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Homework Help: Surface Area of a Sphere by Integration

  1. Sep 9, 2010 #1

    IBY

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

    Say the sphere of radius "a" is made out of various rings with height R(x) and thickness dx. Adding up all of the rings will form a sphere, and in order to do that, I have to integrate.

    2. Relevant equations

    Trigonometric Substitution:
    [tex]\frac{x}{a}=sin \theta[/tex]
    [tex]dx=a cos \theta d\theta[/tex]

    Function of ring height related to position in cartesian plane:
    [tex]R(x)=\sqrt(a^2-x^2)[/tex]

    3. The attempt at a solution
    Set up the integral, I just want to make my life simpler and integrate half a circle:
    [tex]A=\int_0^a 2\pi R(x) dx[/tex]
    Substitute R(x):
    [tex]\int_0^a 2\pi \sqrt(a^2-x^2) dx[/tex]
    Now using trigonometric substitution and factoring out a:
    [tex]\int_0^a 2\pi a \sqrt(1-sin^2 \theta) dx[/tex]
    Using pythagorean trig identity and trig substituting for dx:
    [tex]\int_0^a 2\pi a cos\theta (a cos\theta d\theta)[/tex]
    [tex]\int_0^a 2\pi a^2 cos^2\theta d\theta[/tex]
    Putting out all the constants, and integrating, using tables of integral, I get:
    [tex]2\pi a^2\int_0^a cos^2\theta d\theta[/tex]
    [tex]2\pi a^2\int_0^a cos^2\theta d\theta[/tex]
    [tex]2\pi a^2 (\frac{\theta}{2}+\frac{sin(2\theta)}{4})|_0^a[/tex]

    The problem is, how do I solve the final part of the definite integral? What I know is that the parenthesis should be equal to 1 because the surface area is 4 pi r^2 and I integrated half a sphere.
     
  2. jcsd
  3. Sep 9, 2010 #2
    Don't forget to change your limits of integration when you make your substitution.
     
  4. Sep 9, 2010 #3

    Dick

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    You might also want to take note that integrating ring circumference DOES NOT give you area. You might want to find a real formula for surface area before you put too much work into this.
     
  5. Sep 9, 2010 #4

    IBY

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    @dick
    But isn't (circumference*dx) the tiny little area I need?
     
  6. Sep 9, 2010 #5

    gabbagabbahey

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    Oh, I think it should work just fine. For a circular ribbon of infinitesimal thickness [itex]dx[/itex] and cirumference [itex]2\pi R(x)[/itex], [itex]dA=2\pi R(x)dx[/itex] gives the area of one face of the ribbon.
     
  7. Sep 9, 2010 #6

    Dick

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    Wanna think about that again? It gives you the area of the ribbon if it's vertical (i.e. parallel to the direction dx). If not it ignores the horizontal component. That's not a good enough approximation. Look up the formula for area of a surface of revolution.
     
  8. Sep 9, 2010 #7

    gabbagabbahey

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    Right, whoops... gotcha.
     
  9. Sep 9, 2010 #8

    IBY

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    So first I have to solve the following for rings?
    [tex]dA=2\pi y\sqrt(1+(\frac{dy}{dx})^2)dx[/tex]
     
  10. Sep 9, 2010 #9

    Dick

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    Yes. You have to do some algebra to reduce it to a simple form. But once you get that form the integral is super easy.
     
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