2

Compute the following integral: $\int_D \ln(x^2+y^2)dxdy$ where $D$ is the disc $x^2+y^2 \le1$ in $R$ into polar coordinates.

What I have tried: $r^2\le1 \implies r\le1$ we also have $0 \le\theta\le2\pi$

This produces $$\int_0^{2\pi} d\theta\int_0^1\ln(r)\cdot r dr \implies 2\pi\int \ln(r)\cdot r dr$$

Taking the integral by parts to tackle the $dr$ integral we have $du = \frac{1}{r}$ and $v = \frac{r^2}{2}$

Which gives $$2\pi \left[\frac{\ln(r)r^2}{2}\Biggr|_0^1-\frac{1}{2}\int_0^1 rdr\right]\implies-\frac{2\pi}{4}$$

However, the right answer should be $-\pi$ any idea where I went wrong? Also, the actual question asks me to let "$D_\delta$ be the annulus $\delta^2\le x^2+y^2\le1$ as I'm supposed to show that this integral has a finite limit which is $-\pi$ although I'm unsure on how to proceed with this and would really appreciate the communities support.

Arctic Char
  • 16,007
Stackcans
  • 361
  • 4
    Good work but a silly mistake :) - it should have been $ \ln (x^2+y^2) = \ln (r^2) = 2 \ln r$, which would give double of your answer i.e $- \pi$ – Math Lover Oct 17 '21 at 12:32
  • @MathLover Hammering my head ... simple mistakes. Any idea on how to prevent human errors like this? furthermore, do you have some support on how to correctly compute the limit which uses $\delta^2 \le x^2 +y^2 \le 1$? – Stackcans Oct 17 '21 at 12:34
  • Does this help? https://math.stackexchange.com/questions/434742/finding-lim-limits-x-to0x2-ln-x-without-lhospital – Math Lover Oct 17 '21 at 12:38
  • 1
    @MathLover Ah I think I get it now - I believe I was supposed to integrate r from $\int_{\delta}^1$ then compute the limit as $lim_{\delta \to 0}$? – Stackcans Oct 17 '21 at 12:41
  • On your more difficult question as to how to prevent human errors like this , the problem is that we are all humans :) I wish I had a solution. My daughter loses $10$-$15$ marks every exam due to such mistakes and the only thing I keep telling her is to revise her work, be more focused into the calculation and at ease but so far it has not worked so not sure how I advise the same to others :) – Math Lover Oct 17 '21 at 12:43
  • 1
    Yes you are right about the limits of the integral being $\int_{\delta}^1$ and then compute $\lim_{\delta \to 0}$ – Math Lover Oct 17 '21 at 12:49
  • 1
    @MathLover Thank you for those tips! I can agree with them as they've definitely helped me throughout my learning. Though the 'at ease' part is usually the hardest! However, I do believe the greatest trait for any mathematician is luck. Without it, I wouldn't have been able to meet great people like you that helped me so much towards improving my math skills :) – Stackcans Oct 17 '21 at 12:55
  • The reason you must compute this as a limit is because ln(0) is undefined. – JJ Hoo Oct 17 '21 at 13:07

1 Answers1

2

Using annulus area $D_\delta=\{(x, y)|\delta^2\leq x^2+y^2\leq 1\}$, I'll solve that integral. Let $I_\delta$ as that value.

$$I_\delta=\int_{D_\delta}\ln(x^2+y^2)$$

Then, represent that annulus area as that... $D_\delta=\{(r\cos\theta,r\sin\theta)|\delta\leq r\leq1, 0\leq\theta\lt2\pi\}$ Then the integral $I_\delta$ goes like this. $$I_\delta=\int_0^{2\pi}d\theta\int_\delta^12r\ln rdr=4\pi\int_\delta^1r\ln rdr$$ ($\ln(x^2+y^2)=\ln r^2=2\ln r$) So... $$I_\delta=4\pi \biggl[{\cfrac{r^2(2\ln r-1)}{4}}\biggr]_\delta^1=-\pi(2\delta^2\ln\delta-\delta^2+1)$$ Then, Gave an limit. $$I=\lim \limits_{\delta \to 0+}{-\pi(2\delta^2\ln\delta-\delta^2+1)}=-\pi$$

That is the answer.

MH.Lee
  • 5,568