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So it's been a while since I've had to do any remotely difficult integration, and one of my profs kind of sprung this on us.

$$ \int_{x=0.5}^{x=1.5}\int_{-0.5}^{0.5}\frac{dx\;dy}{\sqrt{x^2+y^2}} $$

If I convert to polar coordinates and integrate, I get r times theta or: $$ \sqrt{x^2+y^2}\cdot \tan^{-1}(\frac{y}{x}) $$ In rectangular coordinates. However, whenever I tired using the limits given, I arrive at an answer of 0.0933, when the expected value is close to 1. Does anybody know what could have gone awry? I think the integration method is correct, but I'm not sure.

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up vote 4 down vote accepted

Computing the integrals numerically I got $\approx 1.038049736$

Polar coordinates are more complicated for a rectangular area (the length must depend of the angle) so let's try more directly : $$\int_{\frac 12}^{\frac 32} dx\int_{-\frac 12}^{\frac 12} \frac{dy}{\sqrt{x^2+y^2}} =2\int_{\frac 12}^{\frac 32} \left[\log\left(y+\sqrt{x^2+y^2}\right)\right]_0^{\frac 12} dx$$ (or $\int \operatorname{argsinh}(\frac yx) dx$ if you prefer) $$ =2\int_{\frac 12}^{\frac 32} \log\left(\frac 12+\sqrt{x^2+\frac 14}\right)- \log(x)\ dx $$ $$=2\left[x\left(\log\left(\frac 12+\sqrt{x^2+\frac 14}\right)-\log(x)\right)+\frac {\operatorname{argsinh}(2x)}2\right]_{\frac 12}^{\frac 32}$$

(with some help from WolframAlpha)

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Hint: Can you get

$$\frac{1}{2}I=\int_0^{\theta_1}\int_{\frac{0.5}{\cos\theta}}^{\frac{1.5}{\cos\theta}}dr d\theta+\int_{\theta_1}^{\pi/4}\int_{\frac{0.5}{\cos\theta}}^{\frac{0.5}{\sin\theta}}drd\theta $$ where $\tan \theta_1=\frac{1}{3}$?

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No, I don't believe that I can. I'm not sure how theta and r can be used to describe a square – John Mar 20 '12 at 23:48
Sketch a picture, try to find the range for $\theta$. And then fix $\theta$, try to figure out what's the range for $r$, which is a function in $\theta$. – Jack Mar 21 '12 at 0:08

According to Maple, the answer is $$\frac12\,\ln \left( \sqrt {2}-1 \right) -\frac32\,\ln \left( 1+\sqrt {2} \right) -\frac32\,\ln \left( -1+\sqrt {10} \right) -\ln \left( -3+\sqrt {10} \right) +\frac32\,\ln \left( 1+\sqrt {10} \right)$$ This agrees with Raymond's answer.

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