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

Question

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?

Answer 1

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