Question

derieve the formula of surface area of sphere

Answer 1

The total surface area of the sphere is four times the area of great circle. To know more about great circle, see properties of a sphere. Given the radius r of the sphere, the total surface area is
 

A=4πr2A=4πr2

 

From the figure, the area of the strip is
dA=2πxdsdA=2πxds
 

Where ds is the length of differential arc which is given by
ds=1+(dydx)2−−−−−−−−−√dx=1+(dxdy)2−−−−−−−−−√dyds=1+(dydx)2dx=1+(dxdy)2dy


 


 

The total area of the sphere is equal to twice the sum of the differential area dA from 0 to r.
A=2(∫r02πxds)A=2(∫0r2πxds)

A=4π∫r0x1+(dydx)2−−−−−−−−−√dxA=4π∫0rx1+(dydx)2dx
 

From the figure,
x2+y2=r2x2+y2=r2

y=r2−x2−−−−−−√y=r2−x2

dydx=−2x2r2−x2−−−−−−√dydx=−2x2r2−x2

dydx=−xr2−x2−−−−−−√dydx=−xr2−x2

(dydx)2=x2r2−x2(dydx)2=x2r2−x2
 

Thus,
A=4π∫r0x1+x2r2−x2−−−−−−−−−−√dxA=4π∫0rx1+x2r2−x2dx

A=4π∫r0x(r2−x2)+x2r2−x2−−−−−−−−−−−−√dxA=4π∫0rx(r2−x2)+x2r2−x2dx

A=4π∫r0xr2r2−x2−−−−−−−√dxA=4π∫0rxr2r2−x2dx
 

Let
x = r sin θ
dx = r cos θ dθ

When x = 0, θ = 0
When x = r, θ = π/2
 

Thus,
A=4π∫π/20rsinθr2r2−r2sin2θ−−−−−−−−−−−√(rcosθdθ)A=4π∫0π/2rsin⁡θr2r2−r2sin2⁡θ(rcos⁡θdθ)

A=4π∫π/20r2sinθcosθr2r2(1−sin2θ)−−−−−−−−−−−√dθA=4π∫0π/2r2sin⁡θcos⁡θr2r2(1−sin2⁡θ)dθ

A=4πr2∫π/20sinθcosθ1cos2θ−−−−−√dθA=4πr2∫0π/2sin⁡θcos⁡θ1cos2⁡θdθ

A=4πr2∫π/20sinθcosθ(1cosθ)dθA=4πr2∫0π/2sin⁡θcos⁡θ(1cos⁡θ)dθ

A=4πr2∫π/20sinθdθA=4πr2∫0π/2sin⁡θdθ

A=4πr2[−cosθ]π/20A=4πr2[−cos⁡θ]0π/2

A=4πr2[−cos12π+cos0]A=4πr2[−cos⁡12π+cos⁡0]

A=4πr2[−0+1]A=4πr2[−0+1]

A=4πr2A=4πr2           okay!
 

Answer 2

Answer:

Derivation of Formula for Total Surface Area of the Sphere by Integration

The total surface area of the sphere is four times the area of great circle. To know more about great circle, see properties of a sphere. Given the radius r of the sphere, the total surface area is

A=4πr2

From the figure, the area of the strip is

dA=2πxds

Where ds is the length of differential arc which is given by

ds=1+(dydx)2−−−−−−−−−√dx=1+(dxdy)2−−−−−−−−−√dy

Figure for the Derivation of Formula for Surface Area of the Sphere by Integration

See Length of Arc in Integral Calculus for more information about ds.

The total area of the sphere is equal to twice the sum of the differential area dA from 0 to r.

A=2(∫r02πxds)

A=4π∫r0x1+(dydx)2−−−−−−−−−√dx

From the figure,

x2+y2=r2

y=r2−x2−−−−−−√

dydx=−2x2r2−x2−−−−−−√

dydx=−xr2−x2−−−−−−√

(dydx)2=x2r2−x2

Thus,

A=4π∫r0x1+x2r2−x2−−−−−−−−−−√dx

A=4π∫r0x(r2−x2)+x2r2−x2−−−−−−−−−−−−√dx

A=4π∫r0xr2r2−x2−−−−−−−√dx

Let

x = r sin θ

dx = r cos θ dθ

When x = 0, θ = 0

When x = r, θ = π/2

Thus,

A=4π∫π/20rsinθr2r2−r2sin2θ−−−−−−−−−−−√(rcosθdθ)

A=4π∫π/20r2sinθcosθr2r2(1−sin2θ)−−−−−−−−−−−√dθ

A=4πr2∫π/20sinθcosθ1cos2θ−−−−−√dθ

A=4πr2∫π/20sinθcosθ(1cosθ)dθ

A=4πr2∫π/20sinθdθ

A=4πr2[−cosθ]π/20

A=4πr2[−cos12π+cos0]

A=4πr2[−0+1]

A=4πr2

okay!