Mixing
$int_C ydx+z^2dy+xdz$ on a specific curve
$begingroup$
Evaluate
$$int_C ydx+z^2dy+xdz$$
on a specific curve, the intersection between $z=sqrt{x^2+y^2}$ and $z=6-(x^2+y^2)$.
I don't know how to parametrize this, also it seems wrong to me if I subtitute $z^2$ in the second one i get the equation $z=6-z^2implies z=2$ since $zgeq0$. what do I do with this..? Also is there some kind of way to always approach these questions where you have to integrate between 2 surfaces?
integration curves
edited Jan 26 at 8:42
Robert Z
101k1070143
asked Jan 26 at 8:32
C. CristiC. Cristi
1,639218
$endgroup$
add a comment |
$begingroup$
Evaluate
$$int_C ydx+z^2dy+xdz$$
on a specific curve, the intersection between $z=sqrt{x^2+y^2}$ and $z=6-(x^2+y^2)$.
I don't know how to parametrize this, also it seems wrong to me if I subtitute $z^2$ in the second one i get the equation $z=6-z^2implies z=2$ since $zgeq0$. what do I do with this..? Also is there some kind of way to always approach these questions where you have to integrate between 2 surfaces?
integration curves
edited Jan 26 at 8:42
Robert Z
101k1070143
asked Jan 26 at 8:32
C. CristiC. Cristi
1,639218
$endgroup$
$begingroup$
The hint below should be helpful. Once you figure out the curve given by the equations $x^2 + y^2 = 2^2$ and $z=2$, you can parametrize it by using $r(t)= langle x(t), y(t), z(t) rangle = langle 2 cos t, 2 sin t, 2 rangle$, where $0leq t < 2pi$.
$endgroup$
– Mee Seong Im
Jan 26 at 8:39
add a comment |
$begingroup$
Evaluate
$$int_C ydx+z^2dy+xdz$$
on a specific curve, the intersection between $z=sqrt{x^2+y^2}$ and $z=6-(x^2+y^2)$.
I don't know how to parametrize this, also it seems wrong to me if I subtitute $z^2$ in the second one i get the equation $z=6-z^2implies z=2$ since $zgeq0$. what do I do with this..? Also is there some kind of way to always approach these questions where you have to integrate between 2 surfaces?
integration curves
edited Jan 26 at 8:42
Robert Z
101k1070143
asked Jan 26 at 8:32
C. CristiC. Cristi
1,639218
$endgroup$
Evaluate
$$int_C ydx+z^2dy+xdz$$
on a specific curve, the intersection between $z=sqrt{x^2+y^2}$ and $z=6-(x^2+y^2)$.
I don't know how to parametrize this, also it seems wrong to me if I subtitute $z^2$ in the second one i get the equation $z=6-z^2implies z=2$ since $zgeq0$. what do I do with this..? Also is there some kind of way to always approach these questions where you have to integrate between 2 surfaces?
integration curves
integration curves
edited Jan 26 at 8:42
Robert Z
101k1070143
asked Jan 26 at 8:32
C. CristiC. Cristi
1,639218
edited Jan 26 at 8:42
Robert Z
101k1070143
asked Jan 26 at 8:32
C. CristiC. Cristi
1,639218
edited Jan 26 at 8:42
Robert Z
101k1070143
edited Jan 26 at 8:42
Robert Z
101k1070143
edited Jan 26 at 8:42
Robert Z
101k1070143
101k1070143
asked Jan 26 at 8:32
C. CristiC. Cristi
1,639218
asked Jan 26 at 8:32
C. CristiC. Cristi
1,639218
asked Jan 26 at 8:32
C. CristiC. Cristi
1,639218
1,639218
$begingroup$
The hint below should be helpful. Once you figure out the curve given by the equations $x^2 + y^2 = 2^2$ and $z=2$, you can parametrize it by using $r(t)= langle x(t), y(t), z(t) rangle = langle 2 cos t, 2 sin t, 2 rangle$, where $0leq t < 2pi$.
$endgroup$
– Mee Seong Im
Jan 26 at 8:39
add a comment |
$begingroup$
The hint below should be helpful. Once you figure out the curve given by the equations $x^2 + y^2 = 2^2$ and $z=2$, you can parametrize it by using $r(t)= langle x(t), y(t), z(t) rangle = langle 2 cos t, 2 sin t, 2 rangle$, where $0leq t < 2pi$.
$endgroup$
– Mee Seong Im
Jan 26 at 8:39
$begingroup$
The hint below should be helpful. Once you figure out the curve given by the equations $x^2 + y^2 = 2^2$ and $z=2$, you can parametrize it by using $r(t)= langle x(t), y(t), z(t) rangle = langle 2 cos t, 2 sin t, 2 rangle$, where $0leq t < 2pi$.
$endgroup$
– Mee Seong Im
Jan 26 at 8:39
$begingroup$
The hint below should be helpful. Once you figure out the curve given by the equations $x^2 + y^2 = 2^2$ and $z=2$, you can parametrize it by using $r(t)= langle x(t), y(t), z(t) rangle = langle 2 cos t, 2 sin t, 2 rangle$, where $0leq t < 2pi$.
