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{\bf Exam Question

Topic: Double Integral}

$R$ is the region in the positive quadrant in the $x$-$y$ plane
bounded by $$x=0,\ y=x,\ x^2+y^2=1,\ x^2+y^2=0.$$

Show that $$\int \!\!\! \int_R \frac{\ln(x^2+y^2)}{y}\,
d(x,y)=-2\ln(\sqrt2-1)(\ln27-2).$$

[You may use the result that $\tan(\pi/8)=\sqrt2-1.$]
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{\bf Solution}

In polar coordinates
\begin{eqnarray*}
I&=&\int_{\pi/4}^{\pi/2}\, d\theta\int_1^3
\frac{\ln(r^2)}{r\sin\theta}r\, dr \\ &=& \int_{\pi/4}^{\pi/2}
{\mathrm cosec} \theta \, d\theta\int_1^3 2\ln r\, dr =
\left[\ln\tan\left(\textstyle{\frac{\theta}{2}}\right)\right]_{\pi/4}^{\pi/2}
2\left[r\ln r-r\right]_1^3 \\
&=&\left[\ln\tan\textstyle{\frac{\pi}{4}}-\ln\tan\textstyle{\frac{\pi}{8}}\right]
2[3\ln3-3-\ln1+1] \\ &=&
[\ln1-\ln(\sqrt2-1)]2[3\ln3-2]=-2\ln(\sqrt2-1)(\ln27-2)
\end{eqnarray*}

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