhocdotrhocostheta;text drho&(5)\\
&=&2int_0^2picostheta;text dthetaint_0^3rho^2;text drho&(6)\\
&=&2int_0^2picostheta;text dtheta;left.left(fracrho^33right)right|_0^3&(7)\\
&=&18Big.left(sinthetaright)Big|_0^2pi&(8)\\
textCirculation&=&boxed 0.&(9)
endeqnarray*$$
Is this correct?
Thank you!
vector-analysis surfaces stokes-theorem
asked Aug 2 at 23:46
manooooh
415211
 |Â
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up vote
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Given the curve $C$ of equation $$vec X=(3cos t,3sin t,6cos t),qquad0leq tleq2pi$$ oriented according imposes this parameterization, find the circulation of $vec f$ along $C$ if $vec finmathcal C^1$ in $mathbb R^3$ and $$textrot vec f(x,y,z)=(0,2x-y,z),$$ where $textrot$ means curl.
First I found the curve: $$begincasesx=3cos t\y=3sin t\z=6cos tendcasesequivbegincasesx^2+y^2=9\z=6dfrac x3endcasesequivbegincasesx^2+y^2=9\z-2x=0,endcases$$ so the normal becomes $$vec N=(-2,0,1).$$ Now I have to use the Stokes theorem, that is
$$begineqnarray*
&&ointlimits_Cvec f;text dvec s&(1)\\
&=&iintlimits_SBig.nablatimesvec fBig&(2)\\
&=&iintlimits_P_xy(0,2x-y,underbrace z_2x)cdot(-2,0,1);text dxtext dy&(3)\\
&=&2iintlimits_P_xyx;text dxtext dy&(4)\\
&undersettextUsing polar coordinates=&2int_0^2pitext dthetaint_0^3rhocdotrhocostheta;text drho&(5)\\
&=&2int_0^2picostheta;text dthetaint_0^3rho^2;text drho&(6)\\
&=&2int_0^2picostheta;text dtheta;left.left(fracrho^33right)right|_0^3&(7)\\
&=&18Big.left(sinthetaright)Big|_0^2pi&(8)\\
textCirculation&=&boxed 0.&(9)
endeqnarray*$$
Is this correct?
Thank you!
vector-analysis surfaces stokes-theorem
asked Aug 2 at 23:46
manooooh
415211
1
By “$operatornamerot$†do you mean curl?
– amd
Aug 2 at 23:51
Sorry the misunderstanding. Yes, by $textrot$ I mean the curl.
– manooooh
Aug 2 at 23:53
1
There was no particular reason to substitute $z=2x$. $iint z,dx,dy$ is just as good. Either way, you can stop at that point and immediately conclude that the value is $0$ because you’re integrating one cycle of $cos t$ times some constant.
– amd
Aug 3 at 0:14
Hm, if we are in $xy$-plane why we should not keep $x$ and $y$ variables, and replace $z=z(x)$ or $z=z(y)$ or $z=z(x,y)$?
– manooooh
Aug 3 at 0:17
1
You can certainly proceed that way. On the other hand, since you’ve already got a parameterization of the boundary, in terms of $t$ you can pass directly to the $r$-$t$ plane from $dsigma$ after computing the dot product of $nablatimes f$ and the plane normal. The integrand will be $cos t$ times some function of $r$, and the integral vanishes because you’re integrating over $0le tle2pi$, so there’s no need to even work through all of the details of the pullback.
– amd
Aug 3 at 0:29
 |Â
show 1 more comment
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Given the curve $C$ of equation $$vec X=(3cos t,3sin t,6cos t),qquad0leq tleq2pi$$ oriented according imposes this parameterization, find the circulation of $vec f$ along $C$ if $vec finmathcal C^1$ in $mathbb R^3$ and $$textrot vec f(x,y,z)=(0,2x-y,z),$$ where $textrot$ means curl.
First I found the curve: $$begincasesx=3cos t\y=3sin t\z=6cos tendcasesequivbegincasesx^2+y^2=9\z=6dfrac x3endcasesequivbegincasesx^2+y^2=9\z-2x=0,endcases$$ so the normal becomes $$vec N=(-2,0,1).$$ Now I have to use the Stokes theorem, that is
$$begineqnarray*
&&ointlimits_Cvec f;text dvec s&(1)\\
&=&iintlimits_SBig.nablatimesvec fBig&(2)\\
&=&iintlimits_P_xy(0,2x-y,underbrace z_2x)cdot(-2,0,1);text dxtext dy&(3)\\
&=&2iintlimits_P_xyx;text dxtext dy&(4)\\
&undersettextUsing polar coordinates=&2int_0^2pitext dthetaint_0^3rhocdotrhocostheta;text drho&(5)\\
&=&2int_0^2picostheta;text dthetaint_0^3rho^2;text drho&(6)\\
&=&2int_0^2picostheta;text dtheta;left.left(fracrho^33right)right|_0^3&(7)\\
&=&18Big.left(sinthetaright)Big|_0^2pi&(8)\\
textCirculation&=&boxed 0.&(9)
endeqnarray*$$
Is this correct?
