Welcome to our community

Be a part of something great, join today!

Flux of the vector field F

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
Hello again!!! :)
I am given the following exercise:
Find the flux of the vector field $\overrightarrow{F}=zx \hat{i}+ zy \hat{j}+z^2 \hat{k}$ of the surface that consists of the first octant of the sphere $x^2+y^2+z^2=a^2(x,y,z \geq 0).$

That's what I did so far:

$\hat{n}=\frac{\nabla{G}}{| \nabla{G} |}=\frac{x \hat{i}+ y\hat{j}+ z\hat{k}}{a}$

Flux=$\int_C{\overrightarrow{F} \cdot \hat{n}}ds=\int_C{(zx \hat{i}+zy \hat{j}+z^2 \hat{k}) \frac{x \hat{i}+y \hat{j} +z \hat{k}}{a}}ds=\int_C{\frac{zx^2+zy^2+z^3}{a}}ds$

How can I continue? Do I have to use spherical coordinates? Or can I solve this with these coordinates? :confused:
 
Last edited:

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,776
Hello again!!! :)
I am given the following exercise:
Find the flux of the vector field $\overrightarrow{F}=-y \hat{i}+ x \hat{j}$ of the surface that consists of the first octant of the sphere $x^2+y^2+z^2=a^2(x,y,z \geq 0).$

That's what I did so far:

$\hat{n}=\frac{\nabla{G}}{| \nabla{G} |}=\frac{x \hat{i}+ y\hat{j}+ z\hat{k}}{a}$

Flux=$\int_C{\overrightarrow{F} \cdot \hat{n}}ds=\int_C{(zx \hat{i}+zy \hat{j}+z^2 \hat{k}) \frac{x \hat{i}+y \hat{j} +z \hat{k}}{a}}ds=\int_C{\frac{zx^2+zy^2+z^3}{a}}ds$

How can I continue? Do I have to use spherical coordinates? Or can I solve this with these coordinates? :confused:
Hi!! :rolleyes:

Which $\overrightarrow{F}$ are you supposed to integrate?

Is it $\overrightarrow{F}=-y \hat{i}+ x \hat{j}$, or $\overrightarrow{F}=zx \hat{i}+zy \hat{j}+z^2 \hat{k}$?

In the first case, you can use the $\overrightarrow{F} = -\rho \mathbf{\hat \phi}$ in cylindrical coordinates.
Or $\overrightarrow{F} = -r \sin \theta \mathbf{\hat \phi}$ in spherical coordinates.

A good alternative is to apply Gauss's divergence theorem.
You will have to compensate for the coordinate planes (assuming they are not included), but they are easy to integrate.
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
Hi!! :rolleyes:

Which $\overrightarrow{F}$ are you supposed to integrate?

Is it $\overrightarrow{F}=-y \hat{i}+ x \hat{j}$, or $\overrightarrow{F}=zx \hat{i}+zy \hat{j}+z^2 \hat{k}$?

In the first case, you can use the $\overrightarrow{F} = -\rho \mathbf{\hat \phi}$ in cylindrical coordinates.
Or $\overrightarrow{F} = -r \sin \theta \mathbf{\hat \phi}$ in spherical coordinates.

A good alternative is to apply Gauss's divergence theorem.
You will have to compensate for the coordinate planes (assuming they are not included), but they are easy to integrate.
Oh,sorry!!!! :eek: I accidentally wrote the vector field of an other exercise..I meant this vector field: $\overrightarrow{F}=zx \hat{i}+zy \hat{j}+z^2 \hat{k}$ !!!
Why can I apply the Gauss's divergence theorem?? Is it a closed path??
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,776
Oh,sorry!!!! :eek: I accidentally wrote the vector field of an other exercise..I meant this vector field: $\overrightarrow{F}=zx \hat{i}+zy \hat{j}+z^2 \hat{k}$ !!!
Now that I'm looking a bit better, I can see that $\overrightarrow{F}$ can be written in spherical form fairly easily.
Can you find it?

So a spherical integration seems appropriate.


Why can I apply the Gauss's divergence theorem?? Is it a closed path??
Well, you wrote an integral on a path, but apparently you are supposed to integrate over a surface.

And no, it is not a closed surface, but if you include the coordinate planes, it is a closed surface. Afterward you will have to subtract the integrations over the coordinate planes.

For reference, Gauss's divergence theorem is:
$$\bigcirc\!\!\!\!\!\!\!\!\iint_S \mathbf F \cdot \mathbf{\hat n}dS = \iiint_V \nabla \cdot \mathbf F dV$$
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
Now that I'm looking a bit better, I can see that $\overrightarrow{F}$ can be written in spherical form fairly easily.
Can you find it?

So a spherical integration seems appropriate.

Well, you wrote an integral on a path, but apparently you are supposed to integrate over a surface.

And no, it is not a closed surface, but if you include the coordinate planes, it is a closed surface. Afterward you will have to subtract the integrations over the coordinate planes.

