Surface integral of curl












1












$begingroup$


Let $vec F(x,y,z)=(x^2+y^2+sin(xy)$, $e^x$+$2xy, -yz$). Calculate the flow of $nablatimesvec F$ through the following surface:
$$
S={(x,y,z) x^2 + 2y^2 + 3z^2=10, yge 0}
$$



I figured that if you added $S_{aux} ={(x,y,z)in Bbb R^3 : x^2 + 3z^2 =10}$ to S you had a closed surface and therefore could use the divergence theorem. This would mean that:
$$
iint_{S+S_{aux}}nablatimesvec F mathrm{d}s = iiint_{V} nablacirc(nablatimesvec F) mathrm{d}v.
$$

Since $nablacirc(nablatimesvec F)=0$, then:
$$
iiint_{V} nablacirc(nablatimesvec F)mathrm{d}v = 0.
$$

Therefore
$$iint_{S+S_{aux}}nablatimesvec F mathrm{d}s = 0,
$$
and since
$$iint_{S+S_{aux}}nablatimesvec F mathrm{d}s = iint_Snablatimesvec F mathrm{d}s + iint_{S_{aux}}nablatimesvec F mathrm{d}s = 0$$ we have $$iint_Snablatimesvec F mathrm{d}s = -iint_{S_{aux}}nablatimesvec F mathrm{d}s.
$$



Could you please tell me if my thinking is correct?
Thank you very much!






multivariable-calculus surface-integrals divergence curl







share|cite|improve this question











$endgroup$












  • $begingroup$
    $S={(x,y,z)|x^2+2y^2+3z^2=10,y≥10}?$ If $yge10,$ then $x^2+3z^2le -190$ Wait. Reading further, maybe you meant $yge0?$
    $endgroup$
    – saulspatz
    Dec 2 '18 at 22:11








  • 1




    $begingroup$
    Couldn't you just use Stokes' theorem?
    $endgroup$
    – saulspatz
    Dec 2 '18 at 22:16










  • $begingroup$
    Yes, I meant y≥0. Thanks. Let me change that now
    $endgroup$
    – juan deutsch
    Dec 2 '18 at 22:21
















1












$begingroup$


Let $vec F(x,y,z)=(x^2+y^2+sin(xy)$, $e^x$+$2xy, -yz$). Calculate the flow of $nablatimesvec F$ through the following surface:
$$
S={(x,y,z) x^2 + 2y^2 + 3z^2=10, yge 0}
$$



I figured that if you added $S_{aux} ={(x,y,z)in Bbb R^3 : x^2 + 3z^2 =10}$ to S you had a closed surface and therefore could use the divergence theorem. This would mean that:
$$
iint_{S+S_{aux}}nablatimesvec F mathrm{d}s = iiint_{V} nablacirc(nablatimesvec F) mathrm{d}v.
$$

Since $nablacirc(nablatimesvec F)=0$, then:
$$
iiint_{V} nablacirc(nablatimesvec F)mathrm{d}v = 0.
$$

Therefore
$$iint_{S+S_{aux}}nablatimesvec F mathrm{d}s = 0,
$$
and since
$$iint_{S+S_{aux}}nablatimesvec F mathrm{d}s = iint_Snablatimesvec F mathrm{d}s + iint_{S_{aux}}nablatimesvec F mathrm{d}s = 0$$ we have $$iint_Snablatimesvec F mathrm{d}s = -iint_{S_{aux}}nablatimesvec F mathrm{d}s.
$$



Could you please tell me if my thinking is correct?
Thank you very much!






multivariable-calculus surface-integrals divergence curl







share|cite|improve this question











$endgroup$












  • $begingroup$
    $S={(x,y,z)|x^2+2y^2+3z^2=10,y≥10}?$ If $yge10,$ then $x^2+3z^2le -190$ Wait. Reading further, maybe you meant $yge0?$
    $endgroup$
    – saulspatz
    Dec 2 '18 at 22:11