Thank you!
vector-analysis surfaces stokes-theorem
asked Aug 2 at 23:46
manooooh
415211
Given the curve $C$ of equation $$vec X=(3cos t,3sin t,6cos t),qquad0leq tleq2pi$$ oriented according imposes this parameterization, find the circulation of $vec f$ along $C$ if $vec finmathcal C^1$ in $mathbb R^3$ and $$textrot vec f(x,y,z)=(0,2x-y,z),$$ where $textrot$ means curl.
First I found the curve: $$begincasesx=3cos t\y=3sin t\z=6cos tendcasesequivbegincasesx^2+y^2=9\z=6dfrac x3endcasesequivbegincasesx^2+y^2=9\z-2x=0,endcases$$ so the normal becomes $$vec N=(-2,0,1).$$ Now I have to use the Stokes theorem, that is
$$begineqnarray*
&&ointlimits_Cvec f;text dvec s&(1)\\
&=&iintlimits_SBig.nablatimesvec fBig&(2)\\
&=&iintlimits_P_xy(0,2x-y,underbrace z_2x)cdot(-2,0,1);text dxtext dy&(3)\\
&=&2iintlimits_P_xyx;text dxtext dy&(4)\\
&undersettextUsing polar coordinates=&2int_0^2pitext dthetaint_0^3rhocdotrhocostheta;text drho&(5)\\
&=&2int_0^2picostheta;text dthetaint_0^3rho^2;text drho&(6)\\
&=&2int_0^2picostheta;text dtheta;left.left(fracrho^33right)right|_0^3&(7)\\
&=&18Big.left(sinthetaright)Big|_0^2pi&(8)\\
textCirculation&=&boxed 0.&(9)
endeqnarray*$$
Is this correct?
Thank you!
vector-analysis surfaces stokes-theorem
asked Aug 2 at 23:46
manooooh
415211
edited Aug 2 at 23:56
asked Aug 2 at 23:46
manooooh
415211
asked Aug 2 at 23:46
manooooh
415211
asked Aug 2 at 23:46
manooooh
415211
415211
1
By “$operatornamerot$†do you mean curl?
– amd
Aug 2 at 23:51
Sorry the misunderstanding. Yes, by $textrot$ I mean the curl.
– manooooh
Aug 2 at 23:53
1
There was no particular reason to substitute $z=2x$. $iint z,dx,dy$ is just as good. Either way, you can stop at that point and immediately conclude that the value is $0$ because you’re integrating one cycle of $cos t$ times some constant.
– amd
Aug 3 at 0:14
Hm, if we are in $xy$-plane why we should not keep $x$ and $y$ variables, and replace $z=z(x)$ or $z=z(y)$ or $z=z(x,y)$?
– manooooh
Aug 3 at 0:17
1
You can certainly proceed that way. On the other hand, since you’ve already got a parameterization of the boundary, in terms of $t$ you can pass directly to the $r$-$t$ plane from $dsigma$ after computing the dot product of $nablatimes f$ and the plane normal. The integrand will be $cos t$ times some function of $r$, and the integral vanishes because you’re integrating over $0le tle2pi$, so there’s no need to even work through all of the details of the pullback.
– amd
Aug 3 at 0:29
 |Â
show 1 more comment
1
By “$operatornamerot$†do you mean curl?
– amd
Aug 2 at 23:51
Sorry the misunderstanding. Yes, by $textrot$ I mean the curl.
– manooooh
Aug 2 at 23:53
1
There was no particular reason to substitute $z=2x$. $iint z,dx,dy$ is just as good. Either way, you can stop at that point and immediately conclude that the value is $0$ because you’re integrating one cycle of $cos t$ times some constant.
– amd
Aug 3 at 0:14
Hm, if we are in $xy$-plane why we should not keep $x$ and $y$ variables, and replace $z=z(x)$ or $z=z(y)$ or $z=z(x,y)$?
