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Homework Help: Calculating The Area of a Spherical Cap

  1. Oct 10, 2014 #1

    1. The problem statement, all variables and given/known data

    Hi all, I am working through Gravity by James Hartle and have become stuck on a question asking me to calculate the area of a circle of radius r in the 2D geometry that is the surface of a sphere of radius a.

    A surface element on this sphere can be found to be (adΘ)(asinΘdΦ) (see attached image).

    2. Relevant equations

    3. The attempt at a solution
    Integrating the above surface element over Θ=0 to Θ=r/a, and Φ=0 to Φ=2π, I obtain A=-2πa2cos(r/a). The answer given in the solutions manual is this, plus 2πa2. I don't understand how the additional 2πa2 comes about.

    Attached Files:

  2. jcsd
  3. Oct 10, 2014 #2
    What isthis r you are using; thee is n r in the diagram.

    Do you see problem with getting negative area?
  4. Oct 10, 2014 #3
    r is the radius of the circle inscribed on the sphere, i.e., the distance from the pole of the sphere to the point on the sphere which the circle lies, along a path for which Θ varies but dΦ=0 (definition from the book). I understand of course that area is not negative, but taking the integral of the given surface area element should give the result I got unless I am mistaken (probably).

    Here I have attached also the solution. I have no idea how he goes from the surface area element (adΘ)(asinΘdΦ) to the integral shown for the area A. (There is an extra dΦ and dΘ???)

    Attached Files:

    Last edited: Oct 10, 2014
  5. Oct 10, 2014 #4
    Hi ChemEng92. Welcome to Physics Forums.

    What is a ChE doing studying Relativity? Just for fun?

    The limits on ##\theta## should be from 0 to ##\pi/2##, and the integral of ##sin\theta## is ##-cos\theta##

  6. Oct 14, 2014 #5
    Hi Chet,

    Your limits for Θ are correct, IF we are looking to find the area of a hemisphere. However, the calculation I am looking for is the area of a spherical cap, at any angle Θ<π/2 (see attached image in first post). Also, I did the integral you mentioned in your post, however the answer given in the book is my answer, plus 2πa2 . Furthermore, the procedure given in the solutions for obtaining this answer is, to me, nonsensical.

    If anyone understands why he does the procedure he does in the second attached pic, I would greatly appreciate an explanation.
  7. Oct 14, 2014 #6
    Their answer looks OK to me. If ##\theta## is the angle between the north pole and the given latitude, then A = 0 when ##\theta = 0##, A = ##2\pi a^2## for ##\theta = \pi /2##, and A = ##4\pi a^2## for ##\theta = \pi##. It looks like the parameter r is the arc length.

    Last edited: Oct 14, 2014
  8. Oct 14, 2014 #7
    The answer of course makes sense, it is the procedure that is confusing me. Specifically, the step where he goes from deriving the infinitesimal area element (adΘ)(asinΘdΦ) to the part where he says, "Integrating the element of area above...". It is the integrands that make no sense to me. Why are there two dΦ and dΘ terms? Isn't (adΘ)(asinΘdΦ)=a2sinΘdΘdΦ, then this is integrated?
  9. Oct 14, 2014 #8
    He just wanted to show in the integrand the product of the two sides of the differential "rectangle." He thought it might be more understandable to the reader. Does this make sense?
  10. Oct 15, 2014 #9
    I'm sorry, I'm still not understanding. Here is my issue (maybe in tex it will be clearer, I was learning to use it and this will be my first time):

    Differential rectangle from diagram:
    [tex]1. (ad \theta )(a \sin \theta d \phi )[/tex]

    Then he says, "The circle of radius r lies at [itex] \theta = r/a[/itex]." This part I understand.

    Next, "Integrating the element of area above,"

    [tex]2. A=\int_0^{r/a} d\theta \int_0^{2\pi} d\phi a^2 \sin \theta d\theta d\phi [/tex]

    In going from step one to step two, where does the extra [itex]d\theta[/itex] and [itex]d\phi[/itex] come from? Once this is explained, how am I to integrate a [itex](d\phi)^2[/itex] term?

    Furthermore, isn't the result of [itex]\int_0^{r/a} d\theta[/itex] simply [itex]r/a[/itex]? How does r/a end up inside the cosine term?

    I'm sorry if I'm missing something very obvious, and I appreciate the help I've received so far. This problem has been bothering me for the past week, math tutors at my college haven't been able to explain this to me.
    Last edited: Oct 15, 2014
  11. Oct 15, 2014 #10
    Ah. Now I see what you're saying. You're right, of course. There are some typos in that equation.

  12. Oct 15, 2014 #11


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    Staff: Mentor

    What is meant by that O( ) function in the answer?
  13. Oct 16, 2014 #12
    Good, now I know I am not going crazy looking at this solution haha. Unfortunately there still remains the question of how to obtain the answer he gets.

    My method was to integrate the differential rectangle:

    [tex]\int_0^{r/a}\int_0^{2\pi} (ad \theta )(a \sin \theta d \phi ) [/tex]

    to obtain: [tex] -2\pi a^2 \cos (r/a)[/tex].

    As you can see, the book's answer is [tex] 2 \pi a^2 - 2 \pi a^2 \cos (r/a) [/tex]. So where does that extra [itex] 2 \pi a^2 [/itex] come from?

    And NascentOxygen, I believe the O() represents the rest of the cosine power series from the third term on.
  14. Oct 16, 2014 #13
    The problem is with how you applied the limits of integration. I get:

    ##2\pi a^2[-cos \theta]_0^{r/a}##

    What does this give you?

  15. Oct 16, 2014 #14
    Of course. I knew it would be a obvious bonehead mistake of mine. Thank you! so much for this and all your help sir.
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