1. **Problem statement:** Find the volume of the solid obtained by rotating the area between the curves defined by the equations $$y^2 + x^2 = 4$$ and $$x^2 - 4x + y^2 = -3$$ around the x-axis for $$1 \leq x \leq 2$$. 2. **Rewrite the equations:** - First circle: $$y^2 + x^2 = 4$$ - Second circle: $$x^2 - 4x + y^2 = -3$$ can be rewritten by completing the square for $$x$$: $$x^2 - 4x + y^2 = -3 \implies (x^2 - 4x + 4) + y^2 = -3 + 4$$ $$\implies (x - 2)^2 + y^2 = 1$$ 3. **Interpretation:** - The first circle has center at $$(0,0)$$ and radius $$2$$. - The second circle has center at $$(2,0)$$ and radius $$1$$. 4. **Find the volume of the solid of revolution:** - The volume when rotating around the x-axis is given by the formula: $$V = \pi \int_a^b \left(R_{outer}^2 - R_{inner}^2\right) dx$$ - Here, $$R_{outer}$$ and $$R_{inner}$$ are the distances from the x-axis to the outer and inner curves respectively. 5. **Determine the radii:** - For each $$x$$ in $$[1,2]$$, the upper and lower boundaries of the region are given by the positive and negative values of $$y$$ from each circle. - Since both circles are symmetric about the x-axis, the vertical distance from the x-axis to the curve is $$|y| = \sqrt{r^2 - (x - h)^2}$$ where $$r$$ is radius and $$h$$ is x-coordinate of center. - For the first circle: $$R_1 = \sqrt{4 - x^2}$$ - For the second circle: $$R_2 = \sqrt{1 - (x - 2)^2}$$ 6. **Identify which circle is outer and which is inner in the interval $$[1,2]$$:** - For $$x=1$$: $$R_1 = \sqrt{4 - 1} = \sqrt{3} \approx 1.732$$ $$R_2 = \sqrt{1 - (1 - 2)^2} = \sqrt{1 - 1} = 0$$ - For $$x=2$$: $$R_1 = \sqrt{4 - 4} = 0$$ $$R_2 = \sqrt{1 - (2 - 2)^2} = \sqrt{1} = 1$$ - So between $$x=1$$ and $$x=2$$, the first circle radius decreases from about 1.732 to 0, and the second circle radius increases from 0 to 1. - The region between the two circles is the area inside the first circle but outside the second circle. 7. **Volume integral:** $$V = \pi \int_1^2 \left( (\sqrt{4 - x^2})^2 - (\sqrt{1 - (x - 2)^2})^2 \right) dx = \pi \int_1^2 \left(4 - x^2 - [1 - (x - 2)^2]\right) dx$$ 8. **Simplify the integrand:** $$4 - x^2 - 1 + (x - 2)^2 = 3 - x^2 + (x^2 - 4x + 4) = 3 - x^2 + x^2 - 4x + 4 = 7 - 4x$$ 9. **Set up the integral:** $$V = \pi \int_1^2 (7 - 4x) dx$$ 10. **Integrate:** $$\int_1^2 (7 - 4x) dx = \left[7x - 2x^2\right]_1^2 = (7 \times 2 - 2 \times 2^2) - (7 \times 1 - 2 \times 1^2) = (14 - 8) - (7 - 2) = 6 - 5 = 1$$ 11. **Calculate volume:** $$V = \pi \times 1 = \pi$$ **Final answer:** $$\boxed{\pi}$$