Find the volume of the region in the first octant bounded by the surfaces $x = 0, y = x, y = 2 - x^2, z = 0$ and $z = x^2.$
It is difficult for me recognize the region geometrically. Can someone please give some idea about the sketch of the region?
I am confused with the limits of integration.
Required volume = $\int_{z = ?}^{?}\int_{y = ?}^{?}\int_{x = ?}^{?}dxdydz.$
You can see the desired region in the $xy$-plane in this graph.
And $z$ goes from $0$ to $x^2$, so your desired integral is
$$\int_{x=0}^1 \int_{y=x}^{2-x^2}\int_{z=0}^{x^2} \,dz\,dy\,dx$$ $$=\int_{x=0}^1 \int_{y=x}^{2-x^2} x^2 \,dy\,dx$$
Note that I put the variables in a different order than you do. That seems to fit the problem better.
Firstly, consider the region in the $x,y$ plane bounded by $y=x$, $x=0$ and $y=2-x^2$. We see that, for a given $x$, that $y$ ranges from $x$ to $2-x^2$. If we make a similar diagram for $x,z$, we find that $z$ ranges from $0$ to $x^2$. We have that $x$ ranges from $0$ to $1$, as $y=x$ intersects $y=2-x^2$ at $x=1$. This gives the integral as
$$\int_{x=0}^{x=1} \int_{y=x}^{y=2-x^2} \int_{z=0}^{z=x^2} \mathrm{d}z\mathrm{d}y\mathrm{d}x$$
