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find flux outward a sphere cutted with $yle-4$
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$$D=x^2+y^2+z^2le 25,yle -4. qquad F=z^2,y^2,x^2$$
In order to find the total flux going outward D I need to evaluate two fluxes(or maybe not ? ):
the first one is the flux of the circular region surface $x^2+z^2le 9$ (which is the intersection between the sphere and $yle -4$)
and the second one is the surface created by cutting the sphere
I calculated the first flux by first parametrizing the surface and then by evaluating the double integral over if :
$$iintlimits_S_1 F,ds$$
So I have $r(theta)=(3costheta,-4,3sintheta)$ with $0le theta le 2pi$
and the normal vector is $n=(0,-1,0)$. and eventually the flux will be = $-144pi$
The problems start when I try to calculate the flux of the second surface :
So i have $r(theta,phi)=(psinphisintheta,pcosphi,psinphicostheta)$ with $0le theta le 2pi$,$cos^-1(-4/5)le phi le pi$
The normal shoud be : $n = (-p^2sin^2phisintheta,-p^2sinphicosphi,-p^2sin^2phisin^2theta)$
And when I multiply $F(r(theta,phi))n = -p^4cos^2theta(2sin^4phisintheta+sinphicosphi)$
Is that really What I should put in the integral? (in order to find tge flux of the second region)
What I'd like to get is the total flux! What would you do to?
integration multivariable-calculus surfaces
edited Aug 3 at 17:19
Michael Hardy
204k23185460
asked Aug 3 at 15:37
NPLS
1819
add a comment |Â
up vote
2
down vote
favorite
$$D=x^2+y^2+z^2le 25,yle -4. qquad F=z^2,y^2,x^2$$
In order to find the total flux going outward D I need to evaluate two fluxes(or maybe not ? ):
the first one is the flux of the circular region surface $x^2+z^2le 9$ (which is the intersection between the sphere and $yle -4$)
and the second one is the surface created by cutting the sphere
I calculated the first flux by first parametrizing the surface and then by evaluating the double integral over if :
$$iintlimits_S_1 F,ds$$
So I have $r(theta)=(3costheta,-4,3sintheta)$ with $0le theta le 2pi$
and the normal vector is $n=(0,-1,0)$. and eventually the flux will be = $-144pi$
The problems start when I try to calculate the flux of the second surface :
So i have $r(theta,phi)=(psinphisintheta,pcosphi,psinphicostheta)$ with $0le theta le 2pi$,$cos^-1(-4/5)le phi le pi$
The normal shoud be : $n = (-p^2sin^2phisintheta,-p^2sinphicosphi,-p^2sin^2phisin^2theta)$
And when I multiply $F(r(theta,phi))n = -p^4cos^2theta(2sin^4phisintheta+sinphicosphi)$
Is that really What I should put in the integral? (in order to find tge flux of the second region)
What I'd like to get is the total flux! What would you do to?
integration multivariable-calculus surfaces
edited Aug 3 at 17:19
Michael Hardy
204k23185460
asked Aug 3 at 15:37
NPLS
1819
What is $vec F$?
– gimusi
Aug 3 at 15:53
yeah sorry I edited
– NPLS
Aug 3 at 15:55
note that $div vec F=0$
– gimusi
Aug 3 at 15:56
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
$$D=x^2+y^2+z^2le 25,yle -4. qquad F=z^2,y^2,x^2$$
In order to find the total flux going outward D I need to evaluate two fluxes(or maybe not ? ):
the first one is the flux of the circular region surface $x^2+z^2le 9$ (which is the intersection between the sphere and $yle -4$)
and the second one is the surface created by cutting the sphere
I calculated the first flux by first parametrizing the surface and then by evaluating the double integral over if :
$$iintlimits_S_1 F,ds$$
So I have $r(theta)=(3costheta,-4,3sintheta)$ with $0le theta le 2pi$
and the normal vector is $n=(0,-1,0)$. and eventually the flux will be = $-144pi$
The problems start when I try to calculate the flux of the second surface :
So i have $r(theta,phi)=(psinphisintheta,pcosphi,psinphicostheta)$ with $0le theta le 2pi$,$cos^-1(-4/5)le phi le pi$
The normal shoud be : $n = (-p^2sin^2phisintheta,-p^2sinphicosphi,-p^2sin^2phisin^2theta)$
And when I multiply $F(r(theta,phi))n = -p^4cos^2theta(2sin^4phisintheta+sinphicosphi)$
Is that really What I should put in the integral? (in order to find tge flux of the second region)
What I'd like to get is the total flux! What would you do to?
