Why Do Sphere Surface Areas Differ With Same Approximation?

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The discussion centers on the discrepancies in calculating the surface area of a sphere using two methods that employ the same approximation. One participant notes that their calculations yield different results, questioning the validity of assuming a frustum is equivalent to a cylinder. The confusion arises from a misapplication of formulas, particularly in the transformation process and the introduction of an extra sine term in the area differential. Clarification is sought on the proper application of these mathematical concepts. The conversation highlights the importance of precise assumptions and transformations in geometric calculations.
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i have found the area of sphere in two ways using the same approximation.but i get two different answers ;one the correct value 4∏R^2?how does this happen?
i'm attaching the solution below?please refer to the attachments and give a solution?
 

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Your mistake is where "transform the frustrum to a cylinder". You don't actually do a "tranformation" you just assume frustrum is equal to the cylinder and that is not true.
 


You write dA = 2Pirh and h = rsin(theta)dtheta, but then you write dA = 2Pir^2 sin^2(theta) dtheta. Where does the extra sin(theta) come from?
 

Thread 'Proving that convexity implies second order derivative being positive'

There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...
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