Mixing
Breaking a path, in 2 integrals
$begingroup$
Let $overrightarrow V, Gamma$, be a vectorial field and a path such that:
$$overrightarrow V=-(x^2+y^2)overrightarrow i-(x^2-y^2)overrightarrow j$$
and
$$Gamma={(x,y)inmathbb{R}^2 | x^2+y^2=4, y < 0}cup {(x,y)inmathbb{R^2}| (x-1)^2+y^2=1, y geq0}$$
I know that: $int_{Gamma}overrightarrow V doverrightarrow r=int_{Gamma}Pdx+Qdy$
$P(x,y)=-x^2-y^2$
$Q(x,y)=y^2-x^2$
I'm going to use this parametrization to calculate this:
$x(t)=2cos t$
$y(t)=2sin t$
$tin[pi,2pi]$
But that is only for the first part of the path can I calculate the first part of the path with a parametrization and the second one of the path with another parametrization?
integration vector-fields parametrization
asked Jan 13 at 11:26
C. CristiC. Cristi
1,629218
$endgroup$
add a comment |
$begingroup$
Let $overrightarrow V, Gamma$, be a vectorial field and a path such that:
$$overrightarrow V=-(x^2+y^2)overrightarrow i-(x^2-y^2)overrightarrow j$$
and
$$Gamma={(x,y)inmathbb{R}^2 | x^2+y^2=4, y < 0}cup {(x,y)inmathbb{R^2}| (x-1)^2+y^2=1, y geq0}$$
I know that: $int_{Gamma}overrightarrow V doverrightarrow r=int_{Gamma}Pdx+Qdy$
$P(x,y)=-x^2-y^2$
$Q(x,y)=y^2-x^2$
I'm going to use this parametrization to calculate this:
$x(t)=2cos t$
$y(t)=2sin t$
$tin[pi,2pi]$
But that is only for the first part of the path can I calculate the first part of the path with a parametrization and the second one of the path with another parametrization?
integration vector-fields parametrization
asked Jan 13 at 11:26
C. CristiC. Cristi
1,629218
$endgroup$
add a comment |
$begingroup$
Let $overrightarrow V, Gamma$, be a vectorial field and a path such that:
$$overrightarrow V=-(x^2+y^2)overrightarrow i-(x^2-y^2)overrightarrow j$$
and
$$Gamma={(x,y)inmathbb{R}^2 | x^2+y^2=4, y < 0}cup {(x,y)inmathbb{R^2}| (x-1)^2+y^2=1, y geq0}$$
I know that: $int_{Gamma}overrightarrow V doverrightarrow r=int_{Gamma}Pdx+Qdy$
$P(x,y)=-x^2-y^2$
$Q(x,y)=y^2-x^2$
I'm going to use this parametrization to calculate this:
$x(t)=2cos t$
$y(t)=2sin t$
$tin[pi,2pi]$
But that is only for the first part of the path can I calculate the first part of the path with a parametrization and the second one of the path with another parametrization?
integration vector-fields parametrization
asked Jan 13 at 11:26
C. CristiC. Cristi
1,629218
$endgroup$
Let $overrightarrow V, Gamma$, be a vectorial field and a path such that:
$$overrightarrow V=-(x^2+y^2)overrightarrow i-(x^2-y^2)overrightarrow j$$
and
$$Gamma={(x,y)inmathbb{R}^2 | x^2+y^2=4, y < 0}cup {(x,y)inmathbb{R^2}| (x-1)^2+y^2=1, y geq0}$$
I know that: $int_{Gamma}overrightarrow V doverrightarrow r=int_{Gamma}Pdx+Qdy$
$P(x,y)=-x^2-y^2$
$Q(x,y)=y^2-x^2$
I'm going to use this parametrization to calculate this:
$x(t)=2cos t$
$y(t)=2sin t$
$tin[pi,2pi]$
But that is only for the first part of the path can I calculate the first part of the path with a parametrization and the second one of the path with another parametrization?
