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How can the integral expression be defined to find Area (S) on $x+y+z=1$ and bordered with intersection of the cone ($x^2+y^2-z^2=0$) and the plane ($x+y+z=1$) ?

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The title and body contain two different questions. Do you want to define the area (title) or define an integral expression to find it (body)? Or perhaps you just want to find the area? That would be more straightforward. –  joriki Apr 11 '12 at 22:44
    
If I can define the integral expression , to find area is just calculation. I need help to define integral expression to find the Area (S). –  Mathlover Apr 11 '12 at 22:47
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In this case the intersection of plane and cone is a hyperbola, so the area is infinite. –  Robert Israel Apr 11 '12 at 23:00
    
@Robert: is it not a parabola? –  Théophile Apr 11 '12 at 23:03
    
It is not. Substitute $z = 1 - x - y$ into the equation of the cone and simplify to get $2x + 2y - 2 x y = 1$ or $(x-1)(y-1) = 1/2$. –  Robert Israel Apr 12 '12 at 0:03

2 Answers 2

up vote 2 down vote accepted

Using $x$ and $y$ as parameters on the plane, the parametric representation is ${\bf R} = [x,y,1-x-y]$ and $\frac{\partial {\bf R}}{\partial x} \times \frac{\partial {\bf R}}{\partial z} = [1,1,1]$ has length $\sqrt{3}$, so the area element is $\sqrt{3} \ dx \ dy$.

As I wrote in my comment, the curve of intersection with the double cone has $(x-1)(y-1)=1/2$. Since e.g. the point $(0,0,1)$ is inside the cone and has $(x-1)(y-1)=1 > 1/2$, the region inside the cone is $(x-1)(y-1) \ge 1/2$. This has two components, one where $x > 1$ and $y \ge 1 + \frac{1}{2(x-1)} = \frac{2x-1}{2x-2}$ (and $z < 0$) and one where $x < 1$ and $y \le \frac{2x-1}{2x-2}$ (and $z > 0$).

So if you still want to set up the area as an integral, it is

$$ \int_1^\infty \ dx \int_{\frac{2x-1}{2x-2}}^\infty \ dy \ \sqrt{3} + \int_{-\infty}^1 \ dx \int_{-\infty}^{\frac{2x-1}{2x-2}} \ dy \ \sqrt{3}$$

Both integrals, of course, are $+\infty$.

Here's a picture of part of the cone and plane.

enter image description here

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The angle between the $z$ axis and the cone is the same as the angle between the $z$ axis and a plane normal to $(1,0,1)$. Without explicitly calculating this angle, you can see that it must be less than the angle between the $z$ axis and a plane normal to $(1,1,1)$. Thus, as Robert commented, the intersection is not an ellipse, as your image suggests, but a hyperbola, which has infinite area.

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I understand what you mean. Thanks for answer. How can I define the hyperbola with parameters? –  Mathlover Apr 11 '12 at 23:13

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