Flux across the surface $x^{2} + y^{2} + z^{2} = 1$
$begingroup$
Let $S$ be the oriented surface $x^{2} + y^{2} + z^{2} =1$ with the unit normal $hat{n}$ pointing outward for the vector field $F = xi +yj+ zk$, the value of double integration of $vec{F} . hat{n} dS$ is -
We need to evaluate flux across the given surface, which can easily be done by gauss divergence formula and it comes out to be $4pi$.
I am trying to solve this by below method-
$int int vec{F} . hat{n} dS$
$$= int int vec{F} . frac{grad S}{ | grad S| } frac{|grad S|}{|grad S. hat{p}|} dA$$ ( this integration is over R which is shadow region of surface S, $hat{p}$ is the unit normal vector to R)
So, flux = $$int int frac{F.grad S}{|grad S. hat{p}|} dA$$
$$= intint frac{(xi + yj + zk).(2xi +2yj+2zk)dxdy}{|2z. hat{k}|}$$
$$= intint frac{2x^{2} + 2y^{2} + 2z^{2}}{2z} dxdy$$
$$= int int (1/z) dxdy$$
$$= int int 1/(1 - x^{2} -y^{2})^{1/2}dxdy
$$
further, this integration leads to $2pi$ which is wrong. So, where did I go wrong?
calculus multivariable-calculus
asked Dec 23 '18 at 7:05
MathsaddictMathsaddict
3669
$endgroup$
|
show 4 more comments
$begingroup$
Let $S$ be the oriented surface $x^{2} + y^{2} + z^{2} =1$ with the unit normal $hat{n}$ pointing outward for the vector field $F = xi +yj+ zk$, the value of double integration of $vec{F} . hat{n} dS$ is -
We need to evaluate flux across the given surface, which can easily be done by gauss divergence formula and it comes out to be $4pi$.
I am trying to solve this by below method-
$int int vec{F} . hat{n} dS$
$$= int int vec{F} . frac{grad S}{ | grad S| } frac{|grad S|}{|grad S. hat{p}|} dA$$ ( this integration is over R which is shadow region of surface S, $hat{p}$ is the unit normal vector to R)
So, flux = $$int int frac{F.grad S}{|grad S. hat{p}|} dA$$
$$= intint frac{(xi + yj + zk).(2xi +2yj+2zk)dxdy}{|2z. hat{k}|}$$
$$= intint frac{2x^{2} + 2y^{2} + 2z^{2}}{2z} dxdy$$
$$= int int (1/z) dxdy$$
$$= int int 1/(1 - x^{2} -y^{2})^{1/2}dxdy
$$
further, this integration leads to $2pi$ which is wrong. So, where did I go wrong?
calculus multivariable-calculus
asked Dec 23 '18 at 7:05
MathsaddictMathsaddict
3669
$endgroup$
1
$begingroup$
It's not wrong. $2pi$ is the flux over the upper half of the sphere. $4pi$ is the entire sphere.
$endgroup$
– Dylan
Dec 23 '18 at 7:22
$begingroup$
@Dylan why does the above integration find the flux over the upper half sphere only?
$endgroup$
– Mathsaddict
Dec 23 '18 at 7:27
1
$begingroup$
Both the upper and the lower half have the same projection on the $Bbb R^2$ plane. So the projection integral is equivalent to only one half of the sphere
$endgroup$
– Dylan
Dec 23 '18 at 7:29
1
$begingroup$
It's a little inconvenient to answer when you leave out the bounds of the integration when that's exactly what you're confused about. Probably putting them in would resolve your own confusion.
$endgroup$
– zoidberg
Dec 23 '18 at 7:30
$begingroup$
@Mathsaddict Why did you put a modulus around $vec{nabla S}cdothat p$?
