# Double integral involving partial derivatives: change to polar coordinates

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.

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

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.,

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

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?

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

• 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 $(r \cos(\theta),r \sin(\theta))$? – Ritz Jun 18 '18 at 20:23
• Yes exactly you're right ! – Pagode Jun 19 '18 at 8:44