Originally Posted by

**Shakarri** Since you are just finding a volume and not integrating a function over a volume the integral will be

$\displaystyle \int \int \int dxdydz$ with the limits added

You can do the integration in any order you like, I think the easiest is z then y then x

$\displaystyle \int \int \int dzdydx$

As you rightly pointed out, the limits on the z integral are 0 to 4+x^{2}

The limits on the y integral are got from the equations y= 4-x^{2} and x+y=2 which will be y= 4-x^{2} and y=2-x. Now to decide which is the lower and which is the upper limit, the lower limit will have the smaller value. 4-x^{2} is always less than or equal to 2-x so 2-x should be the upper limit.

The limits on the x integral are also got from the equations y= 4-x^{2} and y=2-x. Putting them equal to each other you find that 4-x^{2}=2-x The solutions to that equations are the limits for x.