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I know this question was asked already, but I would really like to resolve some problems I still have with it.

An area element in cylindrical coordinate is $\quad rd \theta dz \quad $ and $ \quad r = R(1- \frac{z}{h}) $

So why can't I write

$$ S = \int_{0}^{2 \pi} \int_{0}^{h} r(z) d\theta dz = \quad \int_{0}^{2 \pi} \int_{0}^{h} R(1- \frac{z}{h}) d\theta dz = \quad \pi Rh $$

If I divide the cone into small rings with radius $ r_i $, then the area of each ring is

$$ 2\pi r_i \Delta z$$

Taking the sum of all rings, $$ \sum_{i=0}^{N} 2\pi r_i \Delta z$$ And in the limit $N \to \infty $ : $$ 2 \pi \int_{0}^{h} r(z) dz$$

What's wrong about it? I understand that the problem is in the infinitesimal area that was used, that I didn't take into account the slant of the cone.

But, when we integrate $ y = x $ we also don't consider the slant, because for smaller and smaller $\Delta x's$ the area aproaches the correct area. Why is it not the case when dividing the cone into slantless rings?

Setting $y = x$ into $ \int ydx$ is taking into account the slope, the different height in each place, then why is it not enough to set $ r = R(1- \frac{z}{h})$ into the integral? what is the difference between those to examples?

I would really like some intuitive explnation to understand the differences.

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RobPratt
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EB97
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  • There is a problem with your first formula. Instead of $dz$, you should have $dl$, where $l$ is along the surface. See for example https://socratic.org/questions/how-to-get-surface-area-of-a-cone-using-integral-calculus or https://math.stackexchange.com/questions/18226/setting-up-an-integral-to-find-a-cones-surface-area – Andrei Mar 02 '22 at 00:47
  • @Andrei I was refering to it, please read the end of my question. – EB97 Mar 02 '22 at 00:50
  • The difference is that when you calculate area, by making smaller $\Delta x$ you approximate better and better the area under the curve. But try to calculate the length of the segment from $(0,0)$ to $(1,1)$. You can't just integrate $dx$ from $0$ to $1$. In your problem, the ratio between the area of the infinitesimally thin disk and the disk height is different than the ratio between the area of a frustum and the height. – Andrei Mar 02 '22 at 00:57
  • @Andrei Is taking thinner rings won't approximate better the area of each ring to the area of a frustum? Why is it like integrating $dx$ from $0$ to $1$? I'm not using the same $r$ in every ring, so how is it like taking the same height and sum it from $0$ to $1$? Why is the ratio you mentioned important? – EB97 Mar 02 '22 at 01:16
  • Let's assume that you know the area of a frustum: $$A=\pi(r_1+r_2)\sqrt{(r_1-r_2)^2+h^2}$$You can take $h$ outside of the square root:$$A=2\pi\frac{r_1+r_2}2h\sqrt{1+\frac{(r_1-r_2)^2}{h^2}}$$When you take the limit $h\to 0$, $r_1$ is close to $r_2$, but $\lim_{h\to 0}\frac{r_1-r_2}h$ is a constant for a cone (the slope of the cone). – Andrei Mar 02 '22 at 01:46
  • Related: Areas versus volumes of revolution – ryang Mar 02 '22 at 05:49

1 Answers1

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But, when we integrate y=x we also don't consider the slant, because for smaller and smaller Δx′s the area ap[p]roaches the correct area.

Yes, the area of the rectangles under the line approaches the correct area. But the length of the lines formed by the tops of the rectangles doesn't approach the true length of the diagonal line. It's a staircase shape, and will always have a length of $x+y$ (i.e., the combined length of the horizontal segments plus the combined length of the vertical segments) no matter how small you make the "stairs".

It's the exact fallacy committed in this fake proof that $\pi=4$.

Similarly, while the combined volume of the discs does approach the volume of the cone, the combined surface area does not approach the surface area of the cone.

Dan
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