# a) Show that the formula for the surface area of a sphere with radius #r# is #4pir^2#. b) If a portion of the sphere is removed to form a spherical cap of height #h# then then show the curved surface area is #2pihr^2#?

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a) Show that the formula for the surface area of a sphere with radius #r# is #4pir^2# .

b) If a portion of the sphere is removed to form a spherical cap of height #h# then then show the curved surface area is #2pihr^2#

a) Show that the formula for the surface area of a sphere with radius

b) If a portion of the sphere is removed to form a spherical cap of height

##### 2 Answers

A = int dA

#### Explanation:

An area element on a sphere has constant radius r, and two angles. One is longitude

So the area element is

Integrated over the whole sphere gives

In part b,

Then

Note: Every derivation I found of this result uses cylindrical coordinates and is far more involved than this one. Can someone else check?

a)

b)

#### Explanation:

It is easier to use Spherical Coordinates, rather than Cylindrical or rectangular coordinates. This solution looks long because I have broken down every step, but it can be computer in just a few lines of calculation

With spherical coordinates, we can define a sphere of radius

The Jacobian for Spherical Coordinates is given by

And so we can calculate the surface area of a sphere of radius

# A = int int_R \ \ dS \ \ \ #

where

# :. A = int_0^pi \ int_0^(2pi) \ r^2 sin phi \ d theta \ d phi#

If we look at the inner integral we have:

# int_0^(2pi) \ r^2 sin phi \ d theta = r^2sin phi \ int_0^(2pi) \ d theta #

# " " = r^2sin phi [ \ theta \ ]_0^(2pi)#

# " " = (r^2sin phi) (2pi-0)#

# " " = 2pir^2 sin phi#

So our integral becomes:

# A = int_0^pi \ 2pir^2 sin phi \ d phi#

# \ \ \ = -2pir^2 { cos phi ]_0^pi#

# \ \ \ = -2pir^2 (cospi-cos0)#

# \ \ \ = -2pir^2 (-1-1)#

# \ \ \ = -2pir^2 (-2)#

# \ \ \ = 4pir^2 \ \ \ # QED

For the area of the portion of a sphere we have a similar set-up:

By trigonometry

# A = int_arccos(h/r)^(pi/2) \ int_0^(2pi) \ r^2 sin phi \ d theta \ d phi#

# \ \ \ = int_arccos(h/r)^(pi/2) \ (r^2sin phi)(2pi-0) \ d phi#

# \ \ \ = int_arccos(h/r)^(pi/2) \ 2pir^2sin phi \ d phi#

# \ \ \ = -2pir^2[cosphi]_arccos(h/r)^(pi/2) #

# \ \ \ = -2pir^2(cos(pi/2)-cos(arccos(h/r))) #

# \ \ \ = -2pir^2(0-h/r) #

# \ \ \ = -2pir^2(-h/r) #

# \ \ \ = 2pihr^2 \ \ \ # QED