  • 1




    $begingroup$
    Couldn't you just use Stokes' theorem?
    $endgroup$
    – saulspatz
    Dec 2 '18 at 22:16










  • $begingroup$
    Yes, I meant y≥0. Thanks. Let me change that now
    $endgroup$
    – juan deutsch
    Dec 2 '18 at 22:21














1












1








1


1



$begingroup$


Let $vec F(x,y,z)=(x^2+y^2+sin(xy)$, $e^x$+$2xy, -yz$). Calculate the flow of $nablatimesvec F$ through the following surface:
$$
S={(x,y,z) x^2 + 2y^2 + 3z^2=10, yge 0}
$$



I figured that if you added $S_{aux} ={(x,y,z)in Bbb R^3 : x^2 + 3z^2 =10}$ to S you had a closed surface and therefore could use the divergence theorem. This would mean that:
$$
iint_{S+S_{aux}}nablatimesvec F mathrm{d}s = iiint_{V} nablacirc(nablatimesvec F) mathrm{d}v.
$$

Since $nablacirc(nablatimesvec F)=0$, then:
$$
iiint_{V} nablacirc(nablatimesvec F)mathrm{d}v = 0.
$$

Therefore
$$iint_{S+S_{aux}}nablatimesvec F mathrm{d}s = 0,
$$
and since
$$iint_{S+S_{aux}}nablatimesvec F mathrm{d}s = iint_Snablatimesvec F mathrm{d}s + iint_{S_{aux}}nablatimesvec F mathrm{d}s = 0$$ we have $$iint_Snablatimesvec F mathrm{d}s = -iint_{S_{aux}}nablatimesvec F mathrm{d}s.
$$



Could you please tell me if my thinking is correct?
Thank you very much!






multivariable-calculus surface-integrals divergence curl







share|cite|improve this question











$endgroup$




Let $vec F(x,y,z)=(x^2+y^2+sin(xy)$, $e^x$+$2xy, -yz$). Calculate the flow of $nablatimesvec F$ through the following surface:
$$
S={(x,y,z) x^2 + 2y^2 + 3z^2=10, yge 0}
$$



I figured that if you added $S_{aux} ={(x,y,z)in Bbb R^3 : x^2 + 3z^2 =10}$ to S you had a closed surface and therefore could use the divergence theorem. This would mean that:
$$
iint_{S+S_{aux}}nablatimesvec F mathrm{d}s = iiint_{V} nablacirc(nablatimesvec F) mathrm{d}v.
$$

Since $nablacirc(nablatimesvec F)=0$, then:
$$
iiint_{V} nablacirc(nablatimesvec F)mathrm{d}v = 0.
$$

Therefore
$$iint_{S+S_{aux}}nablatimesvec F mathrm{d}s = 0,
$$
and since
$$iint_{S+S_{aux}}nablatimesvec F mathrm{d}s = iint_Snablatimesvec F mathrm{d}s + iint_{S_{aux}}nablatimesvec F mathrm{d}s = 0$$ we have $$iint_Snablatimesvec F mathrm{d}s = -iint_{S_{aux}}nablatimesvec F mathrm{d}s.
$$



Could you please tell me if my thinking is correct?
Thank you very much!






multivariable-calculus surface-integrals divergence curl




multivariable-calculus surface-integrals divergence curl






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 2 '18 at 22:21







juan deutsch

















asked Dec 2 '18 at 21:40









juan deutschjuan deutsch

156




156












  • $begingroup$
    $S={(x,y,z)|x^2+2y^2+3z^2=10,y≥10}?$ If $yge10,$ then $x^2+3z^2le -190$ Wait. Reading further, maybe you meant $yge0?$
    $endgroup$
    – saulspatz
    Dec 2 '18 at 22:11








  • 1




    $begingroup$
    Couldn't you just use Stokes' theorem?
    $endgroup$
    – saulspatz
    Dec 2 '18 at 22:16










