# Calculate surface integral on sphere

In summary, you are supposed to integrate the surface integral A over the sphere with radius R using spherical coordinates. However, you are not sure what the variable e_r is. If you express A in terms of its components in spherical coordinates, you get (1,0,0) which does not provide a resonable answer. You may be confused because thick r is a vector with another r as variable. You should extract the r from A using e_r. Finally, you should change to polar coordinates to do the integration.

Homework Statement
Do surface integral using spherical coordinate system over
$$A = (x, y, z)/(x^2 + y^2 + z^2)^{3/2}$$
Surface is a sphere at origin with radius R.
Relevant Equations
Not gauss
I'm supposed to do the surface integral on A by using spherical coordinates.

$$A = (rsin\theta cos\phi, rsin\theta sin\phi, rcos\theta)/r^{3/2}$$
$$dS = h_{\theta}h_{\phi} d_{\theta}d_{\phi} = r^2sin\theta d_{\theta}d_{\phi}$$

Now I'm trying to do
$$\iint A dS = (rsin\theta cos\phi, rsin\theta sin\phi, rcos\theta)/r^{3/2} * r^2sin\theta d_{\theta}d_{\phi}$$

Which obviously doesn't work because I'm trying to integrate over a vector.

What I suspect I'm supposed to do is
$$\iint A \cdot n dS = \iint A \cdot e_r dS = \iint A_r dS$$
However I'm not sure what e_r is..

If expressed in spherical coordinates it's (1, 0, 0)
But I cant just do
$$(rsin\theta cos\phi, rsin\theta sin\phi, rcos\theta) \cdot (1, 0 ,0)$$

So I'm stuck because I can't extract A_r from A.

If $\mathbf{r} = r\mathbf{e}_r$, then $\mathbf{e}_r = \mathbf{r}/r$.

pasmith said:
If $\mathbf{r} = r\mathbf{e}_r$, then $\mathbf{e}_r = \mathbf{r}/r$.
Is thick r a vector with another r as variable? I always find that extremly confusing..

As I understand it
$$e_r = dA/dr = (sin\theta cos\phi, sin\theta sin\phi, cos\theta)$$
$$\iint A \cdot e_r dS = 2 \pi r \int r^2 sin\theta d\theta = r^3 4\pi$$
But the answer should just be 4 pi.

Even if I had ##e_r = dA/dr /r## I would end up with an r in the solution so I'm not sure what I'm doing wrong.

Does $$A = \frac{(x,y,z)}{(x^2 + y^2 + z^2)^{3/2}} = \frac{\mathbf{r}}{r^3} = \frac{\mathbf{e}_r}{r^2}$$ clarify things?

• Is thick r a vector with another r as variable? I always find that extremally confusing..
The variable ##\mathbf r##, when written in bold typeface (you referred to as "thick"), indicates that ##\mathbf r## is a vector quantity. The variable ##r## in standard typeface is used to refer to the magnitude of the vector ##\mathbf r##.

Alternatively, you may have seen an arrow or harpoon above a variable to indicate a vector quantity. For instance ##\vec r## or ##\vec A## .

Also notice that ##A##, as it is defined in the OP, is a vector quantity. Write as ##\mathbf A## .

As I understand it
$$e_r = dA/dr = (sin\theta cos\phi, sin\theta sin\phi, cos\theta)$$
$$\iint A \cdot e_r dS = 2 \pi r \int r^2 sin\theta d\theta = r^3 4\pi$$
But the answer should just be 4 pi.

Even if I had ##e_r = dA/dr /r## I would end up with an r in the solution so I'm not sure what I'm doing wrong.

Another issue which may be leading to your confusion has to do with coordinate systems versus vector components. For instance, when you write
you are using spherical coordinates, but the components are rectilinear, i.e. they are along ##x,\ y, \text{ and }, \ z## directions. On the other hand, when you wrote ##\mathbf{e}_r = (1,\,0,\,0)##, the components are spherical as well as the coordinates ,

No need to go to spherical components.

Last edited:
Homework Statement:: Do surface integral using spherical coordinate system over
$$A = (x, y, z)/(x^2 + y^2 + z^2)^{3/2}$$
Surface is a sphere at origin with radius R.
Relevant Equations:: Not gauss

I'm supposed to do the surface integral on A by using spherical coordinates.

$$A = (r sin\theta cos\phi, r sin\theta sin\phi, r cos\theta)/r^{3/2}$$
Now, let's look at the OP.

You have an error in that expression for A.

##\displaystyle x^2 + y^2 + z^2 = r^2 ##,

so what you should have for A is:

##\displaystyle \mathbf{A} = \dfrac{(r \sin\theta \cos\phi, r \sin\theta \sin\phi, r \cos\theta)}{r^3}##

What I suspect I'm supposed to do is
$$\iint A \cdot n dS = \iint A \cdot e_r dS$$
However I'm not sure what e_r is..
Don't forget, the integral is over the surface of the sphere of radius ##R##, centered at the origin, so ##r = R##.

You now have all the parts, so ...

Take the scalar product and integrate.

• Its been a while for me, but couldn't we just take the dot product of A (most textbooks have A written as F for vector function) and a unit vector normal to the sphere times Ds? Since the sphere is centered at the origin, we know that a normal unit vector is given by (x,y,z) divided by its magnitude (sorry for no LaTek on a trip and do not have access to a computer) is normal to it.

Now, what is Ds (magnitude of the vector Ds) for a sphere?If you cannot recall, find the surface element on another sheet of paper. Integrate

We can also just find the vector Ds and find the dot product of A and this vector.

Do not forget to change to polar
I glanced at the problem, but did not attempt it on my own. This is the general idea for these types of problems.

Ah yes, a should be divided by r^3 not r^3/2.
I calculated ##e_r = dA/dr * 1/|dA/dr|## which is in normal components.
Then ##A \cdot e_r = 1/r^2## and I solved it.

Thanks!