1. **Problem Statement:** Calculate the double integral $$\iint \frac{dx\,dy}{xu}$$ over the region bounded by the four circles given by the equations: $$x^2 + y^2 = 2x,$$ $$x^2 + y^2 = 4x,$$ $$x^2 + y^2 = 2y,$$ $$x^2 + y^2 = 4y.$$ 2. **Rewrite the circle equations:** Complete the square for each circle to find their centers and radii. - For $$x^2 + y^2 = 2x$$ rewrite as $$x^2 - 2x + y^2 = 0$$ or $$(x-1)^2 + y^2 = 1^2$$ (circle centered at $(1,0)$ radius $1$). - For $$x^2 + y^2 = 4x$$ rewrite as $$(x-2)^2 + y^2 = 2^2$$ (center $(2,0)$ radius $2$). - For $$x^2 + y^2 = 2y$$ rewrite as $$x^2 + y^2 - 2y = 0$$ or $$x^2 + (y-1)^2 = 1^2$$ (center $(0,1)$ radius $1$). - For $$x^2 + y^2 = 4y$$ rewrite as $$x^2 + (y-2)^2 = 2^2$$ (center $(0,2)$ radius $2$). 3. **Region of integration:** The region is bounded between the smaller circles $(x-1)^2 + y^2 = 1$ and $x^2 + (y-1)^2 = 1$ and the larger circles $(x-2)^2 + y^2 = 4$ and $x^2 + (y-2)^2 = 4$. 4. **Change of variables:** Set $$u = x^2 + y^2$$ and $$v = x - y$$ or consider polar coordinates since the circles are centered on axes. 5. **Polar coordinates substitution:** Let $$x = r\cos\theta$$ and $$y = r\sin\theta$$. Then $$x^2 + y^2 = r^2$$. Rewrite the circles: - $$r^2 = 2r\cos\theta \Rightarrow r = 0 \text{ or } r = 2\cos\theta$$ - $$r^2 = 4r\cos\theta \Rightarrow r = 0 \text{ or } r = 4\cos\theta$$ - $$r^2 = 2r\sin\theta \Rightarrow r = 0 \text{ or } r = 2\sin\theta$$ - $$r^2 = 4r\sin\theta \Rightarrow r = 0 \text{ or } r = 4\sin\theta$$ 6. **Bounds for $r$ and $\theta$:** The region is bounded between the circles with radii $2\cos\theta$ and $4\cos\theta$ in the $x$-direction and $2\sin\theta$ and $4\sin\theta$ in the $y$-direction. 7. **Integral in polar coordinates:** The integral becomes $$\iint \frac{dx\,dy}{xu} = \int_{\theta=0}^{\pi/2} \int_{r=2\max(\cos\theta,\sin\theta)}^{4\min(\cos\theta,\sin\theta)} \frac{r\,dr\,d\theta}{r\cos\theta \cdot r^2} = \int_0^{\pi/2} \int_{r_1}^{r_2} \frac{r}{r^3 \cos\theta} dr d\theta = \int_0^{\pi/2} \frac{1}{\cos\theta} \int_{r_1}^{r_2} \frac{1}{r^2} dr d\theta,$$ where $r_1$ and $r_2$ are the inner and outer radii depending on $\theta$. 8. **Evaluate the inner integral:** $$\int_{r_1}^{r_2} \frac{1}{r^2} dr = \left[-\frac{1}{r}\right]_{r_1}^{r_2} = \frac{1}{r_1} - \frac{1}{r_2}.$$ 9. **Final integral:** $$\int_0^{\pi/2} \frac{1}{\cos\theta} \left( \frac{1}{r_1} - \frac{1}{r_2} \right) d\theta.$$ Substitute $r_1$ and $r_2$ accordingly and evaluate numerically or analytically. **Final answer:** The integral reduces to evaluating $$\int_0^{\pi/2} \frac{1}{\cos\theta} \left( \frac{1}{2\max(\cos\theta,\sin\theta)} - \frac{1}{4\min(\cos\theta,\sin\theta)} \right) d\theta.$$