Calculating the surface area of a sphere using dA

In summary, the surface area of a sphere using dA is confusing to me. I don't understand why the other side can't just be rd\phi.
  • #1
annamal
381
33
TL;DR Summary
I am confused by the surface area of a sphere using dA. Not sure why the other side can't just be ##Rd\phi##
Below is an image to calculate the surface area of a sphere using dA. I can see how ##rcos\theta d\phi## works, but I don't understand how that side can't just be ##rd\phi## with a slanted circle representing the arc length. The second part I don't understand is why it is integrated from ##\frac{\pi}{2}## to ##\frac{-\pi}{2}##
Screen Shot 2022-04-23 at 12.59.57 AM.png
 
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  • #2
annamal said:
Summary:: I am confused by the surface area of a sphere using dA. Not sure why the other side can't just be ##Rd\phi##

I can see how rcosθdϕ works,
[tex]r\sin\theta d\phi[/tex]
isn't it ? Usually ##\theta## is measured from z axis, not from xy plane. And the side is
[tex]r d\theta[/tex]
 
  • #3
anuttarasammyak said:
[tex]r\sin\theta d\phi[/tex]
isn't it ? Usually ##\theta## is measured from z axis, not from xy plane. And the side is
[tex]r d\theta[/tex]
That is not answering my question. I am naming them based on the image attached.
 
  • #4
Then apply to post#2 which uses ##\theta_{ordinary} [0,\pi]##, the transformation,
[tex]\theta_{yours}=\frac{\pi}{2}-\theta_{ordinary}[/tex] which would give what you want.
 
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  • #5
anuttarasammyak said:
Then apply to post#2 which uses ##\theta_{ordinary} [0,\pi]##, the transformation,
[tex]\theta_{yours}=\frac{\pi}{2}-\theta_{ordinary}[/tex] which would give what you want.
I am not sure where you got the yours and mine...could you explain in more detail?
 
  • #6
In ordinary polar coordinates,
[tex]dA= r^2 \sin \theta dr d\theta d\phi[/tex]
[tex]\theta[0,\pi],\phi[0,2\pi][/tex]
Substitute here
anuttarasammyak said:
[tex]\theta_{yours}=\frac{\pi}{2}-\theta_{ordinary}[/tex]
Your ##\theta## is same as latitude in global positioning with - for south and + for north.
 
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  • #7
The reason ##\theta## is integrated from ##-\frac{\pi}{2}## to ##\frac{\pi}{2}## is because that's how you define it in the picture, ##\theta## is an angle that goes from south to north pole of the sphere as I see it.

As for your question regarding why do you need a cosine factor in your case, why not just ##rd\phi##, I would suggest this kind of experiment. Imagine you fix ##\theta##, and you want to calculate the surface of one 'ring' that is elevated at ##\theta## from the surface. The surface of that ring, you should get if you don't integrate over ##\theta## but just over ##\phi##, right? But if you do that you will find that the surface of such ring would be ##2r\pi \cdot rd\theta##. And that can't be correct, right, the ring's circumference is not ##2r\pi## above the equator, it is ##2r\pi\cos\theta##. So that's why you need that factor, it includes the change of size of those rings because of the curvature of the sphere.
 
  • #8
Antarres said:
The reason ##\theta## is integrated from ##-\frac{\pi}{2}## to ##\frac{\pi}{2}## is because that's how you define it in the picture, ##\theta## is an angle that goes from south to north pole of the sphere as I see it.

As for your question regarding why do you need a cosine factor in your case, why not just ##rd\phi##, I would suggest this kind of experiment. Imagine you fix ##\theta##, and you want to calculate the surface of one 'ring' that is elevated at ##\theta## from the surface. The surface of that ring, you should get if you don't integrate over ##\theta## but just over ##\phi##, right? But if you do that you will find that the surface of such ring would be ##2r\pi \cdot rd\theta##. And that can't be correct, right, the ring's circumference is not ##2r\pi## above the equator, it is ##2r\pi\cos\theta##. So that's why you need that factor, it includes the change of size of those rings because of the curvature of the sphere.
I drew a picture. It seems like the ring's circumference is ##2r\pi## according to my drawing (in red).

Screen Shot 2022-04-23 at 12.59.57 AM.png
 
  • #10
anuttarasammyak said:
No, it doesn’t fit the curve. Horizontal circle centered on z axis of radius r cos ##\theta## does. Vertical red circles fit the sides. Ref. https://en.m.wikipedia.org/wiki/File:FedStats_Lat_long.svg
I am confused. In the picture the red circle seems to fit the curve.
 
  • #11
スライド1.JPG
 
  • #12
anuttarasammyak said:
Yes, I understand the red and blue circles you drew. My question is why the red circle I drew in my image is not right.
 
