Calculate $int_{Gamma}omega$ when $omega = x^2yzdx+xy^2zdy+xyz^2dz$ and $Gamma$ is the intersection of...
$begingroup$
Calculate $int_{Gamma}omega$, when: $omega = x^2yzdx+xy^2zdy+xyz^2dz$
and $Gamma$ is the intersection of the surfaces: $x=1$, $y^2+z^2=1$.
Can you help me with that?
At first, I thought of the parametrization $$x=x$$ $$y=r cos t$$ $$z=r sin t$$ but I don't think it's a good one, since I don't know how to write the bounds on it. However, I didn't understand very well how it's the procedure when I have the domain the intersection of some things. Can you help me out with this?
integration surfaces parametrization
edited Jan 13 at 7:50
Blue
49.1k870156
asked Jan 13 at 7:45
C. CristiC. Cristi
1,634218
$endgroup$
add a comment |
$begingroup$
Calculate $int_{Gamma}omega$, when: $omega = x^2yzdx+xy^2zdy+xyz^2dz$
and $Gamma$ is the intersection of the surfaces: $x=1$, $y^2+z^2=1$.
Can you help me with that?
At first, I thought of the parametrization $$x=x$$ $$y=r cos t$$ $$z=r sin t$$ but I don't think it's a good one, since I don't know how to write the bounds on it. However, I didn't understand very well how it's the procedure when I have the domain the intersection of some things. Can you help me out with this?
integration surfaces parametrization
edited Jan 13 at 7:50
Blue
49.1k870156
asked Jan 13 at 7:45
C. CristiC. Cristi
1,634218
$endgroup$
$begingroup$
Uhh $x=1$. What is the parameterization of a circle?
$endgroup$
– Zachary Selk
Jan 13 at 7:59
add a comment |
$begingroup$
Calculate $int_{Gamma}omega$, when: $omega = x^2yzdx+xy^2zdy+xyz^2dz$
and $Gamma$ is the intersection of the surfaces: $x=1$, $y^2+z^2=1$.
Can you help me with that?
At first, I thought of the parametrization $$x=x$$ $$y=r cos t$$ $$z=r sin t$$ but I don't think it's a good one, since I don't know how to write the bounds on it. However, I didn't understand very well how it's the procedure when I have the domain the intersection of some things. Can you help me out with this?
integration surfaces parametrization
edited Jan 13 at 7:50
Blue
49.1k870156
asked Jan 13 at 7:45
C. CristiC. Cristi
1,634218
$endgroup$
Calculate $int_{Gamma}omega$, when: $omega = x^2yzdx+xy^2zdy+xyz^2dz$
and $Gamma$ is the intersection of the surfaces: $x=1$, $y^2+z^2=1$.
Can you help me with that?
At first, I thought of the parametrization $$x=x$$ $$y=r cos t$$ $$z=r sin t$$ but I don't think it's a good one, since I don't know how to write the bounds on it. However, I didn't understand very well how it's the procedure when I have the domain the intersection of some things. Can you help me out with this?
integration surfaces parametrization
integration surfaces parametrization
edited Jan 13 at 7:50
Blue
49.1k870156
asked Jan 13 at 7:45
C. CristiC. Cristi
1,634218
edited Jan 13 at 7:50
Blue
49.1k870156
asked Jan 13 at 7:45
C. CristiC. Cristi
1,634218
edited Jan 13 at 7:50
Blue
49.1k870156
edited Jan 13 at 7:50
Blue
49.1k870156
edited Jan 13 at 7:50
Blue
49.1k870156
49.1k870156
asked Jan 13 at 7:45
C. CristiC. Cristi
1,634218
asked Jan 13 at 7:45
C. CristiC. Cristi
1,634218
asked Jan 13 at 7:45
C. CristiC. Cristi
1,634218
1,634218
$begingroup$
Uhh $x=1$. What is the parameterization of a circle?
$endgroup$
– Zachary Selk
Jan 13 at 7:59
add a comment |
$begingroup$
Uhh $x=1$. What is the parameterization of a circle?
$endgroup$
– Zachary Selk
Jan 13 at 7:59
$begingroup$
Uhh $x=1$. What is the parameterization of a circle?
$endgroup$
– Zachary Selk
Jan 13 at 7:59
$begingroup$
Uhh $x=1$. What is the parameterization of a circle?
$endgroup$
– Zachary Selk
Jan 13 at 7:59
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You already have it!:
$$begin{cases}x=1\{}\begin{cases}x=cos t\y=sin tend{cases};,;;tin[0,2pi]end{cases}$$
answered Jan 13 at 8:02
DonAntonioDonAntonio
179k1494233
$endgroup$
$begingroup$
Why $tin [0,2pi ]?$
$endgroup$
– C. Cristi
Jan 13 at 8:09
$begingroup$
Because the plane $;x=1;$ intersects the cylinder $;y^2+z^2=1;$ in a whole circle of radius $;1;$ ....
$endgroup$
– DonAntonio
Jan 13 at 8:16
$begingroup$
Yeah.. but my question actually was, or what I was thinking is: What if there was something more complicated then $x=1$?
$endgroup$
– C. Cristi
Jan 13 at 8:18
$begingroup$
@C.Cristi You shall cross that bridge whenever you arrive to it. It indeed can be a rather tough task to find a reasonable parametrization of some intersecting surfaces...
$endgroup$
– DonAntonio
Jan 13 at 8:23
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You already have it!:
$$begin{cases}x=1\{}\begin{cases}x=cos t\y=sin tend{cases};,;;tin[0,2pi]end{cases}$$
answered Jan 13 at 8:02
DonAntonioDonAntonio
179k1494233
$endgroup$
$begingroup$
Why $tin [0,2pi ]?$
$endgroup$
– C. Cristi
Jan 13 at 8:09
$begingroup$
Because the plane $;x=1;$ intersects the cylinder $;y^2+z^2=1;$ in a whole circle of radius $;1;$ ....
