Volume of a solid bounded by surfaces - is it correct?

Question

Could you check if my calculations and reasoning are correct.

And maybe suggest a nicer way of solving this problem?

We are given a solid bounded by these surfaces:

$y=x^2, \ y=1, \ 2x+y+z = 4, \ z=0$

I drew a picture, and it appears that

$z$ goes from $0$ up to $4-2x-y$ (the region below $z=0$ isn't bounded, here: http://www.wolframalpha.com/input/?i=2x%2By%2Bz%3D4 )

$y$ goes from $0$ to $1$ (due to $y = x^2$)

and $x$ goes from $-\sqrt{y}$ to $\sqrt{y}$

Now is it correct to use thses limits to integrate $\int \int \int dz dx dy$?

I would really appreciate all your help.

Thank you.

Answer 1

That looks pretty much correct. But you don't want to integrate in that order: you need your outermost integral to not depend on the values of inner variables. So a good choice is $$ \int_0^1 \int_{-\sqrt{y}}^\sqrt{y}\int_0^{4-2x-y} \ldots ~dz ~dx~dy $$

Now the value of the upper limit on $z$ is well-defined at the point it's used, etc. If this doesn't make sense, just ask and I'll try to say more.