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.
Best Answer
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.