evaluate triple integral z dV , where E is enclosed by the paraboloid z = 2x^2 + 2y^2 and the plane z = 2

use any coordinates

Question

evaluate triple integral z dV , where E is enclosed by the paraboloid z = 2x^2 + 2y^2 and the plane z = 2

use any coordinates

anonymous · Answer

The intersection of the paraboloid and plane is the unit circle:
$$2=2x^2+2y^2\
x^2+y^2=1$$
For some point within the unit circle, say $(0,0)$, you have the corresponding three-dimensional points $(0,0,2)$ on the plane and $(0,0,0)$ on the paraboloid. So, the point on the plane is higher and can be considered the top portion of the bounded region $E$. In other words, $z$ is bounded below by the paraboloid and above by the plane. 

Write $E$ is the following set of points:
$$\left\{(x,y,z):-1\le x\le1, ~-1\le y\le1, ~x^2+y^2\le z\le2\right\}$$
Thus making your integral
$$\int\int\int_E z~dV=\int_{-1}^1\int_{-1}^1\int_{x^2+y^2}^2z~dz~dy~dx$$
This is where my knowledge of triple and greater integrals gets rusty. I think you'll have to convert to spherical or cylindrical coordinates from here. Maybe someone else will be able to pick up where I left off.