Show that ##\int \frac{d\vec{a}}{d} = \frac{4}{3}\pi \vec{r}##

[happyparticle]

Homework Statement

Show that $\int \frac{d\vec{a}}{d} = \frac{4}{3}\pi \vec{r}$

Relevant Equations

$d\vec{a} = R^2sin\theta d\theta d\phi \hat{r}$

$\hat{r} = sin \theta cos \phi \hat{x} + sin \theta sin \phi \hat{y} + cos \theta \hat{z}$

$d = |\vec{r} - \vec{R}|$ and $\vec{r} = r \hat{z}$

Hey!

I have to show that the integral of the area of a sphere $\int \frac{d\vec{a}}{d} = \frac{4}{3}\pi \vec{r}$, with $d = |\vec{r} - \vec{R}|$ and $\vec{r} = r \hat{z}$

This is what I did.

$d\vec{a} = R^2sin\theta d\theta d\phi \hat{r}$, $\hat{r} = sin \theta cos \phi \hat{x} + sin \theta sin \phi \hat{y} + cos \theta \hat{z}$

Thus,
$\int \frac{d\vec{a}}{d} = \int_0^{2\pi} \int_0^{\pi} \frac{R^2 sin \theta (sin \theta cos \phi \hat{x} + sin \theta sin \phi \hat{y} + cos \theta \hat{z}) d\theta d\phi}{d}$

From here, I can separate this integral into 3 integrals. 2 of 3 = 0 and the last one = $\frac{-2R^2 \pi \hat{z}}{d}$

$= \frac{-2R^2 \pi \hat{z}}{|\vec{r} - \vec{R}|}$

I don't see any error with my integrals. I don't know what I missed. I'm not sure if this is the right way to solve this kind of problem.

 

[haruspex]

You seem to have treated d as constant. It will vary as the surface area element varies across the integral.

 

[happyparticle]

I think you are right, since only R is constant which is the radius of the sphere. However, I don't think it depends of $\theta$ or $\phi$, so I don't see what that changes?

 

[haruspex]

I think you are right, since only R is constant which is the radius of the sphere. However, I don't think it depends of $\theta$ or $\phi$, so I don't see what that changes?

I think it does. You quoted:
$\hat{r} = sin \theta cos \phi \hat{x} + sin \theta sin \phi \hat{y} + cos \theta \hat{z}$
$d = |\vec{r} - \vec{R}|$ and $\vec{r} = r \hat{z}$
For consistency, we need to change the first to $\hat{R} = sin \theta cos \phi \hat{x} + sin \theta sin \phi \hat{y} + cos \theta \hat{z}$, but I don't quite get the second. What is 'r' on the right? Do you mean d consists of the x and y components of $\vec R$? I.e. $\vec{r} = \cos(\theta)\hat{z}$?

 

[happyparticle]

$\hat{r} = sin \theta cos \phi \hat{x} + sin \theta sin \phi \hat{y} + cos \theta \hat{z} = \hat{R}$ ?

Is the r on the right $\vec{r} = r \hat{z}$ ?

I'm a bit confuse.

From what I understand, I have to replace $\vec{r}$ because it is not constant. However, I don't see how $\vec{r}$ is related to $\phi$ or $\theta$.

 

[haruspex]

$\hat{r} = sin \theta cos \phi \hat{x} + sin \theta sin \phi \hat{y} + cos \theta \hat{z} = \hat{R}$ ?

Is the r on the right $\vec{r} = r \hat{z}$ ?

I'm a bit confuse.

From what I understand, I have to replace $\vec{r}$ because it is not constant. However, I don't see how $\vec{r}$ is related to $\phi$ or $\theta$.

In your original post, you used the vector r to mean two different things. In the generic polar coordinate equation ($\hat r=$) it was the vector to a point on the spherical surface. Earlier, R was for the radius of the sphere, and in $d=..$ you introduced $\vec R$, which is presumably also the vector to a point on the spherical surface, and later defined vector r to be something else - but what exactly I am not sure.
You wrote $\vec r=r\hat z$, so clearly it is a vector in the z direction, but since plain "r" on the right is undefined it cannot serve as a definition of the vector r.
My guess is that this $\vec r$ is supposed to be the z component of $\vec R$, $\vec r=\vec z$. If so, $|\vec r-\vec R|=\sqrt{x^2+y^2}$. Another possibility is that $\vec r=R\hat z$.

 

[kuruman]

I think you need to write a fixed vector

$\vec r=R(\sin\theta\cos\phi~\hat x+\sin\theta\sin\phi~\hat y+\cos\theta~\hat z)$

and a variable vector

$\vec r'=R(\sin\theta '\cos\phi '~\hat x+\sin\theta '\sin\phi '~\hat y '+\cos\theta '~\hat z)$

then do the 3 integrals over primed coordinates. To make the integrals simpler, I recommend that you pick $\vec R$ in the $\hat z$ direction, which you can always do without loss of generality, because of the spherical symmetry. Doing it that way got me the given answer.

 

[haruspex]

I think you need to write a fixed vector

$\vec r=R(\sin\theta\cos\phi~\hat x+\sin\theta\sin\phi~\hat y+\cos\phi~\hat z)$

and a variable vector

$\vec r'=R(\sin\theta '\cos\phi '~\hat x+\sin\theta '\sin\phi '~\hat y '+\cos\theta '~\hat z)$

then do the 3 integrals over primed coordinates. To make the integrals simpler, I recommend that you pick $\vec R$ in the $\hat z$ direction, which you can always do without loss of generality, because of the spherical symmetry. Doing it that way got me the given answer.

