wer,HowGan

I 'm having problem of proving the total surface area of sphere = 4pi(a^2) ,
in my workin g, i got surface area = 2pi(a^2) only ....
I 've checked it many times , still cant figure out which part i'm wrong

P/s : the sphere has radius of a .

http://imgur.com/a/N61eT

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3 months ago 5

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  1. Cguy

    .
    You have done everything right except that you need the sum of the top and bottom surface areas. You only have the the half portion of the surface area.
    x² + y² + z² = a²

    z = ±√(a² - x² - y²)
    ⎯⎯⎯➤ upper : z = √(a² - x² - y²) bottom : z = -√(a² - x² - y²)

    z² = a² - x² - y²
    2z δz/δx = -2x ⎯⎯➤ δz/δx = -x/z ⎯⎯➤ δz/δx = -x/√(a² - x² - y²) ⎯⎯➤ (δz/δx)² = x²/(a² - x² - y²)
    2z δz/δy = -2y ⎯⎯➤ δz/δy = -y/z ⎯⎯➤ δz/δy = -y/√(a² - x² - y²) ⎯⎯➤ (δz/δy)² = y²/(a² - x² - y²)
    x²/(a² - x² - y²) + y²/(a² - x² - y²) + 1
    = (x² + y² + a² - x² - y²)/(a² - x² - y²)
    = a²/(a² - x² - y²)
    = a²/[ a² - (x² + y²) ]
    = a² / (a² - r²)

    ∴ √ [a² / (a² - r²) ] = a / √(a² - r²)

    Top surface area = ∫ ∫a / √(a² - r²) r dr dθ from r = 0 to r = a and from θ = 0 to θ = 2π

    double top surface area to get total surface area:

    S = 2 ∫ ∫a / √(a² - r²) r dr dθ from r = 0 to r = a and from θ = 0 to θ = 2π
    S = 2a ∫ ∫r / √(a² - r²) dr dθ from r = 0 to r = a and from θ = 0 to θ = 2π ∫r / √(a² - r²) dr from r = 0 to r = a
    = -√(a² - r²)
    = -√(a² - a²) + √(a² - 0²)
    = a

    S = 2a ∫ ∫r / √(a² - r²) dr dθ from r = 0 to r = a and from θ = 0 to θ = 2π
    S = 2a ∫ a dθ from θ = 0 to θ = 2π
    S = 2a² ∫ dθ from θ = 0 to θ = 2π
    S = 2a²θ
    S = (2a²*2π) - (2a²*0)
    S = 4πa² - 0
    S = 4πa²
    ————

  2. Steve A

    V = (4/3)pi*r^3
    V' = 4pi*r^2
    V' is surface area.

  3. Vaman

    ds = a^2 sin theta d theta dphi.
    phi goes from 0 to 2pi. Theta goes from 0 to pi. Integrate it. You will get the correct answer. Missing 2 comes from the sine integration. a is a constant. Do not integrate it.

  4. ted s

    integral of x in [ - a , a ] of { 2π y ds }......ds = √ ( 1 + [dy/dx]² ) dx ≡ a / √ ( a² - x²) ≡ a / y !!

    thus int over x in [ - a , a ] of { 2π a dx} ≡ 4π a²......comment : the "mathalino " site has an error in its work..

    their integration , using their diagram, should be along the y axis

  5. Douglas

    I have provided a link to the derivation.

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