Volume of a cone covered with a plane

[shakaflaka]

Homework Statement

I need to find the volume of a cone covered with a plane z=h using multiple integrals. The scheme is something like this:

Homework Equations

Formula of the cone x^2-y^2-z^2=0

The Attempt at a Solution

I tried to integrate ∫∫∫(x^2-y^2-z^2)dxdydz in all different kinds of bounds, for example {x,0,L}, {y,-R,R}, {z,-R, H}, but the answer is always negative. I can't really get my head around the error, help please?

 

[voko]

I tried to integrate ∫∫∫(x^2-y^2-z^2)dxdydz

What does this formula even mean?

 

[LCKurtz]

And remember if you are calculating a volume with a triple integral you always use a formula$$
Vol = \iiint_R 1\, dV$$
where $R$ is the region in question and $dV$ is $dxdydz$ in some order and $R$ is described by putting appropriate limits on the integrals. The integrand is always $1$.

 

[haruspex]

Your first step is to choose a way to slice up the volume into laminae whose areas you can hope to calculate. I would suggest slices parallel to the xz plane, so the area will be a function of y.

 

[LCKurtz]

Also notice that either the picture or the equation is incorrect. The cone that is pictured would be a portion of $x^2-y^2+z^2=0$.

 

[klondike]

Your first step is to choose a way to slice up the volume into laminae whose areas you can hope to calculate. I would suggest slices parallel to the xz plane, so the area will be a function of y.

The integral is very nasty if it were to be done by hand. Is it a homework question?