I am taking multivariable calculus and we are on change of variables; a particular problem was posed that doesn't seem solvable to me.
The region is bounded by $xy = 1, xy = 4$; that's easy; set $u = xy$.
But the region is also bounded by $y = \frac x2, y = x^2$. Thus $v$ isn't as simple as $\frac yx$, as many examples do show us (giving us bounds such as $y = x$ and $y = 2x$).
How can I solve for $v$ in terms of $x$ and $y$? After all, the power of one of the bounds differs from its supposed counterpart bound ($y =\frac x2$ is linear while $y = x^2$ is polynomial)?
The farthest I can get is determining that $\frac yx =\frac 12$ and $\frac yx = 2x$; but if $v$ is to have constant boundaries that don't involve itself, this doesn't work. Is this problem doable, or is it some sort of typo?
Related question, but not relevant: here
To follow up on our discussion in the comments under your original post: this problem is a bit tricky because I'm pretty sure we can't make a change with both new variables varying within constant limits.
Next, as a disclaimer: I don't know how to explain any general approach to such problems; it's more of a trial-and-error to me, plus learning various tricks from examples.
Now, let's get to the actual question here. After deciding on the most natural choice for $u=xy$, so $1\le u\le4$, what can we define as our $v$? And the trick here is that we don't have to change both, i.e. we can keep one of the old variables. So let's keep $y$ as the second variable; or rather, we should say that we define $v=y$.
Before we proceed, note that from the graph of the original domain we know that $x\ge0$ and $y\ge0$, and therefore also $u\ge0$. So this is how this substitution is going to work: we're going to multiply the first equation by $y$ and the second equation by $y^2$ in order to create copies of $u$ (so to speak) on the right-hand sides. $$\begin{aligned} &y=\frac{1}{2}x \implies y^2=\frac{1}{2}xy \implies y^2=\frac{1}{2}u \implies y=\sqrt{u/2} \implies v=\sqrt{u/2}; \\ &y=x^2 \implies y^3=x^2y^2 \implies y^3=u^2 \implies y=u^{2/3} \implies y=u^{2/3} \implies v=u^{2/3}. \end{aligned}$$ For $1\le u\le4$, the second graph lies above the first, giving us boundaries of integration for $v$.