To solve the problem of finding the volume of the solid of revolution, we use the disk method, since the solid is generated by revolving the region under the curve $y = \sqrt{x}$ about the $x$-axis from $x = 0$ to $x = 1$.

Step-by-step solution:

1. Volume formula for the disk method: The formula for the volume of a solid of revolution about the $x$-axis is: $V = \pi \int_a^b [f(x)]^2 \, dx$ Here:

   • $f(x) = \sqrt{x}$
   • The bounds are $a = 0$ and $b = 1$.

2. Set up the integral: Substituting $f(x) = \sqrt{x}$ into the formula, we get: $V = \pi \int_0^1 (\sqrt{x})^2 \, dx$

3. Simplify the integrand: Since $(\sqrt{x})^2 = x$, the integral becomes: $V = \pi \int_0^1 x \, dx$

4. Evaluate the integral: The integral of $x$ is: $\int x \, dx = \frac{x^2}{2}$ Applying the bounds $0$ to $1$, we compute: $\left[ \frac{x^2}{2} \right]_0^1 = \frac{1^2}{2} - \frac{0^2}{2} = \frac{1}{2}$

5. Compute the volume: Multiply the result by $\pi$: $V = \pi \cdot \frac{1}{2} = \frac{\pi}{2}$

Final Answer:

The volume of the solid of revolution is: $\boxed{\frac{\pi}{2}}$

Would you like a detailed explanation of any step? 😊

Related Questions:

1. What is the difference between the disk and washer methods in solids of revolution?
2. How would the integral change if the region was rotated about the line $y = 1$ instead of the $x$-axis?
3. Can you compute the volume if the function is $y = x^2$ instead of $y = \sqrt{x}$?
4. What if the bounds were changed from $x = 0$ to $x = 2$? How would the volume calculation change?
5. How do we apply the shell method for revolving the same region?

Tip:

When using the disk method, always remember to square the function $f(x)$ before integrating to account for the area of the circular cross-sections.