How can I write the limits of the integrals below using a solid V in the first octant that is delimited by the parabolic cylinder y= 27/8*x^3 and the plane y = 2 - 2z? I need to sketch and label the volume V as well. int int int dx dy dz int int int dy dx dz int int int dz dy dx If you could please help me, that would be much appreciated!

First octant means xgt=0, ygt=0, zgt=0. The formula y=27/8 x^3 may be written as y=(3/2 x)^3. We need the (simple) inverse formulas, too: x=2/3 root(3)(y), z=1-y/2.
Because zgt=0 we see ylt=2 and therefore xlt=2/3 root(3)(2). Also, z=1-y/2lt=1. Now we are ready to write the integral in different orders of variables.
If x is fixed, y can be between (3/2 x)^3 and 2 in xy plane z=0. For fixed x and y we see z can be between 0 and 1-y/2, i.e. the integral becomes
int_(x=0)^(x=2/3 root(3)(2)) dx int_(y=(3/2 x)^3)^(y=2) dy int_(z=0)^(z=1-y/2) dz(1) = int_(x=0)^(x=2/3 root(3)(2)) int_(y=(3/2 x)^3)^(y=2) int_(z=0)^(z=1-y/2) 1 dz.
If we start from y it can be from 0 to 2, then x can be from 0 to 2/3 root(3)(y) and z can be from 0 to 1-y/2. The integral is
int_(y=0)^(y=2) int_(x=0)^(x=2/3 root(3)(y)) int_(z=0)^(z=1-y/2) 1 dy dx dz.
And if we start with z, it can be from 0 to 1, then y can be from 0 to 2-2z and x from 0 to 2/3 root(3)(y), i.e.
int_(z=0)^(z=1) int_(y=0)^(2-2z) int_(x=0)^(x=2/3 root(3)(y)) 1 dz dy dx.
It is not that simple to sketch V but I tried. :)