# 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 #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

A = int dA

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

By signing up, you agree to our Terms of Service and Privacy Policy

a)

b)

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 \ \ \ # QEDFor 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

By signing up, you agree to our Terms of Service and Privacy Policy

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

- What is a solution to the differential equation #ydy/dx=e^x# with y(0)=4?
- How do you find the volume of the solid if the region in the first quadrant bounded by the curves #x=y-y^2# and the y axis is revolved about the y axis?
- What is a general solution to the differential equation #y=y+2xe^(2x)#?
- If a region is bounded by #y = sqrt(x) + 3#, #y=5#, and the y=axis and it is revolved around the y = 7, how do you find the volume?
- A Sphere of radius #2a# has a hole of radius #a# drilled through the centre. What is the remaining volume?

- 98% accuracy study help
- Covers math, physics, chemistry, biology, and more
- Step-by-step, in-depth guides
- Readily available 24/7