  • $begingroup$
    Yes, I meant y≥0. Thanks. Let me change that now
    $endgroup$
    – juan deutsch
    Dec 2 '18 at 22:21


















  • $begingroup$
    $S={(x,y,z)|x^2+2y^2+3z^2=10,y≥10}?$ If $yge10,$ then $x^2+3z^2le -190$ Wait. Reading further, maybe you meant $yge0?$
    $endgroup$
    – saulspatz
    Dec 2 '18 at 22:11








  • 1




    $begingroup$
    Couldn't you just use Stokes' theorem?
    $endgroup$
    – saulspatz
    Dec 2 '18 at 22:16










  • $begingroup$
    Yes, I meant y≥0. Thanks. Let me change that now
    $endgroup$
    – juan deutsch
    Dec 2 '18 at 22:21
















$begingroup$
$S={(x,y,z)|x^2+2y^2+3z^2=10,y≥10}?$ If $yge10,$ then $x^2+3z^2le -190$ Wait. Reading further, maybe you meant $yge0?$
$endgroup$
– saulspatz
Dec 2 '18 at 22:11






$begingroup$
$S={(x,y,z)|x^2+2y^2+3z^2=10,y≥10}?$ If $yge10,$ then $x^2+3z^2le -190$ Wait. Reading further, maybe you meant $yge0?$
$endgroup$
– saulspatz
Dec 2 '18 at 22:11






1




1




$begingroup$
Couldn't you just use Stokes' theorem?
$endgroup$
– saulspatz
Dec 2 '18 at 22:16




$begingroup$
Couldn't you just use Stokes' theorem?
$endgroup$
– saulspatz
Dec 2 '18 at 22:16












$begingroup$
Yes, I meant y≥0. Thanks. Let me change that now
$endgroup$
– juan deutsch
Dec 2 '18 at 22:21




$begingroup$
Yes, I meant y≥0. Thanks. Let me change that now
$endgroup$
– juan deutsch
Dec 2 '18 at 22:21










1 Answer
1






active

oldest

votes


















1












$begingroup$

I'm certain that the point of the problem is to use Stokes' theorem. $partial S$ is the intersection of the plane $y=0$ with $S$ so it's the ellipse $x^2+3z^2=10, y=0.$ On this curve, $F$ simplifies to $(x^2, e^x, 0)$ When you compute the circulation of $F$ over $partial S$ there will only be a contibution from the first coordinate.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I have just tried that by parametrizing the ellipse as $alpha(theta):(sqrt{10}cos{theta},0,sqrt{frac {10}3}sin{theta})$, then calculated the line integral which ended up giving me zero as well. Does that mean that what I did was valid?
    $endgroup$
    – juan deutsch
    Dec 2 '18 at 23:16












  • $begingroup$
    Yes, I think what you did was correct, but it's been a long time since I worked with this stuff, so I'm not certain.
    $endgroup$
    – saulspatz
    Dec 3 '18 at 0:01











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1 Answer
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active

oldest

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oldest

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1












$begingroup$

I'm certain that the point of the problem is to use Stokes' theorem. $partial S$ is the intersection of the plane $y=0$ with $S$ so it's the ellipse $x^2+3z^2=10, y=0.$ On this curve, $F$ simplifies to $(x^2, e^x, 0)$ When you compute the circulation of $F$ over $partial S$ there will only be a contibution from the first coordinate.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I have just tried that by parametrizing the ellipse as $alpha(theta):(sqrt{10}cos{theta},0,sqrt{frac {10}3}sin{theta})$, then calculated the line integral which ended up giving me zero as well. Does that mean that what I did was valid?
    $endgroup$
    – juan deutsch
    Dec 2 '18 at 23:16












  • $begingroup$
    Yes, I think what you did was correct, but it's been a long time since I worked with this stuff, so I'm not certain.
    $endgroup$
    – saulspatz
    Dec 3 '18 at 0:01
















