Mixing
Flux outward Sphere-Cylinder
Clash Royale CLAN TAG#URR8PPP
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$D=x^2+y^2+z^2le 25,y^2+z^2le 9$
$F=y^2,x^2,z$
I need to calculate the flux outward the boundary of $D$.
I think I can use the divergent theorem, but How can I define the triple integral of the whole surface? Or Do I need to divide the surface into 3 parts ?
integration multivariable-calculus surfaces
asked 15 hours ago
NPLS
1819
add a comment |Â
up vote
0
down vote
favorite
$D=x^2+y^2+z^2le 25,y^2+z^2le 9$
$F=y^2,x^2,z$
I need to calculate the flux outward the boundary of $D$.
I think I can use the divergent theorem, but How can I define the triple integral of the whole surface? Or Do I need to divide the surface into 3 parts ?
integration multivariable-calculus surfaces
asked 15 hours ago
NPLS
1819
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
$D=x^2+y^2+z^2le 25,y^2+z^2le 9$
$F=y^2,x^2,z$
I need to calculate the flux outward the boundary of $D$.
I think I can use the divergent theorem, but How can I define the triple integral of the whole surface? Or Do I need to divide the surface into 3 parts ?
integration multivariable-calculus surfaces
asked 15 hours ago
NPLS
1819
$D=x^2+y^2+z^2le 25,y^2+z^2le 9$
$F=y^2,x^2,z$
I need to calculate the flux outward the boundary of $D$.
I think I can use the divergent theorem, but How can I define the triple integral of the whole surface? Or Do I need to divide the surface into 3 parts ?
integration multivariable-calculus surfaces
asked 15 hours ago
NPLS
1819
asked 15 hours ago
NPLS
1819
asked 15 hours ago
NPLS
1819
asked 15 hours ago
NPLS
1819
1819
add a comment |Â
add a comment |Â
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
Use the Divergence Theorem. That is,
$$
beginalign*
iint_E mathbfFcdot dmathbfS &= iiint_E textdiv (mathbfF) :dV \
&= iiint_E langle partial_x, partial_y, partial_z rangle cdot langle y^2,x^2,z rangle : dV \
&= iiint_E partial_x(y^2) + partial_y(x^2) + partial_z(z) : dV \
&=iiint_E 1 :dV \
&=iiint_E 1 :dx : dy : dz\
&=int_0^2pi int_0^3 int_-sqrt25-r^2^sqrt25-r^2 left|fracpartial(x,y,z)partial(x,r,theta) right| dx : dr : dtheta \
&=int_0^2pi int_0^3 int_-sqrt25-r^2^sqrt25-r^2 big|rbig| : dx : dr : dtheta \
&=int_0^2pi int_0^3 int_-sqrt25-r^2^sqrt25-r^2 r : dx : dr : dtheta hspace4mm
mbox since r geq 0 \
&= 2pi int_0^3 r (2sqrt25-r^2) : dr \
&= - 2pi int_25^16 sqrtu: du hspace4mm
mbox where u = 25-r^2, hspace4mm
mbox so hspace4mm
du = -2r :dr
\
&= frac4pi3left( 5^3 - 4^3 right) \
&= boxedfrac2443pi \
endalign*
$$
since given a parametrization in cylindrical coordinates
$$
mathbfr(x,r,theta)
= langle x(x,r,theta), y(x,r,theta), z(x,r,theta)rangle
= langle x,r cos theta,r sin thetarangle, hspace4mm
r geq 0
hspace4mm
mbox and
hspace4mm
0leq theta leq 2pi,
$$
its Jacobian is
$$
beginalign*
fracpartial(x,y,z)partial(x,r,theta)
&=
det
beginpmatrix
fracpartial xpartial x& fracpartial xpartial r & fracpartial xpartial theta \
fracpartial ypartial x& fracpartial ypartial r & fracpartial ypartial theta \
fracpartial zpartial x & fracpartial zpartial r & fracpartial zpartial theta \
endpmatrix \
&=
det
beginpmatrix
1 & 0& 0 \
0 & cos theta & -r sin theta \
0 &sin theta & r cos theta \
endpmatrix \
&= r.
