Proving the volume of sphere using calculus

[loba333]

Homework Statement

We know the surface area of a sphere is 4pir^2
but how can we proove this using integration.
I roughly know what to do, but i would like some guidance

Homework Equations

The Attempt at a Solution

I have made an attempt to the problem here
http://s1200.photobucket.com/albums/bb327/loba333/?action=view&current=IMG_1447.jpg

If you guys could show me how to do it step by step explaining what u''v done, I am sure it would be a great tutorial for other trying to learn this as well

Regards and thanks in advance

Dave

 

[tiny-tim]

Welcome to PF!

Hi Dave! Welcome to PF!

(have a pi: π and an integral: ∫ and try using the X² icon just above the Reply box )

You have the right idea …

you've divided the surface into strips of width dx, and radius y,

but you haven't accounted for the slope of the surface. ​

 

[LCKurtz]

Additionally, have you had the formula for arc length ds? Rotate a chunk of arc length.

Also, I think you meant "surface area" in the thread title.

 

[loba333]

Hi Dave! Welcome to PF!

(have a pi: π and an integral: ∫ and try using the X² icon just above the Reply box )

You have the right idea …

you've divided the surface into strips of width dx, and radius y,

but you haven't accounted for the slope of the surface. ​

cheers for the symbols ! ill hot key them to my pc.

does the 2π not account for the slope of the surface... as does πr when find the volume of a sphere via methods of integrating

 

[tiny-tim]

nooo …

the great thing about volumes is that the slope at the end only affects the volume at the end, which is tiny compared with the whole 4πr² …

but with areas, the whole 2πr is at the end (where the slope is) …

try it with the easy example of a cone instead of a sphere, and you'll see what I mean!

 

[HallsofIvy]

You don't say whether you are required to Cartesian coordinates but if not, here is how I would find the surface area.

Spherical coordinates are given by x= \rho cos(\theta) sin(\phi), y= \rho sin(\theta)sin(\phi), and z= \rho cos(\phi) with \theta going from 0 to 2\pi and \phi form 0 to \pi.

On a sphere of radius R, we have \rho= R and so can write the surface in terms of the two parameters, \theta and \phi as a vector equation:
\vec{p}(\theta, \phi)= Rcos(\theta)sin(\phi)\vec{i}+ Rsin(\theta)sin(\phi)\vec{j}+ Rcos(\phi)\vec{k}

The derivatives of that with respect to \theta and \phi are tangent vectors to that surface along the respective coordinate axes:
\vec{p}_\theta= -Rsin(\theta)sin(\phi)\vec{i}+ Rcos(\theta)sin(\phi)\vec{j}
and
\vec{p}_\phi= Rcos(\theta)cos(\phi)\vec{i}+ Rsin(\theta)cos(\phi)\vec{j}- Rsin(\phi)\vec{i}

The cross product of those tangent vectors (called the "fundamental vector product" for the surface) is perpendicular to the surface and incorporates all metric information- it gives the "vector differential of surface area" for the surface.

\left|\begin{array}{ccc}\vec{i} & \vec{j} & \vec{k} \\-Rsin(\theta)sin(\phi) & Rcos(\theta)sin(\phi) & 0 \\ Rcos(\theta)cos(\phi) & Rsin(\theta)cos(\phi) & -Rsin(\phi)\end{array}\right|
= -R^2cps)\theta sin^2\phi\vec{i}+ R^2 sin(\theta)sin^2(\phi)\vec{j}- R^2 sin(\phi)cos(\phi)\vec{k}

The length of that vector,
\sqrt{R^4(sin^4(\phi)+ sin^2(\phi)cos^2(\phi))}= \sqrt{R^4sin^2(\phi)}= R^2 sin^2(\phi)[/itex] <br /> gives the &quot;differential of surface area&quot;, R^2 sin^2(\phi)d\theta d\phi.<br /> <br /> The surface area of the sphere is <br /> \int_{\theta=0}^{2\pi}\int_{\phi= 0}^\pi R^2 sin^2(\phi)d\phi d\theta<br /> <br /> To integrate that, use the standard identity sin^2(\phi)= (1/2)(1- cos(\phi)):<br /> R^2/2 \int_{\theta= 0}^{2\pi} [1- cos(2\phi)]d\phi d\theta[/itex]

 

[loba333]

Happy new years guys ! thank you all for your posts however.

Im in first year, i think we are going to cover polar co-ordinates next semester. the method you have posted is quite interesting as i thought this how finding such a function would work, sadly some of the maths is a little out if my league ... for now.
So i was wondering if you could post how to do it using Cartesian co-ordinates ?

Kind Regards

David

 

[tiny-tim]

Hi David! Happy new year to you too!

Hint: the slope is dy/dx (which in this case is |x/y|) …

the area will be the width of the slice divided by cos of the slope.