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August 31st, 2019, 03:31 PM   #1
Joined: Aug 2019
From: Czech Republic

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Could someone please explain how a specific part of the expression for the Riemann su

I'm really sorry if this post is not appropriate for this forum. My math level is quite low and it might have been because of that I could not understand this concept. So, the problem that I have is that I don't understand how a part of this expression is generated:

1. $$S =\sum_{i=1}^n 2 \pi f(x^*_i) \sqrt{1+(f'(t_i))^2} (x_i-x_{i-1}) \tag{1}$$

The confusing part of this for me is that since the surface area of a frustum (which is what I have used to find the area under a curve) is $\pi(R_1+R_2)\sqrt{(R_1-R_2)^2+h^2}$ (excluding the top and bottom circles), the total area under the curve should be:

2. $$\lim_{n \to \infty} \pi \sum_{i=1}^n (f(x_{i-1})+f(x_i))\sqrt{(f(x_i)-f(x_{i-1}))^2+(x_i-x_{i-1})^2} \tag{2}$$

When you use the mean value theorem to simply a section of the first riemann sum, then you get $\pi \sum_{i=1}^n (f(x_{i-1})+f(x_i))\sqrt{\sqrt{1+[f'(t_i)]^2}(x_i-x_{i-1})^2}$ but that does not have the $2\pi$ that the final expression for the surface area has.
So I suppose my question is: how do you get from (2) to (1)

This article goes over this entire problem very nicely so if this is not clear then please refer to the article
I once again apologise for my unclear explanations and the pictures instead of actual expressions, I'm very new to the platform.

Thank you a lot for your time.
Samir 1 is offline  

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