endalign*
$$
answered 14 hours ago
Mee Seong Im
2,5591417
1
exactly! Thanks
– NPLS
14 hours ago
add a comment |Â
up vote
1
down vote
You're going to want to convert to cylindrical coordinates, with $z$ being identified with the $x$-axis. (A diagram helps) The integral becomes $$int_0^2piint_0^3int_-sqrt25-r^2^sqrt25-r^2rcdot textdiv(F) dz dr dtheta$$
Where $r$ is the Jacobian of the transformation $(r,theta,z)mapsto (z,rcos(theta),rsin(theta))$. The integral isn't so hard, so I'll leave it to "u" to finish :)
answered 15 hours ago
Rafay A.
564
1
shouldn't it be $dxdrdtheta$ ? becouse my cylinder is parallel to the x-axis
– NPLS
14 hours ago
@NPLS you can use $dx : dr : dtheta$ or $dz : dr : dtheta$, where we identify the variable $z$ to be the coordinate along the $x$-axis.
– Mee Seong Im
14 hours ago
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Use the Divergence Theorem. That is,
$$
beginalign*
iint_E mathbfFcdot dmathbfS &= iiint_E textdiv (mathbfF) :dV \
&= iiint_E langle partial_x, partial_y, partial_z rangle cdot langle y^2,x^2,z rangle : dV \
&= iiint_E partial_x(y^2) + partial_y(x^2) + partial_z(z) : dV \
&=iiint_E 1 :dV \
&=iiint_E 1 :dx : dy : dz\
&=int_0^2pi int_0^3 int_-sqrt25-r^2^sqrt25-r^2 left|fracpartial(x,y,z)partial(x,r,theta) right| dx : dr : dtheta \
&=int_0^2pi int_0^3 int_-sqrt25-r^2^sqrt25-r^2 big|rbig| : dx : dr : dtheta \
&=int_0^2pi int_0^3 int_-sqrt25-r^2^sqrt25-r^2 r : dx : dr : dtheta hspace4mm
mbox since r geq 0 \
&= 2pi int_0^3 r (2sqrt25-r^2) : dr \
&= - 2pi int_25^16 sqrtu: du hspace4mm
mbox where u = 25-r^2, hspace4mm
mbox so hspace4mm
du = -2r :dr
\
&= frac4pi3left( 5^3 - 4^3 right) \
&= boxedfrac2443pi \
endalign*
$$
since given a parametrization in cylindrical coordinates
$$
mathbfr(x,r,theta)
= langle x(x,r,theta), y(x,r,theta), z(x,r,theta)rangle
= langle x,r cos theta,r sin thetarangle, hspace4mm
r geq 0
hspace4mm
mbox and
hspace4mm
0leq theta leq 2pi,
$$
its Jacobian is
$$
beginalign*
fracpartial(x,y,z)partial(x,r,theta)
&=
det
beginpmatrix
fracpartial xpartial x& fracpartial xpartial r & fracpartial xpartial theta \
fracpartial ypartial x& fracpartial ypartial r & fracpartial ypartial theta \
fracpartial zpartial x & fracpartial zpartial r & fracpartial zpartial theta \
endpmatrix \
&=
det
beginpmatrix
1 & 0& 0 \
0 & cos theta & -r sin theta \
0 &sin theta & r cos theta \
endpmatrix \
&= r.
