Calculate the volume of the solid formed by rotating the area between \( y = \sqrt{x} \) and \( y = 0 \) from \( x = 0 \) to \( x = 1 \) about the line \( y = -1 \) utilizing the washer method.

To calculate the volume of the solid formed by rotating the area between \( y = \sqrt{x} \) and \( y = 0 \) from \( x = 0 \) to \( x = 1 \) about the line \( y = -1 \) using the washer method, we start by identifying the outer and inner radii of our washers. 1. The function \( y = \sqrt{x} \) is the upper curve, and the line \( y = 0 \) is the lower curve. When we rotate around the line \( y = -1 \), we need to find the distance from the line \( y = -1 \) to both the top curve and the bottom curve. 2. The outer radius \( R(x) \) (from \( y = -1 \) to \( y = \sqrt{x} \)) is: \[ R(x) = \sqrt{x} - (-1) = \sqrt{x} + 1. \] 3. The inner radius \( r(x) \) (from \( y = -1 \) to \( y = 0 \)) is: \[ r(x) = 0 - (-1) = 1. \] 4. The volume \( V \) of the solid of revolution can be formulated using the washer method with the following integral from \( x = 0 \) to \( x = 1 \): \[ V = \pi \int_{0}^{1} \left[ R(x)^2 - r(x)^2 \right] \, dx. \] 5. Substituting the outer and inner radii, we have: \[ V = \pi \int_{0}^{1} \left[ (\sqrt{x} + 1)^2 - 1^2 \right] \, dx. \] 6. Expanding the integrand: \[ (\sqrt{x} + 1)^2 = x + 2\sqrt{x} + 1, \] so: \[ (\sqrt{x} + 1)^2 - 1 = x + 2\sqrt{x} + 1 - 1 = x + 2\sqrt{x}. \] 7. Thus, our volume integral becomes: \[ V = \pi \int_{0}^{1} (x + 2\sqrt{x}) \, dx. \] 8. Now we compute the integral: \[ \int (x + 2\sqrt{x}) \, dx = \int x \, dx + 2\int \sqrt{x} \, dx = \frac{x^2}{2} + 2 \cdot \frac{2}{3} x^{3/2} = \frac{x^2}{2} + \frac{4}{3} x^{3/2}. \] 9. Evaluating from 0 to 1 gives: \[ \left[ \frac{x^2}{2} + \frac{4}{3} x^{3/2} \right]_{0}^{1} = \left( \frac{1^2}{2} + \frac{4}{3} \cdot 1^{3/2} \right) - \left( 0 + 0 \right) = \frac{1}{2} + \frac{4}{3}. \] 10. To combine these fractions, we find a common denominator: \[ \frac{1}{2} = \frac{3}{6}, \quad \frac{4}{3} = \frac{8}{6} \Rightarrow \frac{3}{6} + \frac{8}{6} = \frac{11}{6}. \] 11. Therefore, the volume \( V \) is: \[ V = \pi \cdot \frac{11}{6} = \frac{11\pi}{6}. \] Thus, the volume of the solid formed by rotating the area between \( y = \sqrt{x} \) and \( y = 0 \) from \( x = 0 \) to \( x = 1 \) about the line \( y = -1 \) is: \[ \boxed{\frac{11\pi}{6}}. \]