2

Okay so I have an integral of the form

$$\int_0^2\int_x^{\sqrt3 x} f\left(\sqrt{x^2+y^2}\right)dydx$$ and I am asked to change this into polar coordinates firstly by integrating with respect to $\theta$ first and then write it again but with respect to $r$ first.

I managed to find it with respect to $r$ first as $$\int_{\pi/4}^{\pi /3}\int_0^{2\sec \theta} f(r)~rdrd\theta$$

But I am struggling to deduce the limits for the the other way around. I know $0 \leq r \leq 4$ over the region of integration but I am struggling to find the limits for $\theta$ I know the method is to pick an arbitrary $r$ in the interval $0 \leq r \leq 4$ and then see how $\theta$ varies over this choice of $r$ but when I do that it just looks like $\theta$ has the same values as in the first part which clearly isn't right.

Could anyone tell me where I am going wrong please?

Shamus
  • 23

1 Answers1

0

This way around it's a bit more tricky because there are two different cases. For $0\le r\le \sqrt8$ you have the entire range $\frac\pi4\le\theta\le\frac\pi3$, whereas for $\sqrt8\le r\le4$ the lower limit for $\theta$ is $\arccos\frac2r$. When in doubt, draw a diagram :-)

joriki
  • 238,052
  • I did draw a diagram. I didn't clear it up for me however :(. Honestly I'm quite confused I do understand (I think) why the limits are the way you say now. But how would I go about writing the integral now? Would I have to write it as the sum of two double integrals over the ranges you say? – Shamus Mar 25 '16 at 21:12
  • @Shamus: Yes. Generally you could also write it in one double integral with the lower limit something like $\max(\frac\pi4,\arccos\frac2r)$, but that's not defined for $r\lt2$. – joriki Mar 25 '16 at 21:16
  • Okay I just used so plotting software and now I see what is going on you have helped immensely. So would my answer just be $\int_0^{\sqrt 8} \int_{\pi /4}^{\pi /3} f(r) r d\theta dr + \int_{\sqrt 8}^{4} \int_{\arccos (2/r)}^{\pi /3} f(r) r d\theta dr$ – Shamus Mar 25 '16 at 21:22
  • @Shamus: Yes, exactly. – joriki Mar 25 '16 at 21:23
  • Brilliant thank you once again. – Shamus Mar 25 '16 at 21:24