endalign*
$$
answered 14 hours ago
Mee Seong Im
2,5591417
1
exactly! Thanks
– NPLS
14 hours ago
add a comment |Â
up vote
1
down vote
accepted
Use the Divergence Theorem. That is,
$$
beginalign*
iint_E mathbfFcdot dmathbfS &= iiint_E textdiv (mathbfF) :dV \
&= iiint_E langle partial_x, partial_y, partial_z rangle cdot langle y^2,x^2,z rangle : dV \
&= iiint_E partial_x(y^2) + partial_y(x^2) + partial_z(z) : dV \
&=iiint_E 1 :dV \
&=iiint_E 1 :dx : dy : dz\
&=int_0^2pi int_0^3 int_-sqrt25-r^2^sqrt25-r^2 left|fracpartial(x,y,z)partial(x,r,theta) right| dx : dr : dtheta \
&=int_0^2pi int_0^3 int_-sqrt25-r^2^sqrt25-r^2 big|rbig| : dx : dr : dtheta \
&=int_0^2pi int_0^3 int_-sqrt25-r^2^sqrt25-r^2 r : dx : dr : dtheta hspace4mm
mbox since r geq 0 \
&= 2pi int_0^3 r (2sqrt25-r^2) : dr \
&= - 2pi int_25^16 sqrtu: du hspace4mm
mbox where u = 25-r^2, hspace4mm
mbox so hspace4mm
du = -2r :dr
\
&= frac4pi3left( 5^3 - 4^3 right) \
&= boxedfrac2443pi \
endalign*
$$
since given a parametrization in cylindrical coordinates
$$
mathbfr(x,r,theta)
= langle x(x,r,theta), y(x,r,theta), z(x,r,theta)rangle
= langle x,r cos theta,r sin thetarangle, hspace4mm
r geq 0
hspace4mm
mbox and
hspace4mm
0leq theta leq 2pi,
$$
its Jacobian is
$$
beginalign*
fracpartial(x,y,z)partial(x,r,theta)
&=
det
beginpmatrix
fracpartial xpartial x& fracpartial xpartial r & fracpartial xpartial theta \
fracpartial ypartial x& fracpartial ypartial r & fracpartial ypartial theta \
fracpartial zpartial x & fracpartial zpartial r & fracpartial zpartial theta \
endpmatrix \
&=
det
beginpmatrix
1 & 0& 0 \
0 & cos theta & -r sin theta \
0 &sin theta & r cos theta \
endpmatrix \
&= r.
endalign*
$$
answered 14 hours ago
Mee Seong Im
2,5591417
1
exactly! Thanks
– NPLS
14 hours ago
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Use the Divergence Theorem. That is,
$$
beginalign*
iint_E mathbfFcdot dmathbfS &= iiint_E textdiv (mathbfF) :dV \
&= iiint_E langle partial_x, partial_y, partial_z rangle cdot langle y^2,x^2,z rangle : dV \
&= iiint_E partial_x(y^2) + partial_y(x^2) + partial_z(z) : dV \
&=iiint_E 1 :dV \
&=iiint_E 1 :dx : dy : dz\
&=int_0^2pi int_0^3 int_-sqrt25-r^2^sqrt25-r^2 left|fracpartial(x,y,z)partial(x,r,theta) right| dx : dr : dtheta \
&=int_0^2pi int_0^3 int_-sqrt25-r^2^sqrt25-r^2 big|rbig| : dx : dr : dtheta \
&=int_0^2pi int_0^3 int_-sqrt25-r^2^sqrt25-r^2 r : dx : dr : dtheta hspace4mm
mbox since r geq 0 \
&= 2pi int_0^3 r (2sqrt25-r^2) : dr \
&= - 2pi int_25^16 sqrtu: du hspace4mm
mbox where u = 25-r^2, hspace4mm
mbox so hspace4mm
du = -2r :dr
\
&= frac4pi3left( 5^3 - 4^3 right) \
&= boxedfrac2443pi \
endalign*
$$
since given a parametrization in cylindrical coordinates
$$
mathbfr(x,r,theta)
= langle x(x,r,theta), y(x,r,theta), z(x,r,theta)rangle
= langle x,r cos theta,r sin thetarangle, hspace4mm
r geq 0
hspace4mm
mbox and
hspace4mm
0leq theta leq 2pi,
$$
its Jacobian is
$$
beginalign*
fracpartial(x,y,z)partial(x,r,theta)
&=
det
beginpmatrix
fracpartial xpartial x& fracpartial xpartial r & fracpartial xpartial theta \
fracpartial ypartial x& fracpartial ypartial r & fracpartial ypartial theta \
fracpartial zpartial x & fracpartial zpartial r & fracpartial zpartial theta \
endpmatrix \
&=
det
beginpmatrix
1 & 0& 0 \
0 & cos theta & -r sin theta \
0 &sin theta & r cos theta \
endpmatrix \
&= r.
