## Area of sphere.

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 :)
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hi sahil_time!
 Quote by sahil_time … why cannot we obtain the SURFACE AREA using, infinitesimally small cylinders.
they're not cylinders, they're frustrums of a cone!

 Quote by sahil_time Now the question is why cannot we obtain the SURFACE AREA using, infinitesimally small cylinders.

For the same reason this comic makes no sense:

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## 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.
 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. Attached Thumbnails

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hi sahil_time!
 Quote by sahil_time 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.
yes, that's correct, the frustrum area will always be more by a factor secθ, where θ is the half-angle of the cone …

(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 )
 Thanx again :) Attached Thumbnails
 Blog Entries: 27 Recognitions: Gold Member Homework Help Science Advisor yes, that's fine! but you could shorten it by using θ from the start … your first line could be Afrustrum = π(r1 + r2)secθ and then show that the bracket = 2r1 as r1 - r2 -> 0 (btw, archimdedes managed to prove this without modern maths … you may be interested to read this: http://arcsecond.wordpress.com/tag/archimedes/)
 That is ingenious, the way he has proved it :) Thanx alot :)

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