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

• happyparticle
In summary: Yes, I made a mistake. ##\hat{r}## should be ##\hat{R}## in the first equation.And yes, I meant to pick ##\vec r## in the ##\hat z## direction. Sorry for the confusion.
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.

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

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?

EpselonZero said:
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}##?

##
\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##.

EpselonZero said:
##
\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##.

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.

Last edited:
sysprog
kuruman said:
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?

kuruman said:
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

EpselonZero said:
##
\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.

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}
)## ?

EpselonZero said:
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:
EpselonZero said:
##d = |\vec{r} - \vec{R}|## and ##\vec{r} = r \hat{z}##

PhDeezNutz
haruspex said:
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 and haruspex
kuruman said:
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}##

Last edited by a moderator:
EpselonZero said:
shouldn't be ##\vec{r} = r \hat{z}## as in the statement

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

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}##

EpselonZero said:
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##

EpselonZero said:
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.

EpselonZero said:
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.

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

EpselonZero said:
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.

kuruman

## 1. What is the meaning of the equation ##\int \frac{d\vec{a}}{d} = \frac{4}{3}\pi \vec{r}##?

The equation represents the surface area of a sphere with radius ##\vec{r}##. The integral symbol ##\int## indicates that the equation is calculating the total surface area by summing up the surface area of infinitesimally small sections of the sphere.

## 2. How is the equation derived?

The equation is derived using the formula for the surface area of a sphere, which is ##4\pi r^2##. By taking the derivative of both sides and using the chain rule, we can rewrite the equation in the form of an integral, which gives us the equation ##\int \frac{d\vec{a}}{d} = \frac{4}{3}\pi \vec{r}##.

## 3. What is the significance of the constant ##\frac{4}{3}\pi## in the equation?

The constant ##\frac{4}{3}\pi## is the ratio of the surface area of a sphere to its volume. This is a fundamental constant in geometry and is often used in calculations involving spheres.

## 4. Can this equation be used to calculate the surface area of any shape?

No, this equation is specifically for calculating the surface area of a sphere. Other shapes have their own equations for calculating surface area.

## 5. How is this equation used in real-world applications?

This equation is commonly used in physics and engineering to calculate the surface area of spherical objects, such as planets, stars, and bubbles. It is also used in calculations for fluid dynamics, as the surface area of a sphere is important in determining the pressure and flow of fluids.

### Similar threads

• Introductory Physics Homework Help
Replies
2
Views
1K
• Introductory Physics Homework Help
Replies
17
Views
465
• Introductory Physics Homework Help
Replies
10
Views
336
• Introductory Physics Homework Help
Replies
3
Views
267
• Introductory Physics Homework Help
Replies
1
Views
964
• Introductory Physics Homework Help
Replies
5
Views
395
• Introductory Physics Homework Help
Replies
1
Views
243
• Introductory Physics Homework Help
Replies
3
Views
1K
• Introductory Physics Homework Help
Replies
14
Views
3K
• Introductory Physics Homework Help
Replies
7
Views
1K