which class the question comes from?

Log in to add a comment

which class the question comes from?

Log in to add a comment

Certified answers contain reliable, trustworthy information vouched for by a hand-picked team of experts. Brainly has millions of high quality answers, all of them carefully moderated by our most trusted community members, but certified answers are the finest of the finest.

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.