Do you mean pick $\vec r$ in the $\hat z$ direction?

Is that perhaps what $\vec r=r\hat z$ was meant to suggest?

 

[happyparticle]

I think you need to write a fixed vector

$\hat{r} = sin \theta cos \phi \hat{x} + sin \theta sin \phi \hat{y} + cos \theta \hat{z}$ and $d\vec{a}$ are given in the problem

 

[haruspex]

$\hat{r} = sin \theta cos \phi \hat{x} + sin \theta sin \phi \hat{y} + cos \theta \hat{z}$ and $d\vec{a}$ are given in the problem

Are you sure you quoted the rest of the question exactly as given to you? As I have noted, the equations, as a system, don't quite make sense as written.

 

[happyparticle]

It doesn't seems like a made a mistake. Since $\hat{r} = sin \theta cos \phi \hat{x} + sin \theta sin \phi \hat{y} + cos \theta \hat{z}$

$\vec{r} = r( sin \theta cos \phi \hat{x} + sin \theta sin \phi \hat{y} + cos \theta \hat{z} )$ ?

 

[haruspex]

It doesn't seems like a made a mistake. Since $\hat{r} = sin \theta cos \phi \hat{x} + sin \theta sin \phi \hat{y} + cos \theta \hat{z}$

$\vec{r} = r( sin \theta cos \phi \hat{x} + sin \theta sin \phi \hat{y} + cos \theta \hat{z} )$ ?

I mean these equations:

$d = |\vec{r} - \vec{R}|$ and $\vec{r} = r \hat{z}$

 

[kuruman]

Do you mean pick $\vec r$ in the $\hat z$ direction?

Is that perhaps what $\vec r=r\hat z$ was meant to suggest?

Yes, that is what I mean and I suspect that $\vec r=r\hat z$ was meant to suggest that one should put the fixed vector in the $z$-direction to make the integration easy. I prefer the usual EM notation where $\vec r$ is a fixed position vector at some arbitrary point $\{r,\theta,\phi\}$ and $\vec r'$ is a "source" position vector at $\{r',\theta',\phi'\}$. One integrates over the primed to coordinates to find the superposition of vectors at the fixed point.

It seems to me that the notation here is unusual in that $\vec r$ is the fixed vector and $\vec R$ is the source vector. Then if one puts the the fixed vector in the $z$-direction, the prime can be dropped in the coordinates and write the fixed and source vectors as
$\vec r = R~\hat z$
$\vec R=R(\sin\!\theta\cos\!\phi~\hat x+\sin\!\theta\sin\!\phi~\hat y+\cos\!\theta~\hat z)$

It's a bit confusing because it is unconventional but it gives the right answer.

To @EpselonZero : Can you find $d=|\vec r-\vec R|$ using the vectors above and do the integral?

 

[happyparticle]

fixed and source vectors as
r→=R z^

shouldn't be $\vec{r} = r \hat{z}$ as in the statement and since the final answer contains $\vec{r}$?
then $d = \sqrt{R^2 sin^2 \theta cos^2 \phi + R^2 sin^2 \theta sin^2 \phi + R^2 r^2 cos^2 \theta}$

 

[haruspex]

shouldn't be $\vec{r} = r \hat{z}$ as in the statement

Yes, that’s what @kuruman wrote in post #13.

then $d = \sqrt{R^2 sin^2 \theta cos^2 \phi + R^2 sin^2 \theta sin^2 \phi + R^2 r^2 cos^2 \theta}$

No, that's not even dimensionally correct. Try that again from @kuruman's equations.

 

[happyparticle]

I know that I made a mistake here.
If $\vec{r} = r\hat{z}$

$d = |\vec{r} - \vec{R}|$

= $|r \hat{z} - R(sin \theta cos \phi \hat{x} + sin \theta sin \phi \hat{y} + cos \theta \hat{z})|$

= $\sqrt{R^2 sin^2 \theta cos^2 \phi + R^2 sin^2 \theta sin^2 \phi + (r -R cos \theta )^2}$

 

[haruspex]

I know that I made a mistake here.
If $\vec{r} = r\hat{z}$

$d = |\vec{r} - \vec{R}|$

= $|r \hat{z} - R(sin \theta cos \phi \hat{x} + sin \theta sin \phi \hat{y} + cos \theta \hat{z})|$

= $\sqrt{R^2 sin^2 \theta cos^2 \phi + R^2 sin^2 \theta sin^2 \phi + (r -R cos \theta )^2}$

Not quite. As I keep telling you, $\vec{r} = r\hat{z}$ is meaningless since r has not been defined.
Use @kuruman's equations exactly: $\vec{r} = R\hat{z}$.

 

[happyparticle]

It works, when I'm using $\vec{r} = R \hat{z}$. However, I'm not sure to understand why $r = R$

 

[kuruman]

It works, when I'm using $\vec{r} = R \hat{z}$. However, I'm not sure to understand why $r = R$

That puts the point of observation on the surface of the sphere. The statement of the problem is not very clear on this.

 

[haruspex]

It works, when I'm using $\vec{r} = R \hat{z}$. However, I'm not sure to understand why $r = R$

There is no given definition of the scalar r, and you cannot solve the question without knowing what it means. So, if you have quoted the problem exactly, it must contain a mistake. The most obvious correction is to change r to R.

 

[happyparticle]

I see. Indeed, there is no given definition for r. Honestly, I would not have find that without you. Thank you.

 

[haruspex]

I see. Indeed, there is no given definition for r. Honestly, I would not have find that without you. Thank you.

Thank @kuruman for spotting the correct interpretation.