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I'm in the process of learning multivariable calculus, and I've been having trouble with the process of changing variables in order to perform an easier integration. I know how to compute the Jacobian and analyze the region once it is given to me, but the problem of actually determining what my substitution should be has been giving me headaches. It's particularly difficult when the equations aren't linear.

For example, here's a question I've come across and managed to solve:

Use a change of variables to set up (but do not evaluate) a double integral (one only) for the area of the region in the first quadrant bounded by the curves $xy = 1$, $xy = 4$, $y = x$, and $y = 2x$.

My first instinct was to set $u=xy$, so that my two equations for $u$ become $u=1$, and $u=4$. However, I wasn't sure about what my substitution for $v$ should be, so I eventually set it to $v=x$.

In my transformation $uv$-space, I then had the curves $v=\sqrt{u}$ and $v=\sqrt{\frac{u}{2}}$ (the positive components, so my space looked like this (sorry, I can't post images yet):

Transformed space

From there, I calculated the Jacobian to be:

$$\frac{\partial (u,v)}{\partial (x,y)} = v$$

Therefore, the inverse Jacobian (the one I want), was $\frac{1}{v}$. I could then set the integral up as the following:

$$\int_{1}^{4} \int_{\sqrt{\frac{u}{2}}}^{\sqrt{u}} \frac{1}{v} \,dv \,du$$

This is indeed the correct answer, but when I look at the solution, my professor provided, he used the substitutions $u=xy$ and $v=\frac{y}{x}$. The transformed space he created is then a simple rectangle, which is simpler to integrate. It looked like this:

$$\int_{1}^{4} \int_{1}^{2} \frac{1}{2v} \,dv \,du$$

In the end, both approaches worked (as they should), but I was wondering if there were some general guidelines or suggestions that could be given when choosing what substitution should be used? Are there any tricks or strategies for transforming regions like these into rectangles? Any help would be appreciated!

P.S. Also, this is my first post to the community, so thanks for having me.

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  • $\begingroup$ Welcome to the community. Nice first post. My strategy is probably not the one you want to hear. You just need to do a lot of practice to get the muscles strong. If you practice enough problems, the solutions will start to come to you naturally, just like when first learning single-variable integration. $\endgroup$ – The Count Feb 18 '17 at 22:08
  • $\begingroup$ Not a general answer, but it's clear that $v = y/x$ was the right variable to choose so that the lines $y = x$ and $y = 2x$ corresponded to constant values of $v$. In most problems like this that you see in practice, it will be best to try to turn the region of integration into a rectangle if you can. If you're lucky, the integrand won't become too complicated when you do this. $\endgroup$ – user49640 Feb 18 '17 at 22:20
  • $\begingroup$ If you can’t easily turn the region into a rectangle, a triangle often works well. $\endgroup$ – amd Feb 18 '17 at 22:56
  • $\begingroup$ @TheCount I feared that may have been the answer. It's what my professor said as well, so I'm not surprised. I guess I'm going to have to practice a bit more in order to get comfortable with these. $\endgroup$ – Germ Feb 18 '17 at 23:17
  • $\begingroup$ @user49640 Thanks for the insight into this question! At first, I was a bit confused as to why the substitution was correct, but it makes sense to me now when I see that I want $v$ to have constant values. I'll make sure to be on the lookout for these kinds of substitutions. $\endgroup$ – Germ Feb 18 '17 at 23:19

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