endalign*
$$
answered 14 hours ago
Mee Seong Im
2,5591417
Use the Divergence Theorem. That is,
$$
beginalign*
iint_E mathbfFcdot dmathbfS &= iiint_E textdiv (mathbfF) :dV \
&= iiint_E langle partial_x, partial_y, partial_z rangle cdot langle y^2,x^2,z rangle : dV \
&= iiint_E partial_x(y^2) + partial_y(x^2) + partial_z(z) : dV \
&=iiint_E 1 :dV \
&=iiint_E 1 :dx : dy : dz\
&=int_0^2pi int_0^3 int_-sqrt25-r^2^sqrt25-r^2 left|fracpartial(x,y,z)partial(x,r,theta) right| dx : dr : dtheta \
&=int_0^2pi int_0^3 int_-sqrt25-r^2^sqrt25-r^2 big|rbig| : dx : dr : dtheta \
&=int_0^2pi int_0^3 int_-sqrt25-r^2^sqrt25-r^2 r : dx : dr : dtheta hspace4mm
mbox since r geq 0 \
&= 2pi int_0^3 r (2sqrt25-r^2) : dr \
&= - 2pi int_25^16 sqrtu: du hspace4mm
mbox where u = 25-r^2, hspace4mm
mbox so hspace4mm
du = -2r :dr
\
&= frac4pi3left( 5^3 - 4^3 right) \
&= boxedfrac2443pi \
endalign*
$$
since given a parametrization in cylindrical coordinates
$$
mathbfr(x,r,theta)
= langle x(x,r,theta), y(x,r,theta), z(x,r,theta)rangle
= langle x,r cos theta,r sin thetarangle, hspace4mm
r geq 0
hspace4mm
mbox and
hspace4mm
0leq theta leq 2pi,
$$
its Jacobian is
$$
beginalign*
fracpartial(x,y,z)partial(x,r,theta)
&=
det
beginpmatrix
fracpartial xpartial x& fracpartial xpartial r & fracpartial xpartial theta \
fracpartial ypartial x& fracpartial ypartial r & fracpartial ypartial theta \
fracpartial zpartial x & fracpartial zpartial r & fracpartial zpartial theta \
endpmatrix \
&=
det
beginpmatrix
1 & 0& 0 \
0 & cos theta & -r sin theta \
0 &sin theta & r cos theta \
endpmatrix \
&= r.
endalign*
$$
answered 14 hours ago
Mee Seong Im
2,5591417
answered 14 hours ago
Mee Seong Im
2,5591417
answered 14 hours ago
Mee Seong Im
2,5591417
answered 14 hours ago
Mee Seong Im
2,5591417
2,5591417
1
exactly! Thanks
– NPLS
14 hours ago
add a comment |Â
1
exactly! Thanks
– NPLS
14 hours ago
1
1
exactly! Thanks
– NPLS
14 hours ago
exactly! Thanks
– NPLS
14 hours ago
add a comment |Â
up vote
1
down vote
You're going to want to convert to cylindrical coordinates, with $z$ being identified with the $x$-axis. (A diagram helps) The integral becomes $$int_0^2piint_0^3int_-sqrt25-r^2^sqrt25-r^2rcdot textdiv(F) dz dr dtheta$$
Where $r$ is the Jacobian of the transformation $(r,theta,z)mapsto (z,rcos(theta),rsin(theta))$. The integral isn't so hard, so I'll leave it to "u" to finish :)
answered 15 hours ago
Rafay A.
