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I'm wondering how to compute the integral $$ \int_2^3\int_0^\sqrt{3x-x^2}\frac{1}{(x^2+y^2)^{1/2}}\,\mathrm{d}y\mathrm{d}x. $$ Clearly it is too complicated to do it directly, so I'm guessing you have to do some change of variables. But what kind of change? This is not really a sphere, so I don't think that polar coordinates would be so good.

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  • $\begingroup$ Please check your upper limit of the inner integral as it should be in $y$. $\endgroup$ – DeepSea Apr 27 '15 at 14:46
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You'll actually want to use polar coordinates. Set your first integral to y, square that y, and you should see you can move x to the left-hand side. You can then turn this into your r value for polar coordinates. You'll also note that it's mapping a sphere in the upper-right coordinate (which allows you to pick the theta values). From there it's a rather simple integration. I'll purposely leave the work to you, if you need clarification let me know.

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  • $\begingroup$ Also, your dxdy should be dydx due to what your first integration is. $\endgroup$ – PTiger17 Apr 27 '15 at 14:49
  • $\begingroup$ I'm not sure I got it. You means that since $y$ ranges in the interval $0\leq y\leq\sqrt{3x-x^2}$, we can square it and rewrite it as $(x-\frac{3}{2})^2+y^2\leq\frac{9}{4}$. But what do you do to transform the integralen then? I don't really get it $\endgroup$ – user233198 Apr 27 '15 at 15:08
  • $\begingroup$ write the equation of the circle in polar $r=3\cos \theta $. draw the region of integration, it extends from the straight line $x=2$ ( $r \cos \theta =2$ in polar) to the circle $\endgroup$ – Lozenges Apr 27 '15 at 15:41
  • $\begingroup$ As Lozenges said, what you'll do is square y, to get $x^2+y^2=3x$, which becomes $r=3cosx$. So that, to zero, is your limit of integration for r. Since this corresponds to a semi-circle ONLY in the upper right quadrant (where x and y are both positive) you can let theta be $pi/2$ to 0. Those are your limits of integration. $\endgroup$ – PTiger17 Apr 27 '15 at 16:27
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Starting with the integral

\begin{equation*} \int \frac{1}{\sqrt{x^2+y^2}}dx \end{equation*}

Use the substitution $x=y\tan(u),~dx=y\sec^2(u)du.$ The integral becomes

\begin{equation*} y\int \frac{\sec(u)}{y}du=\int\sec(u)du=\int \frac{\sec^2(u)+\tan(u)\sec(u)}{\tan(u)+\sec(u)}du. \end{equation*}

Use the substitution $s=\tan(u)+\sec(u):$

\begin{equation*} \int \frac{1}{s}ds=\log(s). \end{equation*}

Substituting back again and using $\sec(\tan^{-1}(z))\sqrt{z^2+1}$ and $\tan(\tan^{-1}(z))=z$ we get

\begin{equation*} \log(\sqrt{x^2+y^2}+x). \end{equation*}

Can you finish the rest of the integration?

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in polar coordinates, the integral is equivalent to

$$\int _0^{\arctan \left(\left.\sqrt{2}\right/2\right)}\int _{2 \sec \theta }^{3\cos \theta }drd\theta =$$ $$\int_0^{\arctan \left(\left.\sqrt{2}\right/2\right)} (3\cos \theta -2 \sec \theta ) \, d\theta =$$ $$\sqrt{3}-\ln \left(2+\sqrt{3}\right)$$

$r$ goes from the straight line $r=2 sec \theta $ to the circle.

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