integration multivariable-calculus surfaces
edited Aug 3 at 17:19
Michael Hardy
204k23185460
asked Aug 3 at 15:37
NPLS
1819
$$D=x^2+y^2+z^2le 25,yle -4. qquad F=z^2,y^2,x^2$$
In order to find the total flux going outward D I need to evaluate two fluxes(or maybe not ? ):
the first one is the flux of the circular region surface $x^2+z^2le 9$ (which is the intersection between the sphere and $yle -4$)
and the second one is the surface created by cutting the sphere
I calculated the first flux by first parametrizing the surface and then by evaluating the double integral over if :
$$iintlimits_S_1 F,ds$$
So I have $r(theta)=(3costheta,-4,3sintheta)$ with $0le theta le 2pi$
and the normal vector is $n=(0,-1,0)$. and eventually the flux will be = $-144pi$
The problems start when I try to calculate the flux of the second surface :
So i have $r(theta,phi)=(psinphisintheta,pcosphi,psinphicostheta)$ with $0le theta le 2pi$,$cos^-1(-4/5)le phi le pi$
The normal shoud be : $n = (-p^2sin^2phisintheta,-p^2sinphicosphi,-p^2sin^2phisin^2theta)$
And when I multiply $F(r(theta,phi))n = -p^4cos^2theta(2sin^4phisintheta+sinphicosphi)$
Is that really What I should put in the integral? (in order to find tge flux of the second region)
What I'd like to get is the total flux! What would you do to?
integration multivariable-calculus surfaces
edited Aug 3 at 17:19
Michael Hardy
204k23185460
asked Aug 3 at 15:37
NPLS
1819
edited Aug 3 at 17:19
Michael Hardy
204k23185460
edited Aug 3 at 17:19
Michael Hardy
204k23185460
edited Aug 3 at 17:19
Michael Hardy
204k23185460
204k23185460
asked Aug 3 at 15:37
NPLS
1819
asked Aug 3 at 15:37
NPLS
1819
asked Aug 3 at 15:37
NPLS
1819
1819
What is $vec F$?
– gimusi
Aug 3 at 15:53
yeah sorry I edited
– NPLS
Aug 3 at 15:55
note that $div vec F=0$
– gimusi
Aug 3 at 15:56
add a comment |Â
What is $vec F$?
– gimusi
Aug 3 at 15:53
yeah sorry I edited
– NPLS
Aug 3 at 15:55
note that $div vec F=0$
– gimusi
Aug 3 at 15:56
What is $vec F$?
– gimusi
Aug 3 at 15:53
What is $vec F$?
– gimusi
Aug 3 at 15:53
yeah sorry I edited
– NPLS
Aug 3 at 15:55
yeah sorry I edited
– NPLS
Aug 3 at 15:55
note that $div vec F=0$
– gimusi
Aug 3 at 15:56
note that $div vec F=0$
– gimusi
Aug 3 at 15:56
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
By divergence theorem we have
$$iint_S vec F cdot vec n dS=iiint_D operatornamedivvec F dV=iiint_D operatorname 2y, dV$$
and by cylindrical coordinates we have
- $z=rcos theta$
- $x=rsin theta$
- $y=y$
$$iiint_D operatorname 2y, dV=int_0^2pi,dtheta int_-5^-4 2y, dyint_0^sqrt25-y^2r,dr$$
(solution $-frac812 pi$)
answered Aug 3 at 15:55
gimusi
63.7k73480
what are the constrains of $D$ ? because I have problem parametrizing the entire surface
– NPLS
Aug 3 at 16:15
@NPLS The simplest way in that case is to observe that $div(F)=0$ then the flux over any closed surface is equal to zero by divergence theorem.
– gimusi
Aug 3 at 16:17
but the dif(F) isn't equal to 2y ?
– NPLS
Aug 3 at 16:19
@NPLS Opsss...yes sorry of course...I don't know what I read. Anyway it is convenient proceed by that way by divergence theorem.
– gimusi
Aug 3 at 16:21
yep, but I don't know how to parametrize that thing
– NPLS
Aug 3 at 16:23
 |Â
show 5 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
By divergence theorem we have
$$iint_S vec F cdot vec n dS=iiint_D operatornamedivvec F dV=iiint_D operatorname 2y, dV$$
and by cylindrical coordinates we have
- $z=rcos theta$
- $x=rsin theta$
- $y=y$
$$iiint_D operatorname 2y, dV=int_0^2pi,dtheta int_-5^-4 2y, dyint_0^sqrt25-y^2r,dr$$
(solution $-frac812 pi$)
answered Aug 3 at 15:55
gimusi
63.7k73480
what are the constrains of $D$ ? because I have problem parametrizing the entire surface
– NPLS
Aug 3 at 16:15
@NPLS The simplest way in that case is to observe that $div(F)=0$ then the flux over any closed surface is equal to zero by divergence theorem.
– gimusi
Aug 3 at 16:17
but the dif(F) isn't equal to 2y ?
– NPLS
Aug 3 at 16:19
@NPLS Opsss...yes sorry of course...I don't know what I read. Anyway it is convenient proceed by that way by divergence theorem.
