Tripple Integral Calculation: Cube & Paraboloid

[sandy.bridge]

Homework Statement

Calculate the tripple integral $$\int\int\int_D(x^2-z)dV$$ in the doman D which is bounded by the cube $$-1\leq{x}, y, z\leq{1}$$ and lies below the parabloid $$z=1-x^2-y^2$$.

Okay, so we have not yet learned these in class, however, we were wanted to try this using intuition from double integrals. Can someone tell me if my "intuition" is wrong?
Thanks.

The Attempt at a Solution

$$\int\int\int_D(x^2-z)dV=\int\int\int_D(2x^2+y^2-1)dV=2\int_{-1}^1x^2dx\int_{-1}^1dy\int_{-1}^{1}dz+\int_{-1}^1dx\int_{-1}^1y^2dy\int_{-1}^1dz-\int_{-1}^1dx\int_{-1}^1dy\int_{-1}^1dz$$

 

[sandy.bridge]

Also, can someone tell me the trick as to how I can avoid latex skipping a line?

 

[ehild]

I am afraid, you did not understand the problem. The function f(x,y,z)=x^2-z has to be integrated for the volume which is bounded by the paraboloid and the cube. You can not replace z by the equation of the paraboloid: it gives the value of function f on the parabolic surface. But you have a value for all points (x,y,z) inside the integration domain. Make a sketch to find out the boundaries of integration.

ehild