use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by y=sqrt(x-12) and x=16 and y=0 about the line y=-4. round to 4 decimal places


To find the volume of the solid obtained by rotating the region bounded by \( y = \sqrt{x - 12} \), \( x = 16 \), and \( y = 0 \) about the line \( y = -4 \) using the method of cylindrical shells, we follow these steps:

### Step 1: Set up the shell height
The given curve is \( y = \sqrt{x - 12} \). Solving for \( x \) in terms of \( y \), we have:

\[
x = y^2 + 12
\]

The region is bounded by \( x = 16 \) and \( y = 0 \). Thus, the bounds for \( y \) are from 0 to \( \sqrt{16 - 12} = \sqrt{4} = 2 \).

### Step 2: Determine the radius of the shell
Since we are rotating about the line \( y = -4 \), the radius of a shell at a height \( y \) is the distance from \( y \) to \( y = -4 \), which is:

\[
\text{Radius} = y - (-4) = y + 4
\]

### Step 3: Determine the height of the shell
The height of the shell is given by the horizontal distance between the curve \( x = y^2 + 12 \) and the vertical line \( x = 16 \). So, the height is:

\[
\text{Height} = 16 - (y^2 + 12) = 16 - y^2 - 12 = 4 - y^2
\]

### Step 4: Set up the volume integral
Using the method of cylindrical shells, the volume \( V \) is given by the integral:

\[
V = 2\pi \int_{0}^{2} (\text{Radius}) \times (\text{Height}) \, dy
\]

Substitute the expressions for the radius and the height:

\[
V = 2\pi \int_{0}^{2} (y + 4)(4 - y^2) \, dy
\]

### Step 5: Expand the integrand and solve
First, expand the product:

\[
(y + 4)(4 - y^2) = y(4 - y^2) + 4(4 - y^2)
= 4y - y^3 + 16 - 4y^2
\]

Thus, the volume integral becomes:

\[
V = 2\pi \int_{0}^{2} (4y - y^3 + 16 - 4y^2) \, dy
\]

Now integrate each term:

\[
V = 2\pi \left[ \int_{0}^{2} 4y \, dy - \int_{0}^{2} y^3 \, dy + \int_{0}^{2} 16 \, dy - \int_{0}^{2} 4y^2 \, dy \right]
\]

\[
V = 2\pi \left[ \left( 2y^2 \right)_{0}^{2} - \left( \frac{y^4}{4} \right)_{0}^{2} + \left( 16y \right)_{0}^{2} - \left( \frac{4y^3}{3} \right)_{0}^{2} \right]
\]

Now, calculate each part:

- \( \left( 2y^2 \right)_{0}^{2} = 2(2^2) - 2(0^2) = 8 \)
- \( \left( \frac{y^4}{4} \right)_{0}^{2} = \frac{2^4}{4} - \frac{0^4}{4} = 4 \)
- \( \left( 16y \right)_{0}^{2} = 16(2) - 16(0) = 32 \)
- \( \left( \frac{4y^3}{3} \right)_{0}^{2} = \frac{4(2^3)}{3} - \frac{4(0^3)}{3} = \frac{32}{3} \)

So, the volume becomes:

\[
V = 2\pi \left[ 8 - 4 + 32 - \frac{32}{3} \right]
\]

Simplify the expression:

\[
V = 2\pi \left[ 36 - \frac{32}{3} \right] = 2\pi \left[ \frac{108}{3} - \frac{32}{3} \right] = 2\pi \times \frac{76}{3}
\]

\[
V = \frac{152\pi}{3}
\]

### Step 6: Approximate the result
Using \( \pi \approx 3.1416 \), the volume is:

\[
V \approx \frac{152 \times 3.1416}{3} = 159.2388
\]

### Final Answer:
The volume of the solid is approximately \( 159.2388 \) cubic units.

Would you like to ask for any clarification or need further details?

Here are 5 related questions:
1. How does the method of cylindrical shells differ from the disk method?
2. What happens if the axis of rotation is a different line, such as \( y = 4 \)?
3. How do we handle cases where the region is bounded by multiple curves?
4. Can we use the method of cylindrical shells when rotating around a vertical line instead of a horizontal one?
5. How does the volume change if the boundary \( x = 16 \) is replaced with another value?

**Tip**: When solving problems with cylindrical shells, always focus on how the radius and height relate to the axis of rotation.

Use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by y=sqrt(x-12) and x=16 and y=0 about the line y=-4. Round to 4 decimal places.

Learn how to calculate the volume of a solid obtained by rotating the region bounded by y=sqrt(x-12), x=16, and y=0 around y=-4 using the method of cylindrical shells. This problem involves integration and is suitable for students in Calculus I.

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