– manooooh
Aug 3 at 0:17
1
You can certainly proceed that way. On the other hand, since you’ve already got a parameterization of the boundary, in terms of $t$ you can pass directly to the $r$-$t$ plane from $dsigma$ after computing the dot product of $nablatimes f$ and the plane normal. The integrand will be $cos t$ times some function of $r$, and the integral vanishes because you’re integrating over $0le tle2pi$, so there’s no need to even work through all of the details of the pullback.
– amd
Aug 3 at 0:29
1
1
By “$operatornamerot$†do you mean curl?
– amd
Aug 2 at 23:51
By “$operatornamerot$†do you mean curl?
– amd
Aug 2 at 23:51
Sorry the misunderstanding. Yes, by $textrot$ I mean the curl.
– manooooh
Aug 2 at 23:53
Sorry the misunderstanding. Yes, by $textrot$ I mean the curl.
– manooooh
Aug 2 at 23:53
1
1
There was no particular reason to substitute $z=2x$. $iint z,dx,dy$ is just as good. Either way, you can stop at that point and immediately conclude that the value is $0$ because you’re integrating one cycle of $cos t$ times some constant.
– amd
Aug 3 at 0:14
There was no particular reason to substitute $z=2x$. $iint z,dx,dy$ is just as good. Either way, you can stop at that point and immediately conclude that the value is $0$ because you’re integrating one cycle of $cos t$ times some constant.
– amd
Aug 3 at 0:14
Hm, if we are in $xy$-plane why we should not keep $x$ and $y$ variables, and replace $z=z(x)$ or $z=z(y)$ or $z=z(x,y)$?
– manooooh
Aug 3 at 0:17
Hm, if we are in $xy$-plane why we should not keep $x$ and $y$ variables, and replace $z=z(x)$ or $z=z(y)$ or $z=z(x,y)$?
– manooooh
Aug 3 at 0:17
1
1
You can certainly proceed that way. On the other hand, since you’ve already got a parameterization of the boundary, in terms of $t$ you can pass directly to the $r$-$t$ plane from $dsigma$ after computing the dot product of $nablatimes f$ and the plane normal. The integrand will be $cos t$ times some function of $r$, and the integral vanishes because you’re integrating over $0le tle2pi$, so there’s no need to even work through all of the details of the pullback.
– amd
Aug 3 at 0:29
You can certainly proceed that way. On the other hand, since you’ve already got a parameterization of the boundary, in terms of $t$ you can pass directly to the $r$-$t$ plane from $dsigma$ after computing the dot product of $nablatimes f$ and the plane normal. The integrand will be $cos t$ times some function of $r$, and the integral vanishes because you’re integrating over $0le tle2pi$, so there’s no need to even work through all of the details of the pullback.
– amd
Aug 3 at 0:29
 |Â
show 1 more comment
1 Answer
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You’re reached the correct answer, but I think you’ve done more work than you really needed to to get there.
First off, you can compute the surface normal directly from the parametric equation for the curve. Take two convenient points on it, say, at $t=0$ and $t=pi/2$ and compute their cross product: $(3,0,6)times(0,3,0)=(-18,0,9)$, which normalizes to $vec N = frac1sqrt5(-2,0,1)$. I might also verify that $vec Ncdotvec X=0$ for all values of $t$ to be sure that I’m dealing with a planar curve.
You then have $vec Ncdotvec f = frac1sqrt5z=frac6sqrt5cos t$, at which point you’re basically done. The region enclosed by the curve can be parameterized as $rvec X(t)$ for $0le rle 1$, which makes the region of integration in the $r$-$t$ plane the rectangle $[0,1]times[0,2pi]$. The integrand will be some function of $r$ times $cos t$, and its integral over a full cycle of $cos t$ will vanish.
answered Aug 3 at 0:58
amd
25.6k2943
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
accepted
You’re reached the correct answer, but I think you’ve done more work than you really needed to to get there.
First off, you can compute the surface normal directly from the parametric equation for the curve. Take two convenient points on it, say, at $t=0$ and $t=pi/2$ and compute their cross product: $(3,0,6)times(0,3,0)=(-18,0,9)$, which normalizes to $vec N = frac1sqrt5(-2,0,1)$. I might also verify that $vec Ncdotvec X=0$ for all values of $t$ to be sure that I’m dealing with a planar curve.
You then have $vec Ncdotvec f = frac1sqrt5z=frac6sqrt5cos t$, at which point you’re basically done. The region enclosed by the curve can be parameterized as $rvec X(t)$ for $0le rle 1$, which makes the region of integration in the $r$-$t$ plane the rectangle $[0,1]times[0,2pi]$. The integrand will be some function of $r$ times $cos t$, and its integral over a full cycle of $cos t$ will vanish.
answered Aug 3 at 0:58
amd
25.6k2943
add a comment |Â
up vote
0
down vote
accepted
You’re reached the correct answer, but I think you’ve done more work than you really needed to to get there.
First off, you can compute the surface normal directly from the parametric equation for the curve. Take two convenient points on it, say, at $t=0$ and $t=pi/2$ and compute their cross product: $(3,0,6)times(0,3,0)=(-18,0,9)$, which normalizes to $vec N = frac1sqrt5(-2,0,1)$. I might also verify that $vec Ncdotvec X=0$ for all values of $t$ to be sure that I’m dealing with a planar curve.
You then have $vec Ncdotvec f = frac1sqrt5z=frac6sqrt5cos t$, at which point you’re basically done. The region enclosed by the curve can be parameterized as $rvec X(t)$ for $0le rle 1$, which makes the region of integration in the $r$-$t$ plane the rectangle $[0,1]times[0,2pi]$. The integrand will be some function of $r$ times $cos t$, and its integral over a full cycle of $cos t$ will vanish.
answered Aug 3 at 0:58
amd
25.6k2943
add a comment |Â
up vote
0
down vote
accepted
up vote
0
down vote
accepted
You’re reached the correct answer, but I think you’ve done more work than you really needed to to get there.
First off, you can compute the surface normal directly from the parametric equation for the curve. Take two convenient points on it, say, at $t=0$ and $t=pi/2$ and compute their cross product: $(3,0,6)times(0,3,0)=(-18,0,9)$, which normalizes to $vec N = frac1sqrt5(-2,0,1)$. I might also verify that $vec Ncdotvec X=0$ for all values of $t$ to be sure that I’m dealing with a planar curve.
You then have $vec Ncdotvec f = frac1sqrt5z=frac6sqrt5cos t$, at which point you’re basically done. The region enclosed by the curve can be parameterized as $rvec X(t)$ for $0le rle 1$, which makes the region of integration in the $r$-$t$ plane the rectangle $[0,1]times[0,2pi]$. The integrand will be some function of $r$ times $cos t$, and its integral over a full cycle of $cos t$ will vanish.
answered Aug 3 at 0:58
amd
25.6k2943
You’re reached the correct answer, but I think you’ve done more work than you really needed to to get there.
First off, you can compute the surface normal directly from the parametric equation for the curve. Take two convenient points on it, say, at $t=0$ and $t=pi/2$ and compute their cross product: $(3,0,6)times(0,3,0)=(-18,0,9)$, which normalizes to $vec N = frac1sqrt5(-2,0,1)$. I might also verify that $vec Ncdotvec X=0$ for all values of $t$ to be sure that I’m dealing with a planar curve.
You then have $vec Ncdotvec f = frac1sqrt5z=frac6sqrt5cos t$, at which point you’re basically done. The region enclosed by the curve can be parameterized as $rvec X(t)$ for $0le rle 1$, which makes the region of integration in the $r$-$t$ plane the rectangle $[0,1]times[0,2pi]$. The integrand will be some function of $r$ times $cos t$, and its integral over a full cycle of $cos t$ will vanish.
answered Aug 3 at 0:58
amd
25.6k2943
answered Aug 3 at 0:58
amd
25.6k2943
answered Aug 3 at 0:58
amd
25.6k2943
answered Aug 3 at 0:58
amd
25.6k2943
25.6k2943
add a comment |Â
add a comment |Â
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1
By “$operatornamerot$†do you mean curl?
– amd
Aug 2 at 23:51
Sorry the misunderstanding. Yes, by $textrot$ I mean the curl.
– manooooh
Aug 2 at 23:53
1
There was no particular reason to substitute $z=2x$. $iint z,dx,dy$ is just as good. Either way, you can stop at that point and immediately conclude that the value is $0$ because you’re integrating one cycle of $cos t$ times some constant.
– amd
Aug 3 at 0:14
Hm, if we are in $xy$-plane why we should not keep $x$ and $y$ variables, and replace $z=z(x)$ or $z=z(y)$ or $z=z(x,y)$?
– manooooh
Aug 3 at 0:17
1
You can certainly proceed that way. On the other hand, since you’ve already got a parameterization of the boundary, in terms of $t$ you can pass directly to the $r$-$t$ plane from $dsigma$ after computing the dot product of $nablatimes f$ and the plane normal. The integrand will be $cos t$ times some function of $r$, and the integral vanishes because you’re integrating over $0le tle2pi$, so there’s no need to even work through all of the details of the pullback.
– amd
Aug 3 at 0:29