$endgroup$
– DonAntonio
Jan 13 at 8:16
$begingroup$
Yeah.. but my question actually was, or what I was thinking is: What if there was something more complicated then $x=1$?
$endgroup$
– C. Cristi
Jan 13 at 8:18
$begingroup$
@C.Cristi You shall cross that bridge whenever you arrive to it. It indeed can be a rather tough task to find a reasonable parametrization of some intersecting surfaces...
$endgroup$
– DonAntonio
Jan 13 at 8:23
add a comment |
$begingroup$
You already have it!:
$$begin{cases}x=1\{}\begin{cases}x=cos t\y=sin tend{cases};,;;tin[0,2pi]end{cases}$$
answered Jan 13 at 8:02
DonAntonioDonAntonio
179k1494233
$endgroup$
$begingroup$
Why $tin [0,2pi ]?$
$endgroup$
– C. Cristi
Jan 13 at 8:09
$begingroup$
Because the plane $;x=1;$ intersects the cylinder $;y^2+z^2=1;$ in a whole circle of radius $;1;$ ....
$endgroup$
– DonAntonio
Jan 13 at 8:16
$begingroup$
Yeah.. but my question actually was, or what I was thinking is: What if there was something more complicated then $x=1$?
$endgroup$
– C. Cristi
Jan 13 at 8:18
$begingroup$
@C.Cristi You shall cross that bridge whenever you arrive to it. It indeed can be a rather tough task to find a reasonable parametrization of some intersecting surfaces...
$endgroup$
– DonAntonio
Jan 13 at 8:23
add a comment |
$begingroup$
You already have it!:
$$begin{cases}x=1\{}\begin{cases}x=cos t\y=sin tend{cases};,;;tin[0,2pi]end{cases}$$
answered Jan 13 at 8:02
DonAntonioDonAntonio
179k1494233
$endgroup$
You already have it!:
$$begin{cases}x=1\{}\begin{cases}x=cos t\y=sin tend{cases};,;;tin[0,2pi]end{cases}$$
answered Jan 13 at 8:02
DonAntonioDonAntonio
179k1494233
answered Jan 13 at 8:02
DonAntonioDonAntonio
179k1494233
answered Jan 13 at 8:02
DonAntonioDonAntonio
179k1494233
answered Jan 13 at 8:02
DonAntonioDonAntonio
179k1494233
179k1494233
$begingroup$
Why $tin [0,2pi ]?$
$endgroup$
– C. Cristi
Jan 13 at 8:09
$begingroup$
Because the plane $;x=1;$ intersects the cylinder $;y^2+z^2=1;$ in a whole circle of radius $;1;$ ....
$endgroup$
– DonAntonio
Jan 13 at 8:16
$begingroup$
Yeah.. but my question actually was, or what I was thinking is: What if there was something more complicated then $x=1$?
$endgroup$
– C. Cristi
Jan 13 at 8:18
$begingroup$
@C.Cristi You shall cross that bridge whenever you arrive to it. It indeed can be a rather tough task to find a reasonable parametrization of some intersecting surfaces...
$endgroup$
– DonAntonio
Jan 13 at 8:23
add a comment |
$begingroup$
Why $tin [0,2pi ]?$
$endgroup$
– C. Cristi
Jan 13 at 8:09
$begingroup$
Because the plane $;x=1;$ intersects the cylinder $;y^2+z^2=1;$ in a whole circle of radius $;1;$ ....
$endgroup$
– DonAntonio
Jan 13 at 8:16
$begingroup$
Yeah.. but my question actually was, or what I was thinking is: What if there was something more complicated then $x=1$?
$endgroup$
– C. Cristi
Jan 13 at 8:18
$begingroup$
@C.Cristi You shall cross that bridge whenever you arrive to it. It indeed can be a rather tough task to find a reasonable parametrization of some intersecting surfaces...
$endgroup$
– DonAntonio
Jan 13 at 8:23
$begingroup$
Why $tin [0,2pi ]?$
$endgroup$
– C. Cristi
Jan 13 at 8:09
$begingroup$
Why $tin [0,2pi ]?$
$endgroup$
– C. Cristi
Jan 13 at 8:09
$begingroup$
Because the plane $;x=1;$ intersects the cylinder $;y^2+z^2=1;$ in a whole circle of radius $;1;$ ....
$endgroup$
– DonAntonio
Jan 13 at 8:16
$begingroup$
Because the plane $;x=1;$ intersects the cylinder $;y^2+z^2=1;$ in a whole circle of radius $;1;$ ....
$endgroup$
– DonAntonio
Jan 13 at 8:16
$begingroup$
Yeah.. but my question actually was, or what I was thinking is: What if there was something more complicated then $x=1$?
$endgroup$
– C. Cristi
Jan 13 at 8:18
$begingroup$
Yeah.. but my question actually was, or what I was thinking is: What if there was something more complicated then $x=1$?
$endgroup$
– C. Cristi
Jan 13 at 8:18
$begingroup$
@C.Cristi You shall cross that bridge whenever you arrive to it. It indeed can be a rather tough task to find a reasonable parametrization of some intersecting surfaces...
$endgroup$
– DonAntonio
Jan 13 at 8:23
$begingroup$
@C.Cristi You shall cross that bridge whenever you arrive to it. It indeed can be a rather tough task to find a reasonable parametrization of some intersecting surfaces...
$endgroup$
– DonAntonio
Jan 13 at 8:23
add a comment |
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$begingroup$
Uhh $x=1$. What is the parameterization of a circle?
$endgroup$
– Zachary Selk
Jan 13 at 7:59