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Consider a paraboloid $x^2+y^2+2=z$. The task is to find the volum of the body obtained by confining the paraboloid by several planes $x=0, z=0, y=0$ and $x=3, y=3$.

The zero planes cuts out a quarter of the paraboloid and the remaining plans cuts out two pieces with crosssections being area of the parabol.

I know how to calculate the volumes, it isn't my problem. The problem is that the answer given is 72. But the way I see it, the confined body is not finite, it's infinite and the volume in this case should be infinite. To make it finite one needs aditional plane, e.g. z=10. I think I am right and the answer given is wrong, but I need reassurance ("another pair of eyes").

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  • $\begingroup$ How is the question phrased? It might be the object below the paraboloid. $\endgroup$ – KittyL Mar 19 '15 at 9:58
  • $\begingroup$ @KittyL The more exact phrasing would be as follows: What is the volume of the body confined by the planes $x=0,y=0z=0,x=3,y=3$ and the paraboloid $z=x^2+y^2+2$? $\endgroup$ – Tomas Mar 19 '15 at 10:01
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    $\begingroup$ Yes, that confirms my guess. It is confined by all those, that means it is the volume below the paraboloid, and above xy-plane, in the first octant. $\endgroup$ – KittyL Mar 19 '15 at 10:06
  • $\begingroup$ I just realized it! I was concentrating on the "insides" of the paraboloid! Thanks! $\endgroup$ – Tomas Mar 19 '15 at 10:08

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