I would like to find the volume of a region bound by the lines 
y = $$ \sqrt{1-x} $$
x = 0
y = 0
revolving around the line y=3. This is the correct set up to the equation, now solve please

To find the volume of the region bounded by the curves $ y = \sqrt{1-x} $, $ x = 0 $, and $ y = 0 $ when revolved around the line $ y = 3 $, we can use the method of disks or washers.

### Step 1: Set up the Integral

The region is revolved around the line $ y = 3 $. For the washer method, we first determine the outer and inner radii of the washers.

- **Outer radius $ R(y) $**: The distance from the line $ y = 3 $ to the curve $ y = 0 $. Since $ y = 0 $ is the x-axis, the outer radius is $ R(y) = 3 - 0 = 3 $.

- **Inner radius $ r(y) $**: The distance from the line $ y = 3 $ to the curve $ y = \sqrt{1-x} $. We first need to express $ x $ as a function of $ y $:
  \[
  y = \sqrt{1-x} \implies y^2 = 1 - x \implies x = 1 - y^2
  \]
  Thus, the inner radius is:
  \[
  r(y) = 3 - y
  \]

### Step 2: Write the Volume Integral

The volume $ V $ of the solid of revolution can be calculated as:

\[
V = \pi \int_0^1 \left[ R(y)^2 - r(y)^2 \right] \, dy
\]

Substituting the expressions for $ R(y) $ and $ r(y) $:

\[
V = \pi \int_0^1 \left[ 3^2 - (3 - y)^2 \right] \, dy
\]

Expanding and simplifying:

\[
V = \pi \int_0^1 \left[ 9 - (9 - 6y + y^2) \right] \, dy
\]

\[
V = \pi \int_0^1 \left[ 9 - 9 + 6y - y^2 \right] \, dy
\]

\[
V = \pi \int_0^1 \left[ 6y - y^2 \right] \, dy
\]

### Step 3: Evaluate the Integral

Now, evaluate the integral:

\[
V = \pi \left[ \int_0^1 6y \, dy - \int_0^1 y^2 \, dy \right]
\]

Integrating each term separately:

\[
\int_0^1 6y \, dy = 6 \left[ \frac{y^2}{2} \right]_0^1 = 6 \left( \frac{1^2}{2} - \frac{0^2}{2} \right) = 6 \times \frac{1}{2} = 3
\]

\[
\int_0^1 y^2 \, dy = \left[ \frac{y^3}{3} \right]_0^1 = \frac{1^3}{3} - \frac{0^3}{3} = \frac{1}{3}
\]

Thus, the volume $ V $ is:

\[
V = \pi \left( 3 - \frac{1}{3} \right) = \pi \times \frac{9}{3} = \frac{8\pi}{3}
\]

### Final Answer:

The volume of the solid formed by revolving the region bounded by $ y = \sqrt{1-x} $, $ x = 0 $, and $ y = 0 $ around the line $ y = 3 $ is:

\[
\boxed{\frac{8\pi}{3}}
\]

Would you like further details or have any questions? Here are some related questions you might find interesting:

1. How would the volume change if the region was revolved around a different line, such as $ y = 2 $?
2. What if the region was revolved around the x-axis instead of $ y = 3 $?
3. How can we find the surface area of the solid formed in this problem?
4. What if the region was defined by a different curve, say $ y = 1-x $?
5. How can the disk method be used instead of the washer method for similar problems?

**Tip:** Always visualize the problem with a sketch when setting up integrals involving solids of revolution. This helps to correctly identify the radii and limits.

Learn how to find the volume of a solid formed by revolving the region bounded by y = sqrt(1-x), x = 0, and y = 0 around the line y = 3. Detailed step-by-step solution using integral calculus and solid of revolution concepts.

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