I want to compute the volume of the solid that lies between the $xz$-plane, the $yz$-plane , the $xy$-plane, the planes with equations $x=1$ and $y=1$ and the surface with equation $z=x^2+y^4$.
Is it as follows?
It must hold that $0 \leq x \leq 1$, $0 \leq y \leq 1$.
So the volume of the solid described above is $\int_0^1 \int_0^1 (x^2+y^4) dxdy$. Is this right?
Also how can we compute the volume of the solid that lies between the plane with equation $z=16$ and the plane with equation $z=x^2+y^2$ ?
The volume you've given for the first question is correct.
To solve the second, notice that the curves overlap on the circle $z = 16, x^2+y^2=16.$
This means to calculate the area between the curves, simply integrate the following:
$$\int^4_{-4}\int^{\sqrt{16-x^2}}_{-\sqrt{16-x^2}}\left(16-x^2-y^2\right)dydx$$
$16-x^2-y^2$ represents the area between the plane and the paraboloid given. Imagine the original shapes and the shape between them, then compare that to $16-x^2-y^2.$