$endgroup$
– Shubham Johri
Dec 23 '18 at 10:49
|
show 4 more comments
$begingroup$
Let $S$ be the oriented surface $x^{2} + y^{2} + z^{2} =1$ with the unit normal $hat{n}$ pointing outward for the vector field $F = xi +yj+ zk$, the value of double integration of $vec{F} . hat{n} dS$ is -
We need to evaluate flux across the given surface, which can easily be done by gauss divergence formula and it comes out to be $4pi$.
I am trying to solve this by below method-
$int int vec{F} . hat{n} dS$
$$= int int vec{F} . frac{grad S}{ | grad S| } frac{|grad S|}{|grad S. hat{p}|} dA$$ ( this integration is over R which is shadow region of surface S, $hat{p}$ is the unit normal vector to R)
So, flux = $$int int frac{F.grad S}{|grad S. hat{p}|} dA$$
$$= intint frac{(xi + yj + zk).(2xi +2yj+2zk)dxdy}{|2z. hat{k}|}$$
$$= intint frac{2x^{2} + 2y^{2} + 2z^{2}}{2z} dxdy$$
$$= int int (1/z) dxdy$$
$$= int int 1/(1 - x^{2} -y^{2})^{1/2}dxdy
$$
further, this integration leads to $2pi$ which is wrong. So, where did I go wrong?
calculus multivariable-calculus
asked Dec 23 '18 at 7:05
MathsaddictMathsaddict
3669
$endgroup$
Let $S$ be the oriented surface $x^{2} + y^{2} + z^{2} =1$ with the unit normal $hat{n}$ pointing outward for the vector field $F = xi +yj+ zk$, the value of double integration of $vec{F} . hat{n} dS$ is -
We need to evaluate flux across the given surface, which can easily be done by gauss divergence formula and it comes out to be $4pi$.
I am trying to solve this by below method-
$int int vec{F} . hat{n} dS$
$$= int int vec{F} . frac{grad S}{ | grad S| } frac{|grad S|}{|grad S. hat{p}|} dA$$ ( this integration is over R which is shadow region of surface S, $hat{p}$ is the unit normal vector to R)
So, flux = $$int int frac{F.grad S}{|grad S. hat{p}|} dA$$
$$= intint frac{(xi + yj + zk).(2xi +2yj+2zk)dxdy}{|2z. hat{k}|}$$
$$= intint frac{2x^{2} + 2y^{2} + 2z^{2}}{2z} dxdy$$
$$= int int (1/z) dxdy$$
$$= int int 1/(1 - x^{2} -y^{2})^{1/2}dxdy
$$
further, this integration leads to $2pi$ which is wrong. So, where did I go wrong?
calculus multivariable-calculus
calculus multivariable-calculus
asked Dec 23 '18 at 7:05
MathsaddictMathsaddict
3669
asked Dec 23 '18 at 7:05
MathsaddictMathsaddict
3669
edited Dec 23 '18 at 7:19
Mathsaddict
asked Dec 23 '18 at 7:05
MathsaddictMathsaddict
3669
asked Dec 23 '18 at 7:05
MathsaddictMathsaddict
3669
asked Dec 23 '18 at 7:05
MathsaddictMathsaddict
3669
3669
1
$begingroup$
It's not wrong. $2pi$ is the flux over the upper half of the sphere. $4pi$ is the entire sphere.
$endgroup$
– Dylan
Dec 23 '18 at 7:22
$begingroup$
@Dylan why does the above integration find the flux over the upper half sphere only?
$endgroup$
– Mathsaddict
Dec 23 '18 at 7:27
1
$begingroup$
Both the upper and the lower half have the same projection on the $Bbb R^2$ plane. So the projection integral is equivalent to only one half of the sphere
$endgroup$
– Dylan
Dec 23 '18 at 7:29
1
$begingroup$
It's a little inconvenient to answer when you leave out the bounds of the integration when that's exactly what you're confused about. Probably putting them in would resolve your own confusion.
$endgroup$
– zoidberg
Dec 23 '18 at 7:30
$begingroup$
@Mathsaddict Why did you put a modulus around $vec{nabla S}cdothat p$?