– gimusi
Aug 3 at 16:21
yep, but I don't know how to parametrize that thing
– NPLS
Aug 3 at 16:23
 |Â
show 5 more comments
up vote
1
down vote
accepted
By divergence theorem we have
$$iint_S vec F cdot vec n dS=iiint_D operatornamedivvec F dV=iiint_D operatorname 2y, dV$$
and by cylindrical coordinates we have
- $z=rcos theta$
- $x=rsin theta$
- $y=y$
$$iiint_D operatorname 2y, dV=int_0^2pi,dtheta int_-5^-4 2y, dyint_0^sqrt25-y^2r,dr$$
(solution $-frac812 pi$)
answered Aug 3 at 15:55
gimusi
63.7k73480
what are the constrains of $D$ ? because I have problem parametrizing the entire surface
– NPLS
Aug 3 at 16:15
@NPLS The simplest way in that case is to observe that $div(F)=0$ then the flux over any closed surface is equal to zero by divergence theorem.
– gimusi
Aug 3 at 16:17
but the dif(F) isn't equal to 2y ?
– NPLS
Aug 3 at 16:19
@NPLS Opsss...yes sorry of course...I don't know what I read. Anyway it is convenient proceed by that way by divergence theorem.
– gimusi
Aug 3 at 16:21
yep, but I don't know how to parametrize that thing
– NPLS
Aug 3 at 16:23
 |Â
show 5 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
By divergence theorem we have
$$iint_S vec F cdot vec n dS=iiint_D operatornamedivvec F dV=iiint_D operatorname 2y, dV$$
and by cylindrical coordinates we have
- $z=rcos theta$
- $x=rsin theta$
- $y=y$
$$iiint_D operatorname 2y, dV=int_0^2pi,dtheta int_-5^-4 2y, dyint_0^sqrt25-y^2r,dr$$
(solution $-frac812 pi$)
answered Aug 3 at 15:55
gimusi
63.7k73480
By divergence theorem we have
$$iint_S vec F cdot vec n dS=iiint_D operatornamedivvec F dV=iiint_D operatorname 2y, dV$$
and by cylindrical coordinates we have
- $z=rcos theta$
- $x=rsin theta$
- $y=y$
$$iiint_D operatorname 2y, dV=int_0^2pi,dtheta int_-5^-4 2y, dyint_0^sqrt25-y^2r,dr$$
(solution $-frac812 pi$)
answered Aug 3 at 15:55
gimusi
63.7k73480
edited Aug 3 at 16:42
answered Aug 3 at 15:55
gimusi
63.7k73480
answered Aug 3 at 15:55
gimusi
63.7k73480
answered Aug 3 at 15:55
gimusi
63.7k73480
63.7k73480
what are the constrains of $D$ ? because I have problem parametrizing the entire surface
– NPLS
Aug 3 at 16:15
@NPLS The simplest way in that case is to observe that $div(F)=0$ then the flux over any closed surface is equal to zero by divergence theorem.
– gimusi
Aug 3 at 16:17
but the dif(F) isn't equal to 2y ?
– NPLS
Aug 3 at 16:19
@NPLS Opsss...yes sorry of course...I don't know what I read. Anyway it is convenient proceed by that way by divergence theorem.
– gimusi
Aug 3 at 16:21
yep, but I don't know how to parametrize that thing
– NPLS
Aug 3 at 16:23
 |Â
show 5 more comments
what are the constrains of $D$ ? because I have problem parametrizing the entire surface
– NPLS
Aug 3 at 16:15
@NPLS The simplest way in that case is to observe that $div(F)=0$ then the flux over any closed surface is equal to zero by divergence theorem.
– gimusi
Aug 3 at 16:17
but the dif(F) isn't equal to 2y ?
– NPLS
Aug 3 at 16:19
@NPLS Opsss...yes sorry of course...I don't know what I read. Anyway it is convenient proceed by that way by divergence theorem.
– gimusi
Aug 3 at 16:21
yep, but I don't know how to parametrize that thing
– NPLS
Aug 3 at 16:23
what are the constrains of $D$ ? because I have problem parametrizing the entire surface
– NPLS
Aug 3 at 16:15
what are the constrains of $D$ ? because I have problem parametrizing the entire surface
– NPLS
Aug 3 at 16:15
@NPLS The simplest way in that case is to observe that $div(F)=0$ then the flux over any closed surface is equal to zero by divergence theorem.
– gimusi
Aug 3 at 16:17
@NPLS The simplest way in that case is to observe that $div(F)=0$ then the flux over any closed surface is equal to zero by divergence theorem.
– gimusi
Aug 3 at 16:17
but the dif(F) isn't equal to 2y ?
– NPLS
Aug 3 at 16:19
but the dif(F) isn't equal to 2y ?
– NPLS
Aug 3 at 16:19
@NPLS Opsss...yes sorry of course...I don't know what I read. Anyway it is convenient proceed by that way by divergence theorem.
– gimusi
Aug 3 at 16:21
@NPLS Opsss...yes sorry of course...I don't know what I read. Anyway it is convenient proceed by that way by divergence theorem.
– gimusi
Aug 3 at 16:21
yep, but I don't know how to parametrize that thing
– NPLS
Aug 3 at 16:23
yep, but I don't know how to parametrize that thing
– NPLS
Aug 3 at 16:23
 |Â
show 5 more comments
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What is $vec F$?
– gimusi
Aug 3 at 15:53
yeah sorry I edited
– NPLS
Aug 3 at 15:55
note that $div vec F=0$
– gimusi
Aug 3 at 15:56