# Double integral - Change of Variables

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 $D = \{(x,y):x^2+y^2 \le 1, y>0\}$

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

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

$$I=\int_{0}^{\pi} \int_{0}^{-2 \cos \theta} r \cdot r dr d\theta =\int_{0}^{\pi} \int_{0}^{-2 \cos \theta} r^2 dr d\theta$$
• 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. – Vinícius Lopes Simões Dec 7 '15 at 19:24