1
$\begingroup$

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?

$\endgroup$
2
$\begingroup$

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

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ 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 '18 at 20:23
  • $\begingroup$ Yes exactly you're right ! $\endgroup$ – Pagode Jun 19 '18 at 8:44

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.