The region between the curve $$ y=\sqrt{x}, 0 \leq x \leq 9 $$ and the $$ y $$-axis is revolved about the $$ y $$-axis to generate a solid. Find the exact value of its volume. $$ [?] \pi $$

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To find the volume of the solid generated by revolving the region between the curve $$ y = \sqrt{x} $$ and the $$ y $$-axis around the $$ y $$-axis, we can use the method of cylindrical shells or the disk method. Here, we'll use the disk method.

1. **Identify the bounds**: The curve $$ y = \sqrt{x} $$ is defined for $$ 0 \leq x \leq 9 $$. This means $$ y $$ ranges from $$ 0 $$ to $$ 3 $$ (since $$ \sqrt{9} = 3 $$).

2. **Express $$ x $$ in terms of $$ y $$**: From the equation $$ y = \sqrt{x} $$, we can express $$ x $$ as:
   $$
   x = y^2
   $$

3. **Set up the volume integral**: The volume $$ V $$ of the solid of revolution can be calculated using the formula:
   $$
   V = \pi \int_{a}^{b} [f(y)]^2 \, dy
   $$
   where $$ f(y) $$ is the function expressed in terms of $$ y $$. Here, $$ f(y) = y^2 $$, and the bounds $$ a = 0 $$ and $$ b = 3 $$.

4. **Write the integral**:
   $$
   V = \pi \int_{0}^{3} (y^2)^2 \, dy = \pi \int_{0}^{3} y^4 \, dy
   $$

5. **Calculate the integral**:
   $$
   \int y^4 \, dy = \frac{y^5}{5}
   $$
   Evaluating from $$ 0 $$ to $$ 3 $$:
   $$
   \left[ \frac{y^5}{5} \right]_{0}^{3} = \frac{3^5}{5} - \frac{0^5}{5} = \frac{243}{5}
   $$

6. **Multiply by $$ \pi $$**:
   $$
   V = \pi \cdot \frac{243}{5} = \frac{243\pi}{5}
   $$

Thus, the exact value of the volume is $$ \frac{243}{5} \pi $$.
The volume of the solid is $$ \frac{243}{5} \pi $$.

The region between the curve

and the -axis is revolved about
the -axis to generate a solid.
Find the exact value of its volume.

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The region between the curve and the -axis is revolved about the -axis to generate a solid. Find the exact value of its volume.