For reference, Gauss's divergence theorem is:
$$\bigcirc\!\!\!\!\!\!\!\!\iint_S \mathbf F \cdot \mathbf{\hat n}dS = \iiint_V \nabla \cdot \mathbf F dV$$
The spherical coordinates are:
$x=a \cos{\theta} \sin{\phi}$
$y=a \sin{\theta} \sin{\phi}$
$z=a \cos{\phi}$
$0 \leq \theta \leq \frac{\pi}{4}$
$0 \leq \phi \leq \frac{\pi}{4}$

So $\overrightarrow{F}=a^2 \cos{\theta} \sin{\phi}\cos{\phi}\hat{i}+a^2\sin{\theta} \sin{\phi}\cos{\phi}\hat{j}+a^2\cos^2{\phi}\hat{k}$.

$\hat{n}=\frac{a \cos{\theta} \sin{\phi} \hat{i}+a \sin{\theta} \sin{\phi} \hat{j}+a \cos{\phi} \hat{k}}{a}=\cos{\theta} \sin{\phi} \hat{i}+ \sin{\theta} \sin{\phi} \hat{j}+ \cos{\phi} \hat{k}$

Flux=$\int_C{\overrightarrow{F} \cdot \hat{n}}ds=\int_C{(a^2\cos^2\theta \sin^2{\phi}\cos{\phi}+a^2\sin^2{\theta} \sin^2{\phi}\cos{\phi}+a^2\cos^3{\phi})}d\theta d\phi=\int_0^{\frac{\pi}{4}}\int_0^{\frac{\pi}{4}}{a^2\cos{\phi}}d\theta d\phi=\int_0^{\frac{\pi}{4}}{\frac{\pi}{4}a^2\cos{\phi} d \phi}=\frac{\pi}{4}a^2\frac{\sqrt{2}}{2}$

I hope that the calculations are right...

But..is that what I did correct or have I done something wrong?? (Thinking)
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,776
The spherical coordinates are:
$x=a \cos{\theta} \sin{\phi}$
$y=a \sin{\theta} \sin{\phi}$
$z=a \cos{\phi}$
$0 \leq \theta \leq \frac{\pi}{4}$
$0 \leq \phi \leq \frac{\pi}{4}$

So $\overrightarrow{F}=a^2 \cos{\theta} \sin{\phi}\cos{\phi}\hat{i}+a^2\sin{\theta} \sin{\phi}\cos{\phi}\hat{j}+a^2\cos^2{\phi}\hat{k}$.

$\hat{n}=\frac{a \cos{\theta} \sin{\phi} \hat{i}+a \sin{\theta} \sin{\phi} \hat{j}+a \cos{\phi} \hat{k}}{a}=\cos{\theta} \sin{\phi} \hat{i}+ \sin{\theta} \sin{\phi} \hat{j}+ \cos{\phi} \hat{k}$

Flux=$\int_C{\overrightarrow{F} \cdot \hat{n}}ds=\int_C{(a^2\cos^2\theta \sin^2{\phi}\cos{\phi}+a^2\sin^2{\theta} \sin^2{\phi}\cos{\phi}+a^2\cos^3{\phi})}d\theta d\phi=\int_0^{\frac{\pi}{4}}\int_0^{\frac{\pi}{4}}{a^2\cos{\phi}}d\theta d\phi=\int_0^{\frac{\pi}{4}}{\frac{\pi}{4}a^2\cos{\phi} d \phi}=\frac{\pi}{4}a^2\frac{\sqrt{2}}{2}$

I hope that the calculations are right...

But..is that what I did correct or have I done something wrong?? (Thinking)
Almost.

Your calculation of $\mathbf F \cdot \mathbf{\hat n}$ is correct.

But in spherical coordinates the surface element $ds = r^2\sin\phi d\theta d\phi = a^2\sin\phi d\theta d\phi$.
(Formally: the absolute value of the Jacobian is $r^2\sin\phi $.)
You have left out this factor $a^2\sin\phi$.

Oh, and the angle limits of an octant are $\frac \pi 2$ instead of $\frac \pi 4$.



Btw, an easier calculation of the dot product is by using the local spherical basis $(\mathbf{\hat r}, \mathbf{\hat \phi}, \mathbf{\hat \theta})$:
$$\mathbf F = z(x\mathbf{\hat i} + y\mathbf{\hat j} + z\mathbf{\hat k}) = zr\mathbf{\hat r}$$
$$\mathbf{\hat n} = \mathbf{\hat r}$$
Therefore:
$$\mathbf F \cdot \mathbf{\hat n} = zr\mathbf{\hat r} \cdot \mathbf{\hat r} = zr = r\cos\phi \cdot r = a^2 \cos\phi$$
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
Almost.

Your calculation of $\mathbf F \cdot \mathbf{\hat n}$ is correct.

But in spherical coordinates the surface element $ds = r^2\sin\phi d\theta d\phi = a^2\sin\phi d\theta d\phi$.
(Formally: the absolute value of the Jacobian is $r^2\sin\phi $.)
You have left out this factor $a^2\sin\phi$.

Oh, and the angle limits of an octant are $\frac \pi 2$ instead of $\frac \pi 4$.



