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#?

An area element on a sphere has constant radius r, and two angles. One is longitude #phi#, which varies from #0# to #2pi#. The other one is the angle with the vertical. To avoid counting twice, that angle only varies between #0# and #pi#.

So the area element is #dA = r d theta r sin theta d phi = r^2 sin theta d theta d phi# Integrated over the whole sphere gives #A = int_0^pi sin theta d theta int_0^(2pi) dphi r^2 = -cos(theta)|_0^(pi) 2 pi r^2 = 4 pi r^2#

In part b, #cos(theta)# varies between #a/r# and #b/r# which is such that #b-a=h# Then #A= cos theta|_b^a 2pi r^2 = (b/r - a/r) 2 pi r^2 = (h/r) 2pi r^2 = 2 pi r h#

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 #r# by all coordinate points where #0 le phi le pi# (Where #phi# is the angle measured down from the positive #z#-axis), and #0 le theta le 2pi# (just the same as it would be polar coordinates), and #rho=r#).

The Jacobian for Spherical Coordinates is given by #J=rho^2 sin phi #

And so we can calculate the surface area of a sphere of radius #r# using a double integral:

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

where #R={(x,y,z) in RR^3 | x^2+y^2+z^2 = r^2 } #

# :. 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 #cos phi = h/r => phi = arccos(h/r) #, and so we must restrict #phi# to #arccos(h/r) le phi le pi/2#, which gives us:

# 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