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

Answer 1

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.