Limits of integration on double integrals I was given this problem:

Find an integral equal to the volume of the solid bounded by $z=4-2y,z=0,x=y^4,x=1$ and evaluate. 

I understand how to evaluate once my double integral is set up, but I do not know how to find my limits of integration.
I am assuming that my function will be $z=4-2y$ and that using this I should be able to find my limits of integration. I can say that $0=4-2y$ which means that $y=2$. I can then plug that into $x=y^4$ and get $1\leq x\leq 16$ which may be correct, but I still am missing the limits of integration for y.   
Am I thinking about this problem correctly? How can I go about solving this?
 A: Note that $x=y^4$ and $x=1$ intersect at $ (x,y)=(1,\pm1)$. which define the limits for the integration region in the $xy$- plane. Thus, the volume integral is 
$$\int_{-1}^1 \int_{y^4}^1 (4-2y)dxdy =\frac{32}5$$
A: That solid is located above the plane $z=0$ and below the plane $z=4-2y$. The possible values for $x$ belong to the $[0,1]$ interval (the condition $x=y^4$ prevents $x$ from being negative). So, you should compute$$\int_0^1\int_{-\sqrt[4]x}^{\sqrt[4]x}\int_0^{4-2y}1\,\mathrm dz\,\mathrm dy\,\mathrm dx\tag1$$But\begin{align}(1)&=\int_0^1\int_{-\sqrt[4]x}^{\sqrt[4]x}4-2y\,\mathrm dy\,\mathrm dx\\&=\int_0^18\sqrt[4]x\,\mathrm dx\\&=\frac{32}5.\end{align}
A: As you know how to evaluate the integral - and as it has been evaluated in other answers! - I'll concentrate on showing that a unique subset of $\mathbb{R}^3$ is bounded by the four surfaces identified in the question, and describing that subset in such terms that one can write down the triple integral that is to be evaluated.
@saulspatz's comment recommends first drawing a figure, ignoring the $z$ coordinate. I also find this to be the easiest way to think about the question.
The plane $x = 1$ cuts the $(x, y)$ plane in a line, and the surface $x = y^4$ cuts the $(x, y)$ plane in a curve. The line and curve together subdivide the $(x, y)$ plane into five subsets, which correspond to four subsets of $\mathbb{R}^3$:
\begin{align*}
A & = \{ (x, y, z) \colon x \geqslant y^4 \text{ and } x \geqslant 1 \}, \\
B & = \{ (x, y, z) \colon x \leqslant y^4 \text{ and } x \leqslant 1 \}, \\
C & = \{ (x, y, z) \colon x \leqslant y^4 \text{ and } x \geqslant 1 \}, \\
D & = \{ (x, y, z) \colon x \geqslant y^4 \text{ and } x \leqslant 1 \}.
\end{align*}
Each of $A, B, C, D$ is a connected subset of $\mathbb{R}^3,$ but the projection of $C$ on the $(x, y)$ plane has two separate components, corresponding to positive and negative values of $y.$
Each of $A, B, C, D$ is an unbounded subset of $\mathbb{R}^3,$ but the projection of $D$ on the $(x, y)$ plane is bounded. That looks hopeful! In order to be in a position to say something more definite than that, the easiest thing to do next (or so I think) is to look at the projection of the planes $z = 4 - 2y$ and $z = 0$ on the $(y, z)$ plane.
One can see that these two planes between them divide $\mathbb{R}^3$ into four subsets:
\begin{align*}
E & = \{ (x, y, z) \colon
(y \geqslant 2 \text{ and } z \geqslant 0) \text{ or }
(y \leqslant 2 \text{ and } z \geqslant 4 - 2y) \}, \\
F & = \{ (x, y, z) \colon
(y \leqslant 2 \text{ and } z \leqslant 0) \text{ or }
(y \geqslant 2 \text{ and } z \leqslant 4 - 2y) \},
\end{align*}
\begin{align*}
G & = \{ (x, y, z) \colon 4 - 2y \leqslant z \leqslant 0 \}, \\
H & = \{ (x, y, z) \colon 0 \leqslant z \leqslant 4 - 2y \}.
\end{align*}
Subset $E$ contains points with arbitrarily large positive values of $z$ for any value of $y$; and subset $F$ contains points with arbitrarily large negative values of $z$ for any value of $y$; therefore neither $E$ nor $F$ has a bounded intersection with any of $A, B, C, D.$
Subset $G$ only contains points with values of $y \geqslant 2,$ therefore its intersection with $D$ is empty.
Subsets $A, B, C$ all have points with arbitrarily large positive values of $y,$ as do their intersections with $G.$
Therefore the only candidate for a subset of $\mathbb{R}^3$ that is bounded by the four given surfaces - and is bounded (!) - is:
$$
D \cap H = \{ (x, y, z) \colon y^4 \leqslant x \leqslant 1 \text{ and }
0 \leqslant z \leqslant 4 - 2y \}.
$$
This is indeed bounded, and we can evaluate the volume integral by writing:
$$
\int_{D \cap H} 1 = \int_{-1}^1\int_{y^4}^1\int_0^{4 - 2y}\,dz\,dx\,dy.
$$
I'll stop here - approximately where the other answers start. :)
