In: Advanced Math
Calculate the integral of the function f (x, y, z) = xyz on the region bounded by the z = 3 plane from the bottom, z = x ^ 2 + y ^ 2 + 4 paraboloid from the side, x ^ 2 + y ^ 2 = 1 from the top.
The lower bound of the region over which we are to integrate is z=3 at he bottom.
But, the upper bound is given to be the cylinder x2+y2=1 that is actually unbounded above.
Again the bound from the sides is given to be the paraboloid x2+y2+4. The minima of this surface is (0,0,4) Then the lower bound to be z=3 does not make any sence.
THE LIMIT OF THE TRIPLE INTEGRATION NAMELY D IS :-
Z=3 from bottom , x2+y2=1 for the sides , x2+y2+4 at the top.
We use cylindrical coordinates for this integral.
The answer is 0.