I have a function $f(x,y)$ that satisfies $f(r \cos \theta, r \sin \theta) = g(r)$. I would like to write the following double integral in polar coordinates, but I am having trouble with the partial derivatives inside the double integral.

\begin{equation} \iint_A \biggl[ \Bigl(\frac{\partial^2 f(s,y)}{\partial s^2} \Bigg\vert_{s = x} \Bigr)^2 + 2 \Bigl(\frac{\partial f(s,t)}{\partial s \partial t} \Bigg\vert_{s = x, \\ t = y} \Bigr)^2 + \Bigl(\frac{\partial^2 f(x,t)}{\partial t^2} \Bigg\vert_{t = y} \Bigr)^2 \biggr] \, \mathrm{d}A \end{equation}

I am aware that I can write $\mathrm{d}A = r \, \mathrm{d}r \, \mathrm{d}\theta$. Can anyone point out to me how I would write, e.g.,

\begin{equation} \frac{\partial f(s,t)}{\partial s \partial t} \Bigg\vert_{s = x, \\ t = y} \end{equation}

in polar coordinates? I am not sure at what point I should do the substitution $x = r \cos \theta$, $y = r \sin \theta$ and what does $\frac{\partial}{\partial s}$ or $\frac{\partial}{\partial t}$ mean in that context?


1 Answer 1


So for that kind of problem you always use a new name for your function with a variable transfer function

You should also clearly write : " Let $x$ a real ... " (I recommend you to use that formulation to talk clearly even if most of the time we understood

For instance define :

$$ \phi: (r,\theta) \to (r\cos(\theta),r\sin(\theta)) $$

$$ F:(r,\theta)\to (f\circ\phi)(r,\theta) $$

So we can clearly work.

$$(\partial_1F)(r,\theta)=((\partial_1f)\circ\phi)(r,\theta)\cos(\theta)+((\partial_2f)\circ\phi)(r,\theta)\sin(\theta) $$ $$(\partial_2F)(r,\theta)=-((\partial_1f)\circ\phi)(r,\theta)r\sin(\theta)+((\partial_2f)\circ\phi)(r,\theta)r\cos(\theta) $$

I guess $f$ is $ \mathcal{C}^2 $ since you used Schwarz.

So we only calculate one second crossed derivative :

$$ \partial_{12}F(r,\theta)= -r\sin(\theta)[(\partial_{11}f)\circ\phi(r,\theta)\cos(\theta)+((\partial_{12}f)\circ\phi)(r,\theta)\sin(\theta)] + r\cos(\theta)[(\partial_{11}f)\circ\phi)(r,\theta)\cos(\theta)+(\partial_{12}f)\circ\phi)(r,\theta)\sin(\theta)] $$

So you can evalute in your polar coordinate

  • $\begingroup$ Thanks a lot for the answer. However, I am not too familiar with the terminology. What does $((\partial_1 f) \circ \phi)(r,\theta)$ mean? Is that the derivative of $f$ with respect to its first argument evaluated in the point $(r \cos(\theta),r \sin(\theta))$? $\endgroup$
    – Ritz
    Jun 18, 2018 at 20:23
  • $\begingroup$ Yes exactly you're right ! $\endgroup$
    – Pagode
    Jun 19, 2018 at 8:44

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