To find the volume of the solid formed by revolving the region bounded by $y = \sqrt{x}$, $x = 1$, and $x = 4$ about the $y$-axis using the disk method, we proceed as follows:

1. Rewrite the equation $y = \sqrt{x}$ in terms of $x$: $x = y^2$

2. Identify the bounds for $y$: Since $x = 1$ and $x = 4$, we need to find the corresponding $y$-values.

   • When $x = 1$: $y = \sqrt{1} = 1$.
   • When $x = 4$: $y = \sqrt{4} = 2$.

   Thus, the region we are revolving is between $y = 1$ and $y = 2$.

3. Set up the volume integral using the disk method: When revolving around the $y$-axis, the radius of a disk at height $y$ is the distance from the $y$-axis to the curve, which is $x = y^2$.

   The area of each circular disk is: $A(y) = \pi \cdot (\text{radius})^2 = \pi \cdot (y^2)^2 = \pi y^4$

4. Integrate to find the volume: $V = \int_{y=1}^{y=2} A(y) \, dy = \int_{1}^{2} \pi y^4 \, dy$ Factor out $\pi$: $V = \pi \int_{1}^{2} y^4 \, dy$

5. Evaluate the integral: $V = \pi \left[ \frac{y^5}{5} \right]_{1}^{2}$ Substitute the bounds: $V = \pi \left( \frac{2^5}{5} - \frac{1^5}{5} \right) = \pi \left( \frac{32}{5} - \frac{1}{5} \right)$ Simplify: $V = \pi \cdot \frac{31}{5} = \frac{31\pi}{5}$

Thus, the volume of the solid is: $V = \frac{31\pi}{5}$

Would you like a more detailed breakdown, or do you have further questions?

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1. What if the region were revolved around the $x$-axis instead?
2. How would the method change if we used the shell method for this problem?
3. What would be the result if the bounds changed to $x = 0$ and $x = 4$?
4. How does the disk method compare to the washer method in setup?
5. What happens to the volume if we change the function to $y = x^2$?

Tip: When setting up volume integrals, always check if rewriting the function in terms of the variable of integration simplifies the problem.