Surface of Cone

A cone is cut by a plane ($z=d$). ($0<d<c$) The top point of the cone is $A(a,b,c)$

Button of the Cone is a circle on $z=0$ plane and center of it is $O(0,0,0)$.

If $a=b=0$ then

$$\frac{c-d}{c}=\frac{r_1}{r}$$

$$r_1=r\frac{(c-d)}{c}$$

$$S=\pi r_1(\sqrt{r^2_1+(c-d)^2}+r_1)$$

What is the formula for surface of the cut cone If $a^2+b^2\neq 0$ ?

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First, cutting by the plane $z=d$ does not change things much. The resulting cone is similar to the bigger cone with the base in $z=0$. The similarity factor is $(c-d)/c$, so the area is $(c-d)^2/c^2$ times the area of the bigger cone.

So I'm going to consider the cone with the base the circle of radius $r$ centered at O, and the vertex at $(a,b,c)$. The base has area $\pi r^2$, of course. The question is what to do with the lateral surface. Here is the calculus approach (I can't think of another one, actually):

The base circle is parametrized by $(r\cos t, r\sin t,0)$. Using the equation of the line segment from this point to $(a,b,c)$, we arrive at the parametrization of the entire surface: $$\vec r(t,s)=\langle sa+(1-s)r\cos t, sb+(1-s)r\sin t, sc \rangle,\qquad 0\le s\le 1, \quad 0\le t\le 2\pi$$ The area is the integral of $|\vec r_t\times \vec r_s|$. Let's do this: $$\vec r_t= (1-s) \langle -r\sin t, r\cos t, 0 \rangle$$ $$\vec r_s=\langle a-r\cos t, b-r\sin t, c \rangle$$ and the cross product simplifies to $$\vec r_t\times \vec r_s = r(1-s)\langle c\cos t, c\sin t, r-a\cos t-b\sin t \rangle$$ Integration with respect to $s$ yields a factor of $1/2$, hence the lateral area is $$A=\frac{r}{2}\int_0^{2\pi}\sqrt{c^2+(r-a\cos t-b\sin t)^2}\,dt$$ When $a=b=0$, the answer is $A=\pi r\sqrt{c^2+r^2}$. But when either of them is nonzero, we get a complete elliptic integral, which cannot be expressed in terms of elementary functions.

Finally, some informal reasoning to show that this outcome is not surprising. You can think of the skewed circular cone as a right cone with elliptical base. Recall that the perimeter of the base enters the computation of lateral area of a right circular cone. But the length of an ellipse is expressible only in terms of elliptic integrals -- which is why they are called elliptic integrals.

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According to symmetry, I think $A(a,b)=S(\sqrt {a^2+b^2})$ can be. Can we simplify more the elliptic integral that you gave ? thanks a lot for answer –  Mathlover Sep 10 '12 at 14:32
$$A=\frac{r}{2}\int_0^{2\pi}\sqrt{c^2+(r-\sqrt{a^2+b^2}\sin (t+t_0))^2}\,dt$$ $$A=\frac{r}{2}\int_{t_0}^{2\pi+t_0}\sqrt{c^2+(r-\sqrt{a^2+b^2}\sin (u))^2}\,du$$ If the integral periodic So result must be $$A=\frac{r}{2}\int_{0}^{2\pi}\sqrt{c^2+(r-\sqrt{a^2+b^2}\sin (u))^2}\,du$$ Could you please check my idea? Thanks –  Mathlover Sep 10 '12 at 14:45
@Mathlover Yes, the area depends only on $\sqrt{a^2+b^2}$. This can be also inferred from the integral using the identity $a\cos t+b\sin t = \sqrt{a^2+b^2}\cos (t-t_0)$ and the change of variable $\tau=t-t_0$. Or better yet, orient the axes so that the vertex is $(a,0,c)$ to begin with. Either way, this does not change anything essential. Here's a concrete example "simplified" with Maple: link . –  user31373 Sep 10 '12 at 14:48
I see what you mean. I will search more about elliptic integrals. It generates nice results but I really expected to see a simple elementary result of the area expression. –  Mathlover Sep 10 '12 at 14:55