  • #13
annamal said:
My question is why the red circle I drew in my image is not right.
Imagine you cut globe or watermelon with knife at near its north pole and get a small circle intersection.
Do you still think that red circle fits this small cross section circle ?
 
  • #14
anuttarasammyak said:
Imagine you cut globe or watermelon with knife at near its north pole and get a small circle intersection.
Do you still think that red circle fits this small cross section circle ?
No, but that is not what I am asking. I am asking why that side of the dA cannot be the red circle I drew.
 
  • #15
You admit blue circles fit the top/bottom side of the element. Are you expecting another circle e.g. red one fit these sides also and it is possible ?
 
  • #16
anuttarasammyak said:
You admit blue circles fit the top/bottom side of the element. Are you expecting another circle e.g. red one fit these sides also and it is possible ?
From the drawing it seems possible.
 
  • #17
Can you draw it for the case of post #13 though you said no in #14 ?
 
  • #18
anuttarasammyak said:
Can you draw it for the case of post #13 though you said no in #14 ?
The blue and red circles seem to indicate the same dx
Screen Shot 2022-04-23 at 12.59.57 AM.png
 
  • #19
The blue and the red cross at the point. The do not share the arc side of the element to fit. As I said smaller radius blue circle case or in other words near pole cross section case would show it to us clearly. In the case of equator only they coincide.

If two or more circles fit the same curve, we could not define unique curvature of the curve which is defined as reciprocal of such a circle.
 
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  • #20
anuttarasammyak said:
The blue and the red cross at the point. The do not share the arc side of the element to fit. As I said smaller radius blue circle case or in other words near pole cross section case would show it to us clearly. In the case of equator only they coincide.

If two or more circles fit the same curve, we could not define unique curvature of the curve which is defined as reciprocal of such a circle.
Yes, I know they are not same curve. My thread is asking why I can't just use the red circle's arc as dx.
 
  • #21
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  • #23
Please find attach figure on the coordinates system that I assume your red-circle requirement in #8 could be satisfied. Not mentioned in the figure but you see there exist East pole and West pole as well as NS poles. The curves ##\theta=const.\ \ \phi=const.## do not intersect perpendicularly. It complicates the calculation of dA.
figure.jpg
figure.jpg
 
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  • #24
Your red circle is tilted relative to the ##xy## (##\theta=\frac{\pi}{2}##) plane. But the standard approach to calculate the surface area of a sphere uses surface slices which are all parallel to the ##xy## plane (like anuttarasammyak's blue circle). The infinitesimal surface area of each slice is ##2\pi r^{2}\sin\left(\theta\right)d\theta## and varies according to the height ##r\cos\left(\theta\right)## of the slice above or below the ##xy## plane. As shown below, you then stack up all the parallel slices and add up their areas (by integration over ##\theta##) to get the total surface area ##4\pi r^{2}## of the sphere.
Spherical Coordinates.jpg
 
  • #25
renormalize said:
Your red circle is tilted relative to the ##xy## (##\theta=\frac{\pi}{2}##) plane. But the standard approach to calculate the surface area of a sphere uses surface slices which are all parallel to the ##xy## plane (like anuttarasammyak's blue circle). The infinitesimal surface area of each slice is ##2\pi r^{2}\sin\left(\theta\right)d\theta## and varies according to the height ##r\cos\left(\theta\right)## of the slice above or below the ##xy## plane. As shown below, you then stack up all the parallel slices and add up their areas (by integration over ##\theta##) to get the total surface area ##4\pi r^{2}## of the sphere.
View attachment 300589
Why can't the red circle be tilted though
 
  • #26
anuttarasammyak said:
Please find attach figure on the coordinates system that I assume your red-circle requirement in #8 could be satisfied. Not mentioned in the figure but you see there exist East pole and West pole as well as NS poles. The curves ##\theta=const.\ \ \phi=const.## do not intersect perpendicularly. It complicates the calculation of dA.
View attachment 300588View attachment 300592
Yes, I see and understand that but if we tilted the circle, why can't that be a side to dA?
 
  • #27
annamal said:
if we tilted the circle, why can't that be a side to dA?
It could, if you were able and willing to change all of your formulas so that they did not have built into them the convention that the circles are all lined up a certain way. This convention does not sacrifice any generality: it amounts to lining up the coordinates a certain way on the sphere to make the computation easy.

The diagram you show in your OP does use the convention that the circles are all lined up a certain way, and all of the math that has been discussed in this thread is based on that convention. So you can either use the math you have been shown, which works and is easy, or you can make the problem much, much harder (and have to start over working it from scratch) by refusing to adopt a simple convention that makes the math easy.

Your call.
 
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  • #28
PeterDonis said:
It could, if you were able and willing to change all of your formulas so that they did not have built into them the convention that the circles are all lined up a certain way. This convention does not sacrifice any generality: it amounts to lining up the coordinates a certain way on the sphere to make the computation easy.

