Boundaries in 2D space

Question

Find the volume of figure that is bounded with the three following surfaces:$z=y^2,y=x^2,z=4$. I don't think this has anything to do with polar coordinates etc. So I tried to put this problem in $OXY$ plane and i get parabola $y=x^2$ and $y=0$ but not sure how to read boundaries, because here(in $OXY$) we don't have that some area is closed, only that is bounded with parabola and x-axis, but not sure how to know boundaries for x,y from that. If someone knows how to proceed following example in this way I would be happy to know.

Answer 1

At any given $z$, your surface will be bound by:

$-\sqrt y \leq x \leq \sqrt y \,$ (by parabolic cylinder $y = x^2$)

$0 \leq y \leq \sqrt z \,$ (by parabolic cylinder $z = y^2$).

Bound of $z$ is $0 \leq z \leq 4$.

Integrate in order $dx$ first, $dy$ next and $dz$ last.

EDIT: Here is the integral

$\displaystyle \int_0^4 \bigg [\int_0^{\sqrt {z}} \bigg[\int_{-\sqrt {y}}^{\sqrt{y}} dx\bigg] \, dy \bigg]\, dz$

And the sketch at any given $z \,$:

You can also change the order of integration by writing the limits differently.

Answer 2

$z=4$ and $z=y^2$ intersection projection gives lines $y=\pm 2$ on $OXY$ which together with $y=x^2$ gives figure on $OXY$. Then integral is $$\int\limits_{-\sqrt{2}}^{\sqrt{2}}\int\limits_{x^2}^{2}\int\limits_{y^2}^{4}dzdydx$$