$endgroup$
– Mee Seong Im
Jan 26 at 8:39
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You are on the right track. $z=2$ tells you that the curve $C$ is contained in the plane $z=2$. Moreover $z=6−(x^2+y^2)$ (or $z=sqrt{x^2+y^2}$) implies that $x^2+y^2=6-2=2^2$. What kind of curve is this? It should be easy to get a parametrization now and evaluate
$$int_C ydx+z^2dy+xdz.$$
answered Jan 26 at 8:35
Robert ZRobert Z
101k1070143
$endgroup$
$begingroup$
It's a circle, but when I get one of the variables as a number into the equation I should just think that it is in that specific plane?
$endgroup$
– C. Cristi
Jan 26 at 8:38
$begingroup$
And also $dz = 0$? I just simply take: $$x=2cost$$ $$y=2sint$$ $$z=2$$ ?
$endgroup$
– C. Cristi
Jan 26 at 8:39
$begingroup$
Yes, it is the circle centered at $(0,0,2)$ of radius $2$ contained in the plane $z=2$. Yes, $dz=0$.
$endgroup$
– Robert Z
Jan 26 at 8:40
$begingroup$
Yes, the parametrization is correct with $tin[0,2pi]$.
$endgroup$
– Robert Z
Jan 26 at 8:41
$begingroup$
Okay, thank you so much! Also can you give me some tips for when I have to integrate when I am gives 2 equations like that?
$endgroup$
– C. Cristi
Jan 26 at 8:42
|
show 1 more comment
Your Answer
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1 Answer
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1 Answer
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oldest
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active
oldest
votes
$begingroup$
You are on the right track. $z=2$ tells you that the curve $C$ is contained in the plane $z=2$. Moreover $z=6−(x^2+y^2)$ (or $z=sqrt{x^2+y^2}$) implies that $x^2+y^2=6-2=2^2$. What kind of curve is this? It should be easy to get a parametrization now and evaluate
$$int_C ydx+z^2dy+xdz.$$
answered Jan 26 at 8:35
Robert ZRobert Z
101k1070143
$endgroup$
$begingroup$
It's a circle, but when I get one of the variables as a number into the equation I should just think that it is in that specific plane?
$endgroup$
– C. Cristi
Jan 26 at 8:38
$begingroup$
And also $dz = 0$? I just simply take: $$x=2cost$$ $$y=2sint$$ $$z=2$$ ?
$endgroup$
– C. Cristi
Jan 26 at 8:39
$begingroup$
Yes, it is the circle centered at $(0,0,2)$ of radius $2$ contained in the plane $z=2$. Yes, $dz=0$.
$endgroup$
– Robert Z
Jan 26 at 8:40
$begingroup$
Yes, the parametrization is correct with $tin[0,2pi]$.
$endgroup$
– Robert Z
Jan 26 at 8:41
$begingroup$
Okay, thank you so much! Also can you give me some tips for when I have to integrate when I am gives 2 equations like that?
$endgroup$
– C. Cristi
Jan 26 at 8:42
|
show 1 more comment
$begingroup$
You are on the right track. $z=2$ tells you that the curve $C$ is contained in the plane $z=2$. Moreover $z=6−(x^2+y^2)$ (or $z=sqrt{x^2+y^2}$) implies that $x^2+y^2=6-2=2^2$. What kind of curve is this? It should be easy to get a parametrization now and evaluate
$$int_C ydx+z^2dy+xdz.$$
answered Jan 26 at 8:35
Robert ZRobert Z
101k1070143
$endgroup$
$begingroup$
It's a circle, but when I get one of the variables as a number into the equation I should just think that it is in that specific plane?
$endgroup$
– C. Cristi
Jan 26 at 8:38
$begingroup$
And also $dz = 0$? I just simply take: $$x=2cost$$ $$y=2sint$$ $$z=2$$ ?
$endgroup$
– C. Cristi
Jan 26 at 8:39
$begingroup$
Yes, it is the circle centered at $(0,0,2)$ of radius $2$ contained in the plane $z=2$. Yes, $dz=0$.
$endgroup$
– Robert Z
Jan 26 at 8:40
$begingroup$
Yes, the parametrization is correct with $tin[0,2pi]$.
$endgroup$
– Robert Z
Jan 26 at 8:41
$begingroup$
Okay, thank you so much! Also can you give me some tips for when I have to integrate when I am gives 2 equations like that?
