# Tripple integral

1. Nov 29, 2011

### sandy.bridge

1. The problem statement, all variables and given/known data
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.

3. The attempt at a solution
$$\int\int\int_D(x^2-z)dV=\int\int\int_D(2x^2+y^2-1)dV=2\int_{-1}^1x^2dx\int_{-1}^1dy\int_{-1}^{1}dz+\int_{-1}^1dx\int_{-1}^1y^2dy\int_{-1}^1dz-\int_{-1}^1dx\int_{-1}^1dy\int_{-1}^1dz$$

2. Nov 29, 2011

### sandy.bridge

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

3. Nov 29, 2011

### 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