1. **Problem:** Find the volume of the solid generated by revolving the region bounded by $y = x^2$, $x=1$ to $x=2$, and $y=0$ about the y-axis using the method of cylindrical shells. 2. **Formula:** The volume $V$ using cylindrical shells when revolving around the y-axis is given by: $$V = 2\pi \int_a^b (\text{radius})(\text{height}) \, dx$$ where: - Radius = distance from the y-axis = $x$ - Height = function value = $y = x^2$ - Limits of integration: $x=1$ to $x=2$ 3. **Set up the integral:** $$V = 2\pi \int_1^2 x \cdot x^2 \, dx = 2\pi \int_1^2 x^3 \, dx$$ 4. **Evaluate the integral:** $$\int_1^2 x^3 \, dx = \left[ \frac{x^4}{4} \right]_1^2 = \frac{2^4}{4} - \frac{1^4}{4} = \frac{16}{4} - \frac{1}{4} = \frac{15}{4}$$ 5. **Calculate the volume:** $$V = 2\pi \times \frac{15}{4} = \frac{30\pi}{4} = \frac{15\pi}{2}$$ **Final answer:** $$\boxed{\frac{15\pi}{2}}$$ This is the volume of the solid generated by revolving the given region about the y-axis.