Why is this way of deriving a cone volume formula by integration wrong?

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?

• 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. Apr 23, 2020 at 19:43

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 $$\int_0^h\pi r^2\frac{x^2}{h^2}dx=\frac13\pi r^2h$$.