$endgroup$
– C. Cristi
Jan 26 at 8:42
|
show 1 more comment
$begingroup$
You are on the right track. $z=2$ tells you that the curve $C$ is contained in the plane $z=2$. Moreover $z=6−(x^2+y^2)$ (or $z=sqrt{x^2+y^2}$) implies that $x^2+y^2=6-2=2^2$. What kind of curve is this? It should be easy to get a parametrization now and evaluate
$$int_C ydx+z^2dy+xdz.$$
answered Jan 26 at 8:35
Robert ZRobert Z
101k1070143
$endgroup$
You are on the right track. $z=2$ tells you that the curve $C$ is contained in the plane $z=2$. Moreover $z=6−(x^2+y^2)$ (or $z=sqrt{x^2+y^2}$) implies that $x^2+y^2=6-2=2^2$. What kind of curve is this? It should be easy to get a parametrization now and evaluate
$$int_C ydx+z^2dy+xdz.$$
answered Jan 26 at 8:35
Robert ZRobert Z
101k1070143
edited Jan 26 at 8:43
answered Jan 26 at 8:35
Robert ZRobert Z
101k1070143
answered Jan 26 at 8:35
Robert ZRobert Z
101k1070143
answered Jan 26 at 8:35
Robert ZRobert Z
101k1070143
101k1070143
$begingroup$
It's a circle, but when I get one of the variables as a number into the equation I should just think that it is in that specific plane?
$endgroup$
– C. Cristi
Jan 26 at 8:38
$begingroup$
And also $dz = 0$? I just simply take: $$x=2cost$$ $$y=2sint$$ $$z=2$$ ?
$endgroup$
– C. Cristi
Jan 26 at 8:39
$begingroup$
Yes, it is the circle centered at $(0,0,2)$ of radius $2$ contained in the plane $z=2$. Yes, $dz=0$.
$endgroup$
– Robert Z
Jan 26 at 8:40
$begingroup$
Yes, the parametrization is correct with $tin[0,2pi]$.
$endgroup$
– Robert Z
Jan 26 at 8:41
$begingroup$
Okay, thank you so much! Also can you give me some tips for when I have to integrate when I am gives 2 equations like that?
$endgroup$
– C. Cristi
Jan 26 at 8:42
|
show 1 more comment
$begingroup$
It's a circle, but when I get one of the variables as a number into the equation I should just think that it is in that specific plane?
$endgroup$
– C. Cristi
Jan 26 at 8:38
$begingroup$
And also $dz = 0$? I just simply take: $$x=2cost$$ $$y=2sint$$ $$z=2$$ ?
$endgroup$
– C. Cristi
Jan 26 at 8:39
$begingroup$
Yes, it is the circle centered at $(0,0,2)$ of radius $2$ contained in the plane $z=2$. Yes, $dz=0$.
$endgroup$
– Robert Z
Jan 26 at 8:40
$begingroup$
Yes, the parametrization is correct with $tin[0,2pi]$.
$endgroup$
– Robert Z
Jan 26 at 8:41
$begingroup$
Okay, thank you so much! Also can you give me some tips for when I have to integrate when I am gives 2 equations like that?
$endgroup$
– C. Cristi
Jan 26 at 8:42
$begingroup$
It's a circle, but when I get one of the variables as a number into the equation I should just think that it is in that specific plane?
$endgroup$
– C. Cristi
Jan 26 at 8:38
$begingroup$
It's a circle, but when I get one of the variables as a number into the equation I should just think that it is in that specific plane?
$endgroup$
– C. Cristi
Jan 26 at 8:38
$begingroup$
And also $dz = 0$? I just simply take: $$x=2cost$$ $$y=2sint$$ $$z=2$$ ?
$endgroup$
– C. Cristi
Jan 26 at 8:39
$begingroup$
And also $dz = 0$? I just simply take: $$x=2cost$$ $$y=2sint$$ $$z=2$$ ?
$endgroup$
– C. Cristi
Jan 26 at 8:39
$begingroup$
Yes, it is the circle centered at $(0,0,2)$ of radius $2$ contained in the plane $z=2$. Yes, $dz=0$.
$endgroup$
– Robert Z
Jan 26 at 8:40
$begingroup$
Yes, it is the circle centered at $(0,0,2)$ of radius $2$ contained in the plane $z=2$. Yes, $dz=0$.
$endgroup$
– Robert Z
Jan 26 at 8:40
$begingroup$
Yes, the parametrization is correct with $tin[0,2pi]$.
$endgroup$
– Robert Z
Jan 26 at 8:41
$begingroup$
Yes, the parametrization is correct with $tin[0,2pi]$.
$endgroup$
– Robert Z
Jan 26 at 8:41
$begingroup$
Okay, thank you so much! Also can you give me some tips for when I have to integrate when I am gives 2 equations like that?
$endgroup$
– C. Cristi
Jan 26 at 8:42
$begingroup$
Okay, thank you so much! Also can you give me some tips for when I have to integrate when I am gives 2 equations like that?
$endgroup$
– C. Cristi
Jan 26 at 8:42
|
show 1 more comment
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$begingroup$
The hint below should be helpful. Once you figure out the curve given by the equations $x^2 + y^2 = 2^2$ and $z=2$, you can parametrize it by using $r(t)= langle x(t), y(t), z(t) rangle = langle 2 cos t, 2 sin t, 2 rangle$, where $0leq t < 2pi$.
$endgroup$
– Mee Seong Im
Jan 26 at 8:39