564
1
shouldn't it be $dxdrdtheta$ ? becouse my cylinder is parallel to the x-axis
– NPLS
14 hours ago
@NPLS you can use $dx : dr : dtheta$ or $dz : dr : dtheta$, where we identify the variable $z$ to be the coordinate along the $x$-axis.
– Mee Seong Im
14 hours ago
add a comment |Â
up vote
1
down vote
You're going to want to convert to cylindrical coordinates, with $z$ being identified with the $x$-axis. (A diagram helps) The integral becomes $$int_0^2piint_0^3int_-sqrt25-r^2^sqrt25-r^2rcdot textdiv(F) dz dr dtheta$$
Where $r$ is the Jacobian of the transformation $(r,theta,z)mapsto (z,rcos(theta),rsin(theta))$. The integral isn't so hard, so I'll leave it to "u" to finish :)
answered 15 hours ago
Rafay A.
564
1
shouldn't it be $dxdrdtheta$ ? becouse my cylinder is parallel to the x-axis
– NPLS
14 hours ago
@NPLS you can use $dx : dr : dtheta$ or $dz : dr : dtheta$, where we identify the variable $z$ to be the coordinate along the $x$-axis.
– Mee Seong Im
14 hours ago
add a comment |Â
up vote
1
down vote
up vote
1
down vote
You're going to want to convert to cylindrical coordinates, with $z$ being identified with the $x$-axis. (A diagram helps) The integral becomes $$int_0^2piint_0^3int_-sqrt25-r^2^sqrt25-r^2rcdot textdiv(F) dz dr dtheta$$
Where $r$ is the Jacobian of the transformation $(r,theta,z)mapsto (z,rcos(theta),rsin(theta))$. The integral isn't so hard, so I'll leave it to "u" to finish :)
answered 15 hours ago
Rafay A.
564
You're going to want to convert to cylindrical coordinates, with $z$ being identified with the $x$-axis. (A diagram helps) The integral becomes $$int_0^2piint_0^3int_-sqrt25-r^2^sqrt25-r^2rcdot textdiv(F) dz dr dtheta$$
Where $r$ is the Jacobian of the transformation $(r,theta,z)mapsto (z,rcos(theta),rsin(theta))$. The integral isn't so hard, so I'll leave it to "u" to finish :)
answered 15 hours ago
Rafay A.
564
answered 15 hours ago
Rafay A.
564
answered 15 hours ago
Rafay A.
564
answered 15 hours ago
Rafay A.
564
564
1
shouldn't it be $dxdrdtheta$ ? becouse my cylinder is parallel to the x-axis
– NPLS
14 hours ago
@NPLS you can use $dx : dr : dtheta$ or $dz : dr : dtheta$, where we identify the variable $z$ to be the coordinate along the $x$-axis.
– Mee Seong Im
14 hours ago
add a comment |Â
1
shouldn't it be $dxdrdtheta$ ? becouse my cylinder is parallel to the x-axis
– NPLS
14 hours ago
@NPLS you can use $dx : dr : dtheta$ or $dz : dr : dtheta$, where we identify the variable $z$ to be the coordinate along the $x$-axis.
– Mee Seong Im
14 hours ago
1
1
shouldn't it be $dxdrdtheta$ ? becouse my cylinder is parallel to the x-axis
– NPLS
14 hours ago
shouldn't it be $dxdrdtheta$ ? becouse my cylinder is parallel to the x-axis
– NPLS
14 hours ago
@NPLS you can use $dx : dr : dtheta$ or $dz : dr : dtheta$, where we identify the variable $z$ to be the coordinate along the $x$-axis.
– Mee Seong Im
14 hours ago
@NPLS you can use $dx : dr : dtheta$ or $dz : dr : dtheta$, where we identify the variable $z$ to be the coordinate along the $x$-axis.
– Mee Seong Im
14 hours ago
add a comment |Â
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