Btw, an easier calculation of the dot product is by using the local spherical basis $(\mathbf{\hat r}, \mathbf{\hat \phi}, \mathbf{\hat \theta})$:
$$\mathbf F = z(x\mathbf{\hat i} + y\mathbf{\hat j} + z\mathbf{\hat k}) = zr\mathbf{\hat r}$$
$$\mathbf{\hat n} = \mathbf{\hat r}$$
Therefore:
$$\mathbf F \cdot \mathbf{\hat n} = zr\mathbf{\hat r} \cdot \mathbf{\hat r} = zr = r\cos\phi \cdot r = a^2 \cos\phi$$
A ok.. :eek: So is it like that?

Flux=$\int_C{\overrightarrow{F} \cdot \hat{n}}ds=\int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}}{(a^2 \cos^2{\theta} \sin^2{\phi} \cos{\phi}+a^2 \sin^2{\theta}\sin^2{\phi}\cos{\phi}+a^2 \cos^3{\phi})a^2 \sin{\theta}}d \theta d \phi=\int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}}{a^4 \cos{\phi} \sin{\theta}}d \theta d \phi=\int_0^{\frac{\pi}{2}}{a^4 \cos{\phi}}d\phi=a^4$
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,776
A ok.. :eek: So is it like that?

Flux=$\int_C{\overrightarrow{F} \cdot \hat{n}}ds=\int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}}{(a^2 \cos^2{\theta} \sin^2{\phi} \cos{\phi}+a^2 \sin^2{\theta}\sin^2{\phi}\cos{\phi}+a^2 \cos^3{\phi})a^2 \sin{\theta}}d \theta d \phi=\int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}}{a^4 \cos{\phi} \sin{\theta}}d \theta d \phi=\int_0^{\frac{\pi}{2}}{a^4 \cos{\phi}}d\phi=a^4$
Almost.

You appear to have chosen $\phi$ as the angle with the z-axis (which surprises me actually, but conventions vary).
With that choice the extra factor must be $a^2\sin\phi$ instead of $a^2\sin\theta$. :eek:
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
Almost.

You appear to have chosen $\phi$ as the angle with the z-axis (which surprises me actually, but conventions vary).
With that choice the extra factor must be $a^2\sin\phi$ instead of $a^2\sin\theta$. :eek:
I changed this:
Flux=$\int_C{\overrightarrow{F} \cdot \hat{n}}ds=\int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}}{(a^2 \cos^2{\theta} \sin^2{\phi} \cos{\phi}+a^2 \sin^2{\theta}\sin^2{\phi}\cos{\phi}+a^2 \cos^3{\phi})a^2 \sin{\phi}}d \theta d \phi=\int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}}{a^4 \cos{\phi} \sin{\phi}}d \theta d \phi=\int_0^{\frac{\pi}{2}}{a^4 \frac{\pi}{2}\cos{\phi} \sin{\phi}}d\phi=a^4\frac{\pi}{4}$

Is it right now or have I done something wrong? :eek:
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,776
I changed this:
Flux=$\int_C{\overrightarrow{F} \cdot \hat{n}}ds=\int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}}{(a^2 \cos^2{\theta} \sin^2{\phi} \cos{\phi}+a^2 \sin^2{\theta}\sin^2{\phi}\cos{\phi}+a^2 \cos^3{\phi})a^2 \sin{\phi}}d \theta d \phi=\int_0^{\frac{\pi}{2}} \int_0^{\frac{\pi}{2}}{a^4 \cos{\phi} \sin{\phi}}d \theta d \phi=\int_0^{\frac{\pi}{2}}{a^4 \frac{\pi}{2}\cos{\phi} \sin{\phi}}d\phi=a^4\frac{\pi}{4}$

Is it right now or have I done something wrong? :eek:
It is right now!!! (Wink)
 

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720

evinda

Well-known member
MHB Site Helper
Apr 13, 2013
3,720
It is right now!!! (Wink)
I am looking again at the exercise... Is it maybe like that:

$$ds= \frac{|\nabla{f}|}{|\nabla{f} \cdot \hat{k}|}$$

? Or am I wrong? :confused: (Blush)

I found that $\displaystyle{ \frac{|\nabla{f}|}{|\nabla{f} \cdot \hat{k}|}=\frac{a}{z}}$..
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,776
I am looking again at the exercise... Is it maybe like that:

$$ds= \frac{|\nabla{f}|}{|\nabla{f} \cdot \hat{k}|}$$

? Or am I wrong? :confused: (Blush)

I found that $\displaystyle{ \frac{|\nabla{f}|}{|\nabla{f} \cdot \hat{k}|}=\frac{a}{z}}$..
What you write is correct, but I don't quite see how you might solve the problem with it.

I think you are referring to the formula:
$$\iint g(\overrightarrow R)d\sigma = \iint g(\overrightarrow R) \frac{|\nabla{f}|}{|\nabla{f} \cdot \hat{k}|}dA$$
But this formula only works for integrating a scalar function $g(\overrightarrow R)$ over a surface given by $f(\overrightarrow R)=0$.

Instead, in this problem, a vector function has to be integrated. (Doh)