Proving Areas, Surfaces, and Volumes Using Integral Methods

[mathusers]

hi how can the following be proved using integral methods:

a) prove surface area of sphere, radius a, is 4 \pi a^2
b) prove area of a disk, radius a, is \pi a^2
c) prove volume of ball, radius a, is \frac{4}{3} \pi a^3
d) prove volume of axisymmetric cone of height h and base with radius a, is \frac{1}{3}\pi a^2 h

......... ...
i think my working of (a) is correct:

working of (a):
use spherical co-ordinates:
|S| = \int\int_{D} ||\frac{dr}{d\theta} X \frac{dr}{d\phi}|| dA

||\frac{dr}{d\theta} X \frac{dr}{d\phi}|| = a^2sin \phi

so:
|S| = \int^{2\pi}_{0}\int^{\pi}_{0} a^2sin \phi d\phi d\theta = \int^{2\pi}_{0} \left[ -a^2cos\phi \right]^{\pi}_0 d\theta = \int^{2\pi}_{0} 2a^2 d\theta = 4\pi a^2

.........
how can i do the rest please. and what integration methods should I be using for each? thnx xxxx

 

[Dick]

You have done the hardest one (a surface area) first in a way that doesn't make it clear how to do the others. Let's start with the second one. The area of a disk is the integral of the (circumference of a circle of radius r)*dr over r ranging from 0 to a. Can you draw a geometric picture to show you why this is so? For the last two divide the ball and the cone into disks.

 

[CompuChip]

And when you've done all that, you may try to divide the area of the sphere into disks and do it similarly to the other three :)

 

[mathusers]

And when you've done all that, you may try to divide the area of the sphere into disks and do it similarly to the other three :)

are we talkinga bout using washer's method here?

just wanted to ask. would washer's method actually be considered a "proof" though? or is it simply just a formula?

 

[mathusers]

i wanted to acutally know. is there a way to solve them using integral methods such as divergence, or greens, or stoke's??

or simply using spherical/polar/cartesian co-ordinates??

for example, i worked out (A) using spherical co-ordinates
where r = a

so then i used the parameter r(\theta,\phi) = (asin\phi cos\theta)i + (asin\phi sin\theta)j + (2cos\phi)k

now cross product of \frac{dr}{d\theta}X \frac {dr}{d\phi} = -(a^2sin^2\phi cos\theta)i - (a^2sin^2\phi sin\theta)j - (a^2sin\phi cos\phi)k

the magnitude is therefore:

||\frac{dr}{d\theta} X \frac{dr}{d\phi}|| = a^2sin \phi

so:
|S| = \int^{2\pi}_{0}\int^{\pi}_{0} a^2sin \phi d\phi d\theta = \int^{2\pi}_{0} \left[ -a^2cos\phi \right]^{\pi}_0 d\theta = \int^{2\pi}_{0} 2a^2 d\theta = 4\pi a^2

so that is how i managed to work out the surface area of the sphere.
......

do you have a similar simple way of solving the others instead of washer's method please?

 

[HallsofIvy]

That is NOT what is normally meant by "integral methods" for finding area and volume! But you certainly can use Stokes' or Green's or the divergence area: A volume is normally a triple integral. Which of those theorems equates a triple integral to on the boundary? What do you get if you apply that theorem to \int\int\int dxdydz?

 

[mathusers]

A volume is normally a triple integral. Which of those theorems equates a triple integral to on the boundary? What do you get if you apply that theorem to \int\int\int dxdydz?

are you referring to the divergence theorem..

if so, could you show me an example from b,c, or d on how to solve it, so that i can solve the rest :)

thanks:)

 

[mathusers]

hi i managed to get the one for (b):
using Green's theorem : \frac{1}{2}\oint_{C}[xdy-ydx]

c(t)=(rcos(t), rsin(t))

c'(t)=(-rsin(t),rcos(t))

\frac{1}{2}\int_{0}^{2\pi}\left[(rcos(t))(rcos(t))-(rsin(t))(-rsin(t))\right]dt

so we can deduce:

\frac{1}{2}\int_{0}^{2\pi}r^{2}dt={\pi}r^{2}

.......

but i don't know how to start on with (C) and (D).. any help would be greatly appreciated.

 

[Dick]

Persistant, aren't you? :) Ok, you can do (C) by realizing that the exterior derivative of the form w=xdydz is dw=dxdydz which is the volume form. So integrating dw over the interior of the sphere (volume) is the same as integrating w over the boundary. Do it in spherical coordinates. Same basic idea for (D). I think I'd really prefer using washers...