I was looking to formulate a general solution for a vertical slice of a cone. After failing to do this by integrating for the area of an ever shrinking chord segment, I eventually came up with a more geometric solution by subtracting the triangular part from the area of a sector which was easier to integrate to get the volume and appears to work fine. See my solution sketch. However, I would still like to formulate the integration area method of the chord segment.
From the information in the sketch, integrating along a $y$ axis, the area of the segment would be:
$$A_c = 2\int_0^{\sqrt(R^2-a^2)} \sqrt(R^2 – y^2) – a \ dy$$
How do I accomplish a second integration along a vertical $x$ axis to integrate these areas into a volume with $R$ decreasing in accordance with $$r = (R – Rx/h)\text{ from } 0 \text{ to } (h – ah/R)$$
I've tried a number of ways but now I'm going cross-eyed.
Best Answer
A cone of radius $R$ and height $H$
would be:
$z = H - \frac {H}{R} \sqrt {x^2 + y^2}$
And we are going to slice it at some line $y = a$
$v = \int_a^R\int_{-\sqrt{R^2 - x^2}}^{\sqrt {R^2- x^2}} (H - \frac {H}{R} \sqrt {x^2 + y^2}) \ dx \ dy$
Convert to polar:
$x = r\cos \theta\\ y = r\sin \theta\\ z = z\\ dy\ dx = r \ dr\ d\theta$
$2\int_{\arcsin\frac {a}{R}}^{\frac {\pi}{2}}\int_{a\csc \theta}^{R} Hr -\frac{H}{R} r^2 \ dr d\theta\\ 2\int_{\arcsin\frac {a}{R}}^{\frac {\pi}{2}} (\frac 16 HR^2 - \frac 12 Ha^2\csc^2\theta + \frac 13 \frac {H}{R}a^3\csc^3\theta \ d\theta\\ 2(\frac 16 HR^2 \theta + \frac 12 Ha^2\cot\theta - \frac 16 \frac {H}{R}a^3\csc\theta\cot\theta + \frac 16\frac {H}{R}a^3(\ln|\csc\theta + \cot\theta|)|_{\arcsin\frac {a}{R}}^{\frac \pi 2}\\ 2(\frac 16 HR^2 \arccos \frac {a}{R} - \frac 13 Ha \sqrt {R^2-a^2}+ \frac 16\frac {H}{R}a^3(\ln|\frac {R}{a} + \frac {\sqrt {R^2-a^2}}{a}|))\\$