The diagram you show in your OP does use the convention that the circles are all lined up a certain way, and all of the math that has been discussed in this thread is based on that convention. So you can either use the math you have been shown, which works and is easy, or you can make the problem much, much harder (and have to start over working it from scratch) by refusing to adopt a simple convention that makes the math easy.

Your call.
Ok, that makes sense. Why aren't we integrating that from 0 to 2*pi? But -pi/2 to pi/2?
 
  • #29
annamal said:
Why aren't we integrating that from 0 to 2*pi? But -pi/2 to pi/2?
We are integrating over ##\theta##, i.e., latitude, which goes from ##- \pi / 2## at the south pole to ##+ \pi / 2## at the north pole.
 
  • #30
PeterDonis said:
We are integrating over ##\theta##, i.e., latitude, which goes from ##- \pi / 2## at the south pole to ##+ \pi / 2## at the north pole.
To be a full circle, doesn't it have to be 2pi? See image.
Screen Shot 2022-04-26 at 8.19.31 PM.png
 
  • #31
Perhaps this bideo from 3blue1brown will give you a different perspective on the sphere

 
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  • #32
annamal said:
To be a full circle
When you go from the South Pole to the North Pole, you only go through half a circle, not a full circle. That is what the integral we are talking about is doing. It is adding up the areas of infinitesimal circular "slices" of the Earth's surface, each "slice" taken at a different latitude, over the full range of latitudes from the South Pole to the North Pole. The area of each infinitesimal slice is the circumference of the circle at latitude ##\theta##, which is ##2 \pi r \cos \theta##, times the infinitesimal width of the slice, which is ##r d \theta##. So we are integrating ##2 \pi r^2 \cos \theta d \theta## from ##\theta = - \pi / 2##, the South Pole, to ##\theta = \pi / 2##, the North Pole.
 
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  • #33
annamal said:
To be a full circle, doesn't it have to be 2pi?

If you wonder about polar coordinates that includes a modified verion of measuring ##\theta## from xy plane not from usual z axis, you can see many explanations in school texts and web contents. Peter also has done a good explanation in the previous post.

If you question about coordinates system that I said in post #23 the figure of which you quoted, imagine as follows.
Imagine all the curves mentioning constant longitude spanning between the North pole and the South pole in your mind.
Imagine the equator circle is made of sponge wet with red ink. Turn it gradually along the East pole - the West pole axis. You see those curves are getting painted red left after the equator circle passes. How much angle should the equator circle turn to paint all parts of those curves ?
And how about it when not the full equator circle but only a half equator circle between EW axis sponge is wet ?
Colored red points on the globe surface have been shown already appointed by the coordinates.
 
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  • #34
anuttarasammyak said:
If you wonder about polar coordinates that includes a modified verion of measuring ##\theta## from xy plane not from usual z axis, you can see many explanations in school texts and web contents. Peter also has done a good explanation in the previous post.

If you question about coordinates system that I said in post #23 the figure of which you quoted, imagine as follows.
Imagine all the curves mentioning constant longitude spanning between the North pole and the South pole in your mind.
Imagine the equator circle is made of sponge wet with red ink. Turn it gradually along the East pole - the West pole axis. You see those curves are getting painted red left after the equator circle passes. How much angle should the equator circle turn to paint all parts of those curves ?
And how about it when not the full equator circle but only a half equator circle between EW axis sponge is wet ?
Colored red points on the globe surface have been shown already appointed by the coordinates.
I get it now. Thank you!
 
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Related to Calculating the surface area of a sphere using dA

1. How do you calculate the surface area of a sphere using dA?

The surface area of a sphere can be calculated using the formula dA = 4πr2, where dA represents the differential area, or small section of the sphere's surface, and r is the radius of the sphere.

2. What is the purpose of using dA in the surface area formula for a sphere?

The use of dA in the formula allows for the calculation of the surface area of a sphere by breaking it down into smaller, manageable sections. This makes the calculation more accurate and easier to understand.

3. Can the surface area of a sphere be calculated without using dA?

No, the use of dA is necessary in order to accurately calculate the surface area of a sphere. Without it, the formula would not take into account the curvature of the sphere's surface.

4. How do you determine the value of dA in the surface area formula for a sphere?

The value of dA can be determined by dividing the surface area of the sphere into smaller sections, such as triangles or rectangles, and then summing up the areas of these sections. As the size of the sections approaches zero, the value of dA becomes more accurate.

5. Can the surface area of a sphere be calculated using a different formula?

Yes, there are other formulas that can be used to calculate the surface area of a sphere, such as A = 4πr2, where A represents the total surface area. However, using dA allows for a more precise calculation and is commonly used in mathematical and scientific contexts.

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