Proving Areas, Surfaces, and Volumes Using Integral Methods

In summary, to prove the surface area of a sphere with radius a is 4πa², you can use spherical coordinates and the integral method to divide the sphere into disks and solve for the surface area using the cross product of the derivatives. Similarly, you can use Green's theorem to prove that the area of a disk with radius a is πa². To prove the volume of a ball with radius a is (4/3)πa³, you can use the divergence theorem to equate a triple integral to the boundary and solve using spherical coordinates. Finally, to prove the volume of an axisymmetric cone with height h and base with radius a is (1/3)πa²h, you can also use
  • #1
mathusers
47
0
hi how can the following be proved using integral methods:

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

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

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

[tex]||\frac{dr}{d\theta} X \frac{dr}{d\phi}|| = a^2sin \phi[/tex]

so:
[tex]|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 [/tex]

.........
how can i do the rest please. and what integration methods should I be using for each? thnx xxxx
 
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  • #2
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
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
CompuChip said:
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?
 
  • #5
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 [tex]r(\theta,\phi) = (asin\phi cos\theta)i + (asin\phi sin\theta)j + (2cos\phi)k[/tex]

now cross product of [tex]\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[/tex]

the magnitude is therefore:

[tex]||\frac{dr}{d\theta} X \frac{dr}{d\phi}|| = a^2sin \phi[/tex]

so:
[tex]|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 [/tex]

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?
 
  • #6
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 [itex]\int\int\int dxdydz[/itex]?
 
  • #7
HallsofIvy said:
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 [itex]\int\int\int dxdydz[/itex]?


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

[itex]c(t)=(rcos(t), rsin(t))[/itex]

[itex]c'(t)=(-rsin(t),rcos(t))[/itex]

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

so we can deduce:

[itex]\frac{1}{2}\int_{0}^{2\pi}r^{2}dt={\pi}r^{2}[/itex]

.......

but i don't know how to start on with (C) and (D).. any help would be greatly appreciated.
 
  • #9
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...
 

1. What is the difference between area, surface, and volume?

Area refers to the measurement of a two-dimensional space, such as the size of a flat surface. Surface is the outer layer of an object or the boundary of a three-dimensional shape. Volume is the amount of space that a three-dimensional object occupies.

2. How do you calculate the area of a shape?

The formula for calculating the area of a shape depends on the shape itself. For example, the area of a square is found by multiplying the length of one side by itself (length x length). The area of a circle is calculated by multiplying pi (3.14) by the square of the radius (pi x radius^2). Other shapes have their own specific formulas.

3. What is the importance of understanding surface area?

Understanding surface area is important in various fields of science, including physics, chemistry, and biology. It helps in determining the amount of material needed to cover or coat an object, the rate of heat transfer, and the amount of surface area available for chemical reactions to occur.

4. How does volume affect the buoyancy of an object?

The volume of an object plays a crucial role in determining its buoyancy, or the ability to float in a liquid. According to Archimedes' principle, the buoyant force exerted on an object is equal to the weight of the fluid it displaces. Therefore, the larger the volume of an object, the greater its buoyant force and the more likely it is to float.

5. Can you explain the relationship between surface area and volume?

The relationship between surface area and volume is known as the surface-to-volume ratio. As the volume of an object increases, its surface area also increases, but not at the same rate. This means that smaller objects have a higher surface-to-volume ratio than larger objects. This concept is important in understanding how different organisms adapt to their environment, as well as in engineering and design.

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