# Areas, surfaces, volumes

1. Aug 7, 2007

### 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 $$\frac{1}{3}\pi a^2 h$$

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

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

2. Aug 7, 2007

### 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.

3. Aug 8, 2007

### 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 :)

4. Aug 8, 2007

### mathusers

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?

5. Aug 8, 2007

### 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 similiar simple way of solving the others instead of washer's method please?

6. Aug 8, 2007

### HallsofIvy

Staff Emeritus
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 $\int\int\int dxdydz$?

7. Aug 8, 2007

### mathusers

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:)

8. Aug 8, 2007

### 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:

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

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

9. Aug 8, 2007

### 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...