$endgroup$
– Shubham Johri
Dec 23 '18 at 10:49
|
show 4 more comments
1
$begingroup$
It's not wrong. $2pi$ is the flux over the upper half of the sphere. $4pi$ is the entire sphere.
$endgroup$
– Dylan
Dec 23 '18 at 7:22
$begingroup$
@Dylan why does the above integration find the flux over the upper half sphere only?
$endgroup$
– Mathsaddict
Dec 23 '18 at 7:27
1
$begingroup$
Both the upper and the lower half have the same projection on the $Bbb R^2$ plane. So the projection integral is equivalent to only one half of the sphere
$endgroup$
– Dylan
Dec 23 '18 at 7:29
1
$begingroup$
It's a little inconvenient to answer when you leave out the bounds of the integration when that's exactly what you're confused about. Probably putting them in would resolve your own confusion.
$endgroup$
– zoidberg
Dec 23 '18 at 7:30
$begingroup$
@Mathsaddict Why did you put a modulus around $vec{nabla S}cdothat p$?
$endgroup$
– Shubham Johri
Dec 23 '18 at 10:49
1
1
$begingroup$
It's not wrong. $2pi$ is the flux over the upper half of the sphere. $4pi$ is the entire sphere.
$endgroup$
– Dylan
Dec 23 '18 at 7:22
$begingroup$
It's not wrong. $2pi$ is the flux over the upper half of the sphere. $4pi$ is the entire sphere.
$endgroup$
– Dylan
Dec 23 '18 at 7:22
$begingroup$
@Dylan why does the above integration find the flux over the upper half sphere only?
$endgroup$
– Mathsaddict
Dec 23 '18 at 7:27
$begingroup$
@Dylan why does the above integration find the flux over the upper half sphere only?
$endgroup$
– Mathsaddict
Dec 23 '18 at 7:27
1
1
$begingroup$
Both the upper and the lower half have the same projection on the $Bbb R^2$ plane. So the projection integral is equivalent to only one half of the sphere
$endgroup$
– Dylan
Dec 23 '18 at 7:29
$begingroup$
Both the upper and the lower half have the same projection on the $Bbb R^2$ plane. So the projection integral is equivalent to only one half of the sphere
$endgroup$
– Dylan
Dec 23 '18 at 7:29
1
1
$begingroup$
It's a little inconvenient to answer when you leave out the bounds of the integration when that's exactly what you're confused about. Probably putting them in would resolve your own confusion.
$endgroup$
– zoidberg
Dec 23 '18 at 7:30
$begingroup$
It's a little inconvenient to answer when you leave out the bounds of the integration when that's exactly what you're confused about. Probably putting them in would resolve your own confusion.
$endgroup$
– zoidberg
Dec 23 '18 at 7:30
$begingroup$
@Mathsaddict Why did you put a modulus around $vec{nabla S}cdothat p$?
$endgroup$
– Shubham Johri
Dec 23 '18 at 10:49
$begingroup$
@Mathsaddict Why did you put a modulus around $vec{nabla S}cdothat p$?
$endgroup$
– Shubham Johri
Dec 23 '18 at 10:49
|
show 4 more comments
1 Answer
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$begingroup$
Note that $vec F=xhat i+yhat j+zhat k=vec r,hat n=hat r thereforevec Fcdotvec r=r=sqrt{x^2+y^2+z^2}=1$
$dS=rdthetacdot rsintheta dphi=r^2sintheta dtheta dphi=sintheta dtheta dphi$
$displaystylethereforeintintvec Fcdothat n dS=int_0^{2pi}int_0^{pi}sintheta dtheta dphi=4pi$.
