how to derive the formula for surface area of a sphere
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See diagram.
Let us take the top half of the sphere of Radius R, located with center at origin.
Let us us consider a small ring of radius r on the surface of the sphere, such that, its plane is parallel to the x-y plane. Its height above x-y plane be z. Then its radius will be r = √(R² - z²) as, r² + z² = R²
equation of the circle (of the ring) with radius r : x² + y² = r² = R² - z²
Its circumference is 2 π r = 2 π √(R² - z²)
Let the width of the ring along the surface of sphere be ds.
ds = approximated as = √ [(dz)² + (dr)² ] - by Pythagoras theorem.
ds = √ [1 + (dr/dz)² ] * dz
z² + r² = R² , Differentiating we get,
2 z + 2 r dr/dz = 0 => dr/dz = -z/r
ds = √(1 + z²/r²) * dz = √ [ (r² + z²) / r² ] * dz = R / r * dz
Area of the small infinitesimal ring = 2π r ds = 2π r R / r dz = 2 π R dz
Total surface area of Sphere = 2 * surface area of hemisphere z = 0 to R
Hence answer.
Let us take the top half of the sphere of Radius R, located with center at origin.
Let us us consider a small ring of radius r on the surface of the sphere, such that, its plane is parallel to the x-y plane. Its height above x-y plane be z. Then its radius will be r = √(R² - z²) as, r² + z² = R²
equation of the circle (of the ring) with radius r : x² + y² = r² = R² - z²
Its circumference is 2 π r = 2 π √(R² - z²)
Let the width of the ring along the surface of sphere be ds.
ds = approximated as = √ [(dz)² + (dr)² ] - by Pythagoras theorem.
ds = √ [1 + (dr/dz)² ] * dz
z² + r² = R² , Differentiating we get,
2 z + 2 r dr/dz = 0 => dr/dz = -z/r
ds = √(1 + z²/r²) * dz = √ [ (r² + z²) / r² ] * dz = R / r * dz
Area of the small infinitesimal ring = 2π r ds = 2π r R / r dz = 2 π R dz
Total surface area of Sphere = 2 * surface area of hemisphere z = 0 to R
Hence answer.
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