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I'm new to calculus.
I saw a proof for volume of cone using integral. They taken the cone's vertex at $(0,0,0)$, it's base centre at $(h,0,0)$ and it's radius is $r$
$$V=\int_0^h \pi \left(\frac{rx}{h}\right)^2 \, dx =\frac{\pi r^2h}{3}$$
$R(x)=\frac{rx}{h}$. $R(x)$ gives the radius of the circle on cone at $x.$

Using the same idea I tried to workout the C.S.A of the cone,
Using integral I sum up all the circumference of those circle which must give the C.S.A of cone but, $$S=\int_0^h 2\pi \left(\frac{rx}{h}\right) \, dx =\pi rh$$ where I am doing mistake?

Michael Hardy
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Sunil kumar
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  • It's a subtle distinction but basically adding up parallel slices will get you a volume since the "error" goes away, however the same cannot be said for surface areas or anything to do with arc length. You have to rotate the arc length of the curve to account for the up and down motion. Right now you have only accounted for the horizontal motion ($dx$). – Ninad Munshi Feb 17 '20 at 11:27
  • @NinadMunshi How to do that? – Sunil kumar Feb 17 '20 at 11:36
  • Arc length formula is $\sqrt{1+(f')^2}$. Multiply that by $2\pi f$ then integrate. – Ninad Munshi Feb 17 '20 at 11:37
  • @NinadMunshi Thank you. I get it. – Sunil kumar Feb 17 '20 at 11:41
  • @NinadMunshi But how did you derive that $\sqrt{1+(f')^2}$ – Sunil kumar Feb 17 '20 at 11:56
  • @NinadMunshi or tell me any source to lern about that. – Sunil kumar Feb 17 '20 at 11:58
  • @Sunil : If $y=f(x)$ then $$ \sqrt{1+(f'(x))^2} ,,dx = \sqrt{(dx)^2+(dy)^2}. $$ If you think of $dx$ as an infinitely small increment of $x$ and $dy$ as the corresponding infinitely small increment of $y,$ then the Pythagorean theorem shows that $ \sqrt{1+(f'(x))^2} ,,dx$ is the corresponding infinitely small increment of arc length. $\qquad$ – Michael Hardy Feb 17 '20 at 19:40
  • See related answer with more details: https://math.stackexchange.com/a/1692595/72031 – Paramanand Singh Feb 18 '20 at 02:56

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When you integrate for volume the $dx$ part is the infinitesimally small vertical height. However, when you want the curved surface area, you will want a small component $dl$ to be along the length of the slope instead.

This has an easy fix. $x$ and $l$ are related as the the following due to the similar triangles: $$\frac xh=\frac lL$$ Hence, we can substitute as the following-

$$S=\int_{0}^{L} 2\pi \biggr(\frac{rl}{L}\biggr) dl$$ $$S=2\pi\frac rL\frac {L^2}2$$ $$S=\pi RL$$

Sam
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