integration vector-fields parametrization
integration vector-fields parametrization
asked Jan 13 at 11:26
C. CristiC. Cristi
1,629218
asked Jan 13 at 11:26
C. CristiC. Cristi
1,629218
asked Jan 13 at 11:26
C. CristiC. Cristi
1,629218
asked Jan 13 at 11:26
C. CristiC. Cristi
1,629218
asked Jan 13 at 11:26
C. CristiC. Cristi
1,629218
1,629218
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your path is in fact the union of two different ones: the lower semicircle $;x^2+y^2=4;$ and the upper semicircle $;(x-1)^2+y^2=1;$ . For the first one you can parametrize (in the positive direction)
$$Gamma_1: r_1(t)=(2cos t,,2sin t);,;;tin[pi,2pi];$$
and the second one
$$Gamma_2: r_2(t)=(1+cos t,,sin t);,;;tin[0,pi]$$
Usually paths given are path connected, yet this one isn't....anyway, now you can integrate on each separatedly.
answered Jan 13 at 11:58
DonAntonioDonAntonio
178k1494230
$endgroup$
add a comment |
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$begingroup$
Your path is in fact the union of two different ones: the lower semicircle $;x^2+y^2=4;$ and the upper semicircle $;(x-1)^2+y^2=1;$ . For the first one you can parametrize (in the positive direction)
$$Gamma_1: r_1(t)=(2cos t,,2sin t);,;;tin[pi,2pi];$$
and the second one
$$Gamma_2: r_2(t)=(1+cos t,,sin t);,;;tin[0,pi]$$
Usually paths given are path connected, yet this one isn't....anyway, now you can integrate on each separatedly.
answered Jan 13 at 11:58
DonAntonioDonAntonio
178k1494230
$endgroup$
add a comment |
$begingroup$
Your path is in fact the union of two different ones: the lower semicircle $;x^2+y^2=4;$ and the upper semicircle $;(x-1)^2+y^2=1;$ . For the first one you can parametrize (in the positive direction)
$$Gamma_1: r_1(t)=(2cos t,,2sin t);,;;tin[pi,2pi];$$
and the second one
$$Gamma_2: r_2(t)=(1+cos t,,sin t);,;;tin[0,pi]$$
Usually paths given are path connected, yet this one isn't....anyway, now you can integrate on each separatedly.
answered Jan 13 at 11:58
DonAntonioDonAntonio
178k1494230
$endgroup$
add a comment |
$begingroup$
Your path is in fact the union of two different ones: the lower semicircle $;x^2+y^2=4;$ and the upper semicircle $;(x-1)^2+y^2=1;$ . For the first one you can parametrize (in the positive direction)
$$Gamma_1: r_1(t)=(2cos t,,2sin t);,;;tin[pi,2pi];$$
and the second one
$$Gamma_2: r_2(t)=(1+cos t,,sin t);,;;tin[0,pi]$$
Usually paths given are path connected, yet this one isn't....anyway, now you can integrate on each separatedly.
answered Jan 13 at 11:58
DonAntonioDonAntonio
178k1494230
$endgroup$
Your path is in fact the union of two different ones: the lower semicircle $;x^2+y^2=4;$ and the upper semicircle $;(x-1)^2+y^2=1;$ . For the first one you can parametrize (in the positive direction)
$$Gamma_1: r_1(t)=(2cos t,,2sin t);,;;tin[pi,2pi];$$
and the second one
$$Gamma_2: r_2(t)=(1+cos t,,sin t);,;;tin[0,pi]$$
Usually paths given are path connected, yet this one isn't....anyway, now you can integrate on each separatedly.
answered Jan 13 at 11:58
DonAntonioDonAntonio
178k1494230
answered Jan 13 at 11:58
DonAntonioDonAntonio
178k1494230
answered Jan 13 at 11:58
DonAntonioDonAntonio
178k1494230
answered Jan 13 at 11:58
DonAntonioDonAntonio
178k1494230
178k1494230
add a comment |
add a comment |
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