
#1
Dec1612, 10:19 AM

P: 108

We know that we calculate the volume of sphere by taking infinitesimally small cylinders.
∫ ∏x^2dh Limits are from R→0 x is the radius of any randomly chosen circle dh is the height of the cylindrical volume. x^2 + h^2 = R^2 So we will get 4/3∏R^3 Now the question is why cannot we obtain the SURFACE AREA using, infinitesimally small cylinders. Where ∫ 2∏xdh Limits are from R→0 x is the radius of any randomly chosen circle dh is the height of the cylindrical volume. x^2 + h^2 = R^2. I have a certain explanation for this which works well, but i would like to know if there is an unambiguous answer. Thankyou :) 



#2
Dec1712, 01:44 PM

Sci Advisor
HW Helper
Thanks
P: 26,167

hi sahil_time!




#3
Dec1812, 12:28 AM

P: 16

For the same reason this comic makes no sense: 



#4
Dec1812, 03:57 AM

HW Helper
P: 2,155

Area of sphere.
For the integral to work the approximation must match well enough, like the above comic. Two shapes can have equal volume and very nearly the sam shape, but very different surface area.




#5
Dec1812, 07:45 AM

P: 108

Thankyou for all the replies. :)
I would just like you to look at the attatchment, where ive tried to convince myself. If we compute the surface area by using CYLINDERS we end up getting a LESSER area than 4∏R^2 .The reason why cylinders do not work, is because "for an infinitesimally small height dh" the area of the ACTUAL surface of the sphere (which represents a conical frustum, i have taken CONE in this case) will always be greater than the surface area of the CYLINDER enclosing it. 



#6
Dec1812, 08:41 AM

Sci Advisor
HW Helper
Thanks
P: 26,167

hi sahil_time!
(but your diagram doesn't really work, it needs to show a proper frustrum, rather than one that goes up to the apex of the cone ) 



#7
Dec2112, 03:58 AM

P: 108

Thanx again :)




#8
Dec2112, 05:40 AM

Sci Advisor
HW Helper
Thanks
P: 26,167

yes, that's fine!
but you could shorten it by using θ from the start … your first line could be A_{frustrum} = π(r_{1} + r_{2})secθ(btw, archimdedes managed to prove this without modern maths … you may be interested to read this: http://arcsecond.wordpress.com/tag/archimedes/) 



#9
Dec2112, 01:02 PM

P: 108

That is ingenious, the way he has proved it :)
Thanx alot :) 


Register to reply 
Related Discussions  
Arc area of a sphere??? (a piece of r^2*sinθ*ΔrΔθΔφ)  Precalculus Mathematics Homework  1  
Surface Area of Sphere  Calculus & Beyond Homework  4  
area of a sphere  General Math  9  
Area of a sphere  General Math  16  
Area of a Sphere.  Calculus  1 