One more question--Triple integrals

[Daniel_Feldman]

I need to find the region E for which the triple integral of (6-3x^2-2y^2-2z^2)dV is a maximum.

I know that triple integrals give volume, and this region would be bounded by some surface, but I'm not quite sure how to go beyond that.

 

[Count Iblis]

I need to find the region E for which the triple integral of (6-3x^2-2y^2-2z^2)dV is a maximum.

I know that triple integrals give volume, and this region would be bounded by some surface, but I'm not quite sure how to go beyond that.

Suppose you start with some region E. If there is region where
6-3x^2-2y^2-2z^2>0 that is not contained in E you could add that region to E and then the value of the integral will increase. If there is a region where
6-3x^2-2y^2-2z^2<0 that is contained inside E, then deleting that region from E will make the integral larger. So, we can conclude that the integral is maximal if E is the set of points at which 6-3x^2-2y^2-2z^2>0.

 

[Unco]

I know that triple integrals give volume

This is the case when the function being integrated is equal to a constant. Otherwise such a "volume" is 4-dimensional.