0
$\begingroup$

To derive a formula for the volume of the cone I used integration with respect to slant. I wanted to sum the areas of circles at all heights of the slant (as in standard way with respect to height). Let $x$ be the specific length of the slant measured from the top of the cone. Then the area of the circle at height $x$ is equal to $\frac{\pi r^2 x^2}{l^2}$ where $r$ is the base circle radius and $l$ is the slant height. So the volume of the cone is equal to: $$V_{cone} = \int_0^l{\frac{\pi r^2 x^2}{l^2} dx} = \frac{\pi r^2 l}{3}$$ which is obviously wrong. What is the reason of it?

$\endgroup$
1
  • $\begingroup$ Slant height is a synthetic length, $L^2=r^2+h^2$. The central axis of the cone is along the x axis, x=L is arbitrary, not the bound you want, likely outside of the cone. First integrate from 0 to H then substitute slant height. $\endgroup$
    – jamie
    Apr 23, 2020 at 19:43

1 Answer 1

0
$\begingroup$

If by "slant height" you mean the length along the slant, the circular cross-section of thickness $dx$ has area $\pi r^2L(x)^2/l^2$, where $l$ is the distance along the slant to that cross-section. Thus $L(x)\ne x$; in fact $L(x)=lx/h$, with $h$ the cone height. So the integral becomes $\int_0^h\pi r^2\frac{x^2}{h^2}dx=\frac13\pi r^2h$.

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .