The question is stated as follows:
Given the function $f(x,y) = 3-x^2-y^2$ and plane $z=1-2x-2y$, find the volume of the solid created by their intersection.
The graph of the paraboloid and plane
Initially, I tried to find the total volume by first calculating the total volume of the paraboloid and then subtracting the part that is intersected by the plane. The process took a huge amount of effort and I don't think I got the correct result anyways, so I am wondering if there is a simple solution I am missing.
The paraboloid intersects the plane $z=1-2x-2y$ at a minimum of $z = -3-4\sqrt2$, intersecting at the point $(1+\sqrt2, 1+\sqrt2, -3-4\sqrt2)$
Graph with the plane and point of minimum intersection
Can someone help me with approaching this problem? I know it has a very simple and straightforward solution but I can't seem to find it.
Thank you so much!
For definition:
$$ ||\Omega|| = \iiint\limits_{\Omega} 1\,\text{d}x\,\text{d}y\,\text{d}z $$
and since the solid in question is:
$$ \Omega = \left\{(x,y,z) \in \mathbb{R}^3 : 1-2x-2y \le z \le 3-x^2-y^2\right\} $$
it's clear that it's convenient to integrate for wires parallel to the z-axis:
$$ ||\Omega|| = \iint\limits_D \text{d}x\,\text{d}y \int_{1-2x-2y}^{3-x^2-y^2} 1\,\text{d}z = \iint\limits_D \left(2-x^2-y^2+2x+2y\right)\text{d}x\,\text{d}y $$
where:
$$ D = \left\{(x,y) \in \mathbb{R}^2 : 1-2x-2y \le 3-x^2-y^2\right\}. $$
On the other hand, completing the squares, we have:
$$ ||\Omega|| = \iint\limits_{(x-1)^2+(y-1)^2\le 4} \left(4-(x-1)^2-(y-1)^2\right)\text{d}x\,\text{d}y $$
that through a transformation of coordinates into polar coordinates with pole $(1,1)$:
$$ ||\Omega|| = \int_0^{2\pi} \text{d}\theta \int_0^2 \left(4-\rho^2\right)\rho\,\text{d}\rho = 8\pi $$
which is what was wanted.