Problem:
Find the volume of the solid generated by revolving the region bounded by \( y = \sqrt{x} \), \( x=0 \), and \( y=1 \) about the \( x \)-axis.
Solution:
Step 1: Identify the limits of integration
\( y=1 \implies \sqrt{x} = 1 \implies x=1 \)
Step 2: Write the volume integral using the disk method
\[ V = \pi \int_0^1 \left( \sqrt{x} \right)^2 dx = \pi \int_0^1 x \, dx \]Step 3: Compute the integral
\[ V = \pi \left[ \frac{x^2}{2} \right]_0^1 = \pi \left( \frac{1^2}{2} - \frac{0^2}{2} \right) \]Step 4: Simplify the result
\[ V = \frac{\pi}{2} \]Final Answer:
\[ \boxed{ V = \frac{\pi}{2} } \]