Surface area of a cut cone

To make the algebra neater, translate everything by $-20$ in the $z$ -direction, so that the cone's apex is at $O$ and the plane passes through $0,0,-10$ .

The equation of the plane is $2x+3y+5z=-50$ . In cylindrical coordinates, this is $2r\cos\theta+3r\sin\theta+5z=50$

The equation of the cone is $r^2=\frac{z^2}{4}$ ; bearing in mind we're interested in negative $z$ , this becomes $2r+z=0$

Solving these, we find $r=\frac{50}{10-3\sin\theta-2\cos\theta}\;\;\;\;z=-\frac{100}{10-3\sin\theta-2\cos\theta}$

The distance of this point from $O$ is given by $R=\frac{50\sqrt5}{10-3\sin\theta-2\cos\theta}$

In order to find the area, we need to evaluate $A=k\int_0^{2\pi} \frac12 R^2 \; d\theta$

Note this is similar to the usual area integral in polar coordinates, but with a constant, $k$ . We're approximating the area by isosceles triangles with equal sides $R(\theta$ - the same as in polars - but we need to convert from $\theta$ (which measures around the base of the cone) to the angle $\phi=k\theta$ at the apex of the cone.

To find this $k$ , just unroll the cone; its net is a sector of a circle, with radius equal to the slant height of the cone (which is $10\sqrt5$ ). The circumference of the base of the cone is $20\pi$ ; so this is the arc length of its net. Hence the angle of the sector is $\frac{20\pi}{10\sqrt5}=\frac{2\pi}{\sqrt5}$ and we need $k=\frac{1}{\sqrt5}$ .

Putting everything together, $A=\frac{1}{2\sqrt5}\int_0^{2\pi} \frac{12500}{(10-3\sin\theta-2\cos\theta)^2} \; d\theta=\frac{25000\sqrt5}{87\sqrt{87}}\pi \approx \boxed{216.4193}$

Using the formulas given, we have,

$\theta = \tan^{-1}(\frac{1}{2})$ , $\phi = \cos^{-1} \dfrac{5}{\sqrt{38}}$ , and $z_0 = 10$

Therefore, the lengths of the semi-axes are,

$\text{Semi-minor axis length} = a = 5.360562674$ , and $\text{Semi-major axis length }= b = 7.085533337$

Hence, the area of the elliptical section is $A = \pi a b = 119.3253718$

Now comes the trick of this problem. We will project the ellipse area onto the $xy$ plane. The area of the projected elliptical section is simply

$A_{\text{Projected}}= A \cos \phi = 96.78565699$

And finally, we note that projecting the area of the conical surface (which is above the cutting plane) onto the $xy$ plane results in the same elliptical region in the $xy$ plane as the elliptical region resulting from projecting the elliptical conic section (the one that is on the cutting plane as well). Thus if $S$ is the surface area then

$S \sin \theta = A_{\text{Projected}}$

Therefore, $S = \dfrac{A \cos \phi}{\sin \theta} = 216.41930... \approx \boxed{216.4}$