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I am trying to evaluate this double integral, but I don't see any good change of variables (I tried polar, but it got really hairy):

$$ \iint_D \sqrt{(x-1)^2+y^2} \, dx \, dy $$ given $D = \{(x,y):x^2+y^2 \le 1, y>0\}$

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Use this change of variable

$$\begin{array}{} x = r \cos \theta +1 \\ y = r \sin \theta \end{array} $$

and the Jacobian will be

$$ J= \begin{vmatrix} \frac{\partial x}{\partial r} & \frac{\partial x}{\partial \theta} \\ \frac{\partial y}{\partial r} & \frac{\partial y}{\partial \theta} \end{vmatrix} = \begin{vmatrix} \cos \theta & -r \sin \theta \\ \sin \theta & r \cos \theta \end{vmatrix} = r $$

The equation of the semi-circle in new coordinates will be

$$\begin{array}{} x^2 + y^2 = 1 \\ (r \cos \theta + 1)^2 + (r \sin \theta)^2 =1 \\ r^2 + 2 r \cos \theta + 1 = 1 \\ r=-2 \cos \theta \end{array}$$

Also, the description of your domain $D$ in new coordinates will be

$$D = \{(r,\theta): 0 \le \theta \le \pi , 0 \le r \le -2 \cos \theta \}$$

and hence your integral becomes

$$I=\int_{0}^{\pi} \int_{0}^{-2 \cos \theta} r \cdot r dr d\theta =\int_{0}^{\pi} \int_{0}^{-2 \cos \theta} r^2 dr d\theta$$

I think you can go on now. :)

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  • $\begingroup$ But what about the limits of integration? $x$ goes from $0$ to $1$ and so does $y$, and I couldn't find a appropriate limits. $\endgroup$ – Vinícius Lopes Simões Dec 7 '15 at 19:24
  • $\begingroup$ Yes, the limit are interesting! :) Shall I write the whole answer? :) $\endgroup$ – Hosein Rahnama Dec 7 '15 at 19:25
  • $\begingroup$ That would be helpful :) $\endgroup$ – Vinícius Lopes Simões Dec 7 '15 at 19:27
  • $\begingroup$ @ViníciusLopesSimões: Take a look at the new answer. :) $\endgroup$ – Hosein Rahnama Dec 7 '15 at 19:41
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    $\begingroup$ Note that this change of variable amounts to first translating to place the center of the circle at the origin and then converting to polar. $\endgroup$ – amd Dec 7 '15 at 22:11

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