Surface area of a cone under a plane How do I find the surface area of of the cone $z^{2}=x^{2}+y^{2}$, $z\geq0$ under the plane $x+y+z=2a$, $a>0$?
I tried to use cylindrical coordinates as follows:
$$\left\{\begin{array}{l} x=r\cos\theta, \\ y=r\sin\theta, \\ z=z \end{array}\right.$$
And to find the interval for $r$, I assumed that
$$z=2a-x-y$$
and hence
$$(2a-x-y)^{2}=x^{2}+y^{2}$$
But I didn't get anything that I could consider useful.
 A: My Calculus III is rusty at best so you might want to take this with a grain of salt (verify yourself), but just by looking at the graph on this website, it seems like your surface integral is unbounded. For my example I will use the case when $a=0$, which you say not to use, but for these purposes it doesn't make a difference. That gives $z_1 = \sqrt{x^2+y^2}$ and $z_2 = -x-y$
We can show that the area underneath the plane will occur when
$$
z_2 \geq z_1
$$
with the equality portion signifying where you integrate from. We get
$$
-x-y \geq \sqrt{x^2+y^2} \implies x^2+2xy+y^2 \geq x^2+y^2 \implies 2xy \geq 0
$$
This shows that the plane will either be greater when $x,y < 0$ or $x,y > 0$. By inspecting the graph we can see that this part is when $x,y < 0$. We also notice that in our case there is an intersection at $(x,y) = (0,0)$
Now to partly show that they will never intersect ever again to close off your surface, we take the gradient of each.
$$
\nabla z_1 = <\dfrac{2x}{\sqrt{x^2+y^2}}, \dfrac{2y}{\sqrt{x^2+y^2}}>
$$
$$
\nabla z_2 = <-1, -1> 
$$
So our plane is actually growing faster than our cone, which means that the two will never intersect again, therefore your surface area is unbounded 
