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
$$\int_{-1}^1 \int_{y^4}^1 (4-2y)dxdy =\frac{32}5$$
The general change of variables formula for double integrals is:
$$
\int_{\varphi(U)}f(x,y)dxdy=\int_Uf(\varphi(u,v))|\det(D\varphi)(u,v)|\;dudv\tag{1}
$$
where we assume nice properties of $f$, $U$ and $\varphi$ that this formula holds.
Calculating the new integrand $f(\varphi(u,v))|\det(D\varphi)(u,v)|$ under the change of variables is very straightforward.
In order to set up the integral properly, you need to figure out the corresponding $\varphi$, $U$, and $\varphi(U)$; this is the difficult step to calculate explicitly if it is possible at all in general.
Let us look at your specific example. Set $\displaystyle\varphi(u,v)=(\frac{\sqrt{2}\tan v}{u},\frac{u^2\cos^2v}{2})$. Then
$$
\varphi(U)=\{(x,y)\mid \alpha(y)\le x\le \beta(y), a\le y\le b\}, \quad
\alpha(y)=1+b-y,\quad \beta(y)=1+b+y\;.
$$
Now what you need is to find $U$ by applying the inverse of $\varphi$ to the region $\varphi(U)$.
Note that your region $\varphi(U)$ depends on the parameters $a$ and $b$. I don't see an explicit way to express $U$ except some easy observations that
$$
\sin^2v=x^2y,\quad u^2=\frac{2y}{1-x^2}
$$
which in principle tells you what $U$ is.
In general, this step is not easy if it is possible at all. This is the reason why "each case is different" and there is no one-fits-all procedure (to find $U$). If you recall, one needs special geometry of the region in order to apply the technique of polar coordinates for double integrals. Also, in order to write a double integral in terms of iterated integrals, one must have "normal domains".
For examples that one can work out $U$ explicitly, see this page and the references therein.
Best Answer
When integrating a function $f(x,y)$, $$\int_A^B\int_{C(y)}^{D(y)} f(x,y) dxdy$$ you can rewrite the integration region as $$ \{(x,y): A\le y \le B , C(y) \le x \le D(y) \}$$ For reasonable functions $C,D$, this is equivalent to $$ \{ (x,y) : \inf C \le x \le \sup D , D^{-1}(x)<y<C^{-1}(x)\}$$ which gets you $$\int_{\inf C}^{\sup D} \int_{D^{-1}(x)}^{C^{-1}(x)} f(x,y) dydx$$