As for your method, the flux through the upper half is $2pi$. Add to that the $2pi$ of the lower half.
answered Dec 23 '18 at 7:24
Shubham JohriShubham Johri
5,204718
$endgroup$
add a comment |
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$begingroup$
Note that $vec F=xhat i+yhat j+zhat k=vec r,hat n=hat r thereforevec Fcdotvec r=r=sqrt{x^2+y^2+z^2}=1$
$dS=rdthetacdot rsintheta dphi=r^2sintheta dtheta dphi=sintheta dtheta dphi$
$displaystylethereforeintintvec Fcdothat n dS=int_0^{2pi}int_0^{pi}sintheta dtheta dphi=4pi$.
As for your method, the flux through the upper half is $2pi$. Add to that the $2pi$ of the lower half.
answered Dec 23 '18 at 7:24
Shubham JohriShubham Johri
5,204718
$endgroup$
add a comment |
$begingroup$
Note that $vec F=xhat i+yhat j+zhat k=vec r,hat n=hat r thereforevec Fcdotvec r=r=sqrt{x^2+y^2+z^2}=1$
$dS=rdthetacdot rsintheta dphi=r^2sintheta dtheta dphi=sintheta dtheta dphi$
$displaystylethereforeintintvec Fcdothat n dS=int_0^{2pi}int_0^{pi}sintheta dtheta dphi=4pi$.
As for your method, the flux through the upper half is $2pi$. Add to that the $2pi$ of the lower half.
answered Dec 23 '18 at 7:24
Shubham JohriShubham Johri
5,204718
$endgroup$
add a comment |
$begingroup$
Note that $vec F=xhat i+yhat j+zhat k=vec r,hat n=hat r thereforevec Fcdotvec r=r=sqrt{x^2+y^2+z^2}=1$
$dS=rdthetacdot rsintheta dphi=r^2sintheta dtheta dphi=sintheta dtheta dphi$
$displaystylethereforeintintvec Fcdothat n dS=int_0^{2pi}int_0^{pi}sintheta dtheta dphi=4pi$.
As for your method, the flux through the upper half is $2pi$. Add to that the $2pi$ of the lower half.
answered Dec 23 '18 at 7:24
Shubham JohriShubham Johri
5,204718
$endgroup$
Note that $vec F=xhat i+yhat j+zhat k=vec r,hat n=hat r thereforevec Fcdotvec r=r=sqrt{x^2+y^2+z^2}=1$
$dS=rdthetacdot rsintheta dphi=r^2sintheta dtheta dphi=sintheta dtheta dphi$
$displaystylethereforeintintvec Fcdothat n dS=int_0^{2pi}int_0^{pi}sintheta dtheta dphi=4pi$.
As for your method, the flux through the upper half is $2pi$. Add to that the $2pi$ of the lower half.
answered Dec 23 '18 at 7:24
Shubham JohriShubham Johri
5,204718
answered Dec 23 '18 at 7:24
Shubham JohriShubham Johri
5,204718
answered Dec 23 '18 at 7:24
Shubham JohriShubham Johri
5,204718
answered Dec 23 '18 at 7:24
Shubham JohriShubham Johri
5,204718
5,204718
add a comment |
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$begingroup$
It's not wrong. $2pi$ is the flux over the upper half of the sphere. $4pi$ is the entire sphere.
$endgroup$
– Dylan
Dec 23 '18 at 7:22
$begingroup$
@Dylan why does the above integration find the flux over the upper half sphere only?
$endgroup$
– Mathsaddict
Dec 23 '18 at 7:27
1
$begingroup$
Both the upper and the lower half have the same projection on the $Bbb R^2$ plane. So the projection integral is equivalent to only one half of the sphere
$endgroup$
– Dylan
Dec 23 '18 at 7:29
1
$begingroup$
It's a little inconvenient to answer when you leave out the bounds of the integration when that's exactly what you're confused about. Probably putting them in would resolve your own confusion.
$endgroup$
– zoidberg
Dec 23 '18 at 7:30
$begingroup$
@Mathsaddict Why did you put a modulus around $vec{nabla S}cdothat p$?
$endgroup$
– Shubham Johri
Dec 23 '18 at 10:49