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It's been a while since I've done calc 3 work, and I'm a bit stuck on how to convert the integral $\iint f_{xyz}(x,y,z)\,dy\,dz$ into polar form.

I would like to use $y = r \cos \theta$ and $z = r \sin \theta$. From this, I tried to derive $dy$ and $dz$, but I'm pretty sure I'm not doing this correctly.

$$ \frac{dy}{dr} = \cos \theta \\ \therefore dy = \cos \theta\,dr \\ \frac{dz}{dr} = \sin \theta \\ \therefore dz = \sin \theta\,dr $$

If I plug this into my integral I get

$$ \iint f_{xyz}(x, r\cos \theta, r \sin \theta) \cos \theta \sin \theta \,dr\,dr $$

Which doesn't look right at all. I'm pretty sure the polar form should be $$ \iint f_{xyz}(x, r\cos \theta, r \sin \theta)\,r\,dr\,d\theta $$

Would the correct approach be to do something like $$ \frac{dz}{d\theta} = r \cos \theta \\ \therefore dz = r \cos \theta\,d\theta $$

Then the integral looks like

$$ \iint f_{xyz}(x, r\cos \theta, r \sin \theta) \cos^2 \theta\,r\,dr\,d\theta $$

But this doesn't look right either because of the $\cos^2 \theta$. Where am I going wrong with my thought process?

Also, I'm aware that we have $$ dA = dy\,dz = r\,dr\,d\theta $$

But here, I'm interested in understanding where I'm going wrong with my approach.

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  • $\begingroup$ A similar question (which is about to be closed as a duplicate, and the chain continues...). $\endgroup$
    – metamorphy
    Oct 9, 2020 at 13:52

2 Answers 2

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The problem is that there is not just a one variable way to convert $dy$ and $dz$ separately into $dr$ and $d\theta$. You have to do it "together". The tool for that is called the Jacobian determinant: in the case of ordinary polar coordinates (usually written in $x,y$) you get $dx dy = r dr d\theta$ (or the opposite order, it doesn't matter). Naturally nothing different really happens if you do this in $y,z$ instead.

What breaks your approach is basically that $y,z$ depend on both $r$ and $\theta$ so that in fact $dy=\cos(\theta) dr - r \sin(\theta) d \theta$ and $dz=\sin(\theta) dr + r \cos(\theta) d \theta$. You can in principle use this idea to develop the Jacobian determinant "from scratch", but it takes some work, and the way it actually works out is a bit counterintuitive as well.

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  • $\begingroup$ But if I do something like $dy/dr$, aren't I making the assumption that $\theta$ is fixed/constant? $\endgroup$
    – student010101
    Oct 9, 2020 at 13:24
  • $\begingroup$ @student010101 If you fix $\theta$ then you can only integrate on a line segment with one end at the origin. To ultimately integrate over some 2D region you need to allow $\theta$ to change too. $\endgroup$
    – Ian
    Oct 9, 2020 at 13:30
  • $\begingroup$ @student010101 For a kind of physics way of thinking about $dy=\cos(\theta) dr - r \sin(\theta) d \theta$, consider a path that moves over time. Then the path has some $\frac{dr}{dt}$ and $\frac{d\theta}{dt}$. In this situation, this equation between two differentials corresponds to the chain rule $\frac{dy}{dt} = \cos(\theta) \frac{dr}{dt} - r \sin(\theta) \frac{d\theta}{dt}$. Now on a path with $\frac{d\theta}{dt}=0$, you can do what you did and everything is fine, but you can't integrate over a 2D region while holding $\theta$ fixed. $\endgroup$
    – Ian
    Oct 9, 2020 at 13:33
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Performing arithmetic on differentials does not extend into two or more dimensions. What you need to do is construct the Jacobian matrix of partial derivatives of each equation in the transformation. We can ignore $x$ since it's unchanged: $$\begin{pmatrix} \frac{\partial y}{\partial r}&\frac{\partial z}{\partial r}\\ \frac{\partial y}{\partial\theta}&\frac{\partial z}{\partial\theta}\end{pmatrix}=\begin{pmatrix} \cos\theta&\sin\theta\\ -r\sin\theta&r\cos\theta\end{pmatrix}$$ The (absolute value of the) determinant of this matrix gives us the factor we need to multiply by when converting to the target coordinates. In this case it's just $r$.

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  • $\begingroup$ Isn't the determinant here always positive? What is the purpose of the absolute value? $\endgroup$
    – student010101
    Oct 9, 2020 at 13:29
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    $\begingroup$ @student010101 That determinant is positive, but if you order the variables the other way around (which is 100% valid mathematically, regardless of how badly it violates convention) it becomes negative. (In general it won't change sign within a region where the transformation is defined, though.) $\endgroup$
    – Ian
    Oct 9, 2020 at 13:30

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