1












$begingroup$

I'm certain that the point of the problem is to use Stokes' theorem. $partial S$ is the intersection of the plane $y=0$ with $S$ so it's the ellipse $x^2+3z^2=10, y=0.$ On this curve, $F$ simplifies to $(x^2, e^x, 0)$ When you compute the circulation of $F$ over $partial S$ there will only be a contibution from the first coordinate.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I have just tried that by parametrizing the ellipse as $alpha(theta):(sqrt{10}cos{theta},0,sqrt{frac {10}3}sin{theta})$, then calculated the line integral which ended up giving me zero as well. Does that mean that what I did was valid?
    $endgroup$
    – juan deutsch
    Dec 2 '18 at 23:16












  • $begingroup$
    Yes, I think what you did was correct, but it's been a long time since I worked with this stuff, so I'm not certain.
    $endgroup$
    – saulspatz
    Dec 3 '18 at 0:01














1












1








1





$begingroup$

I'm certain that the point of the problem is to use Stokes' theorem. $partial S$ is the intersection of the plane $y=0$ with $S$ so it's the ellipse $x^2+3z^2=10, y=0.$ On this curve, $F$ simplifies to $(x^2, e^x, 0)$ When you compute the circulation of $F$ over $partial S$ there will only be a contibution from the first coordinate.






share|cite|improve this answer









$endgroup$



I'm certain that the point of the problem is to use Stokes' theorem. $partial S$ is the intersection of the plane $y=0$ with $S$ so it's the ellipse $x^2+3z^2=10, y=0.$ On this curve, $F$ simplifies to $(x^2, e^x, 0)$ When you compute the circulation of $F$ over $partial S$ there will only be a contibution from the first coordinate.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 2 '18 at 22:49









saulspatzsaulspatz

14.5k21329




14.5k21329












  • $begingroup$
    I have just tried that by parametrizing the ellipse as $alpha(theta):(sqrt{10}cos{theta},0,sqrt{frac {10}3}sin{theta})$, then calculated the line integral which ended up giving me zero as well. Does that mean that what I did was valid?
    $endgroup$
    – juan deutsch
    Dec 2 '18 at 23:16












  • $begingroup$
    Yes, I think what you did was correct, but it's been a long time since I worked with this stuff, so I'm not certain.
    $endgroup$
    – saulspatz
    Dec 3 '18 at 0:01


















  • $begingroup$
    I have just tried that by parametrizing the ellipse as $alpha(theta):(sqrt{10}cos{theta},0,sqrt{frac {10}3}sin{theta})$, then calculated the line integral which ended up giving me zero as well. Does that mean that what I did was valid?
    $endgroup$
    – juan deutsch
    Dec 2 '18 at 23:16












  • $begingroup$
    Yes, I think what you did was correct, but it's been a long time since I worked with this stuff, so I'm not certain.
    $endgroup$
    – saulspatz
    Dec 3 '18 at 0:01
















$begingroup$
I have just tried that by parametrizing the ellipse as $alpha(theta):(sqrt{10}cos{theta},0,sqrt{frac {10}3}sin{theta})$, then calculated the line integral which ended up giving me zero as well. Does that mean that what I did was valid?
$endgroup$
– juan deutsch
Dec 2 '18 at 23:16






$begingroup$
I have just tried that by parametrizing the ellipse as $alpha(theta):(sqrt{10}cos{theta},0,sqrt{frac {10}3}sin{theta})$, then calculated the line integral which ended up giving me zero as well. Does that mean that what I did was valid?
$endgroup$
– juan deutsch
Dec 2 '18 at 23:16














$begingroup$
Yes, I think what you did was correct, but it's been a long time since I worked with this stuff, so I'm not certain.
$endgroup$
– saulspatz
Dec 3 '18 at 0:01




$begingroup$
Yes, I think what you did was correct, but it's been a long time since I worked with this stuff, so I'm not certain.
$endgroup$
– saulspatz
Dec 3 '18 at 0:01


















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