Determine the volume generated by revolving the area between $$ y = \sqrt{x} $$ and the x-axis from $$ x = 0 $$ to $$ x = 4 $$ about the y-axis with the washer method.

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To determine the volume generated by revolving the area between $$ y = \sqrt{x} $$ and the x-axis from $$ x = 0 $$ to $$ x = 4 $$ about the y-axis using the washer method, we will follow these steps:

1. **Identify the function and the limits of integration**: 
   - The function is $$ y = \sqrt{x} $$.
   - The limits of integration are from $$ x = 0 $$ to $$ x = 4 $$.

2. **Express $$ x $$ in terms of $$ y $$**: 
   - Since we are revolving around the y-axis, we need to express $$ x $$ as a function of $$ y $$. From $$ y = \sqrt{x} $$, we can square both sides to get:
     $$
     x = y^2
     $$

3. **Determine the limits for $$ y $$**: 
   - When $$ x = 0 $$, $$ y = \sqrt{0} = 0 $$.
   - When $$ x = 4 $$, $$ y = \sqrt{4} = 2 $$.
   - Therefore, the limits for $$ y $$ are from $$ y = 0 $$ to $$ y = 2 $$.

4. **Set up the volume integral using the washer method**: 
   - The volume $$ V $$ of the solid of revolution is given by the integral:
     $$
     V = \pi \int_{a}^{b} [R(y)^2 - r(y)^2] \, dy
     $$
   - In this case, since we are revolving around the y-axis and there is no inner radius (the area is bounded by the x-axis), we have:
     - Outer radius $$ R(y) = y^2 $$
     - Inner radius $$ r(y) = 0 $$
   - Thus, the volume integral simplifies to:
     $$
     V = \pi \int_{0}^{2} (y^2)^2 \, dy = \pi \int_{0}^{2} y^4 \, dy
     $$

5. **Calculate the integral**:
   - We will now compute the integral:
     $$
     V = \pi \int_{0}^{2} y^4 \, dy
     $$

Now, let's calculate the integral.

$$
\int y^4 \, dy = \frac{y^5}{5}
$$

Evaluating this from $$ 0 $$ to $$ 2 $$:

$$
\left[ \frac{y^5}{5} \right]_{0}^{2} = \frac{2^5}{5} - \frac{0^5}{5} = \frac{32}{5} - 0 = \frac{32}{5}
$$

Thus, the volume $$ V $$ is:

$$
V = \pi \cdot \frac{32}{5} = \frac{32\pi}{5}
$$

Therefore, the volume generated by revolving the area between $$ y = \sqrt{x} $$ and the x-axis from $$ x = 0 $$ to $$ x = 4 $$ about the y-axis is

$$
\boxed{\frac{32\pi}{5}}.
$$
The volume is $$ \frac{32\pi}{5} $$.

Determine the volume generated by revolving the area between and the x-axis from to about the y-axis with the washer method.

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