Questions: x=1+y^2, x=0, y=1, y=2 x=sqrt(y), x=0, y=1 y=x^3, y=8, x=0

Solution Steps

The region bounded by the curves \(x = 1 + y^2\), \(x = 0\), \(y = 1\), and \(y = 2\) is identified. This region will be rotated about the \(x\)-axis to form a solid.

Using the method of cylindrical shells, the volume \(V\) of the solid is given by the integral: \[ V = \int_{1}^{2} 2\pi \cdot \text{radius} \cdot \text{height} \, dy \] Here, the radius is \(y\) and the height is \(1 + y^2\). Thus, the integral becomes: \[ V = \int_{1}^{2} 2\pi \cdot y \cdot (1 + y^2) \, dy \]

The integral can be simplified and evaluated: \[ V = 2\pi \int_{1}^{2} (y + y^3) \, dy \] Calculating this integral yields: \[ V = 2\pi \left[ \frac{y^2}{2} + \frac{y^4}{4} \right]_{1}^{2} \] Evaluating the definite integral results in: \[ V = 2\pi \left( \left( \frac{2^2}{2} + \frac{2^4}{4} \right) - \left( \frac{1^2}{2} + \frac{1^4}{4} \right) \right) \] This simplifies to: \[ V = 2\pi \left( 2 + 4 - \frac{1}{2} - \frac{1}{4} \right) = 2\pi \left( 6 - \frac{3}{4} \right) = 2\pi \cdot \frac{21}{4} \] Thus, the final volume is: \[ V = \frac{21\pi}{2} \]

Final Answer