find values of solid generated by rotating regions bounded by y = x^1/2 and line y =1, x = 4 about line y =1

To find the volume of the solid generated by rotating the region bounded by the curve \( y = \sqrt{x} \), the line \( y = 1 \), and \( x = 4 \), about the line \( y = 1 \), we will use the method of **cylindrical shells** or **washer method**.

### Step-by-step approach using the Washer Method:
We rotate the region about the line \( y = 1 \). First, let's describe the boundaries of the region:

1. **Curve:** \( y = \sqrt{x} \), so \( x = y^2 \).
2. **Line:** \( y = 1 \).
3. **Vertical boundary:** \( x = 4 \), which gives \( y = \sqrt{4} = 2 \).

Thus, the region is bounded between \( y = 0 \) and \( y = 2 \).

Now, when we rotate the region about the line \( y = 1 \), we will calculate the volume using the washer method. The formula for the volume is:

\[
V = \pi \int_{a}^{b} \left( R_{\text{outer}}^2 - R_{\text{inner}}^2 \right) \, dy
\]

Where:

- \( R_{\text{outer}} \) is the distance from the line of rotation to the curve \( y = 1 \),
- \( R_{\text{inner}} \) is the distance from the line \( y = 1 \) to the curve \( y = \sqrt{x} \).

For this problem:
- The **outer radius** is from \( y = 1 \) to \( y = 0 \), which is \( 1 \).
- The **inner radius** is \( 1 - y \), because the distance between \( y = 1 \) and the curve \( y = \sqrt{x} \).

The limits of integration are from \( y = 0 \) to \( y = 2 \). Therefore, the volume is:

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

### Simplify the integrand:
\[
1^2 = 1
\]
\[
(1 - y)^2 = 1 - 2y + y^2
\]
Thus, the integrand becomes:
\[
1 - (1 - 2y + y^2) = 2y - y^2
\]

### Set up the integral:
\[
V = \pi \int_0^2 (2y - y^2) \, dy
\]

### Solve the integral:
\[
V = \pi \left[ y^2 - \frac{y^3}{3} \right]_0^2
\]
Evaluating this at the limits:
\[
V = \pi \left[ \left( 2^2 - \frac{2^3}{3} \right) - \left( 0^2 - \frac{0^3}{3} \right) \right]
\]
\[
V = \pi \left( 4 - \frac{8}{3} \right)
\]
\[
V = \pi \left( \frac{12}{3} - \frac{8}{3} \right)
\]
\[
V = \pi \times \frac{4}{3}
\]

Thus, the volume of the solid is:

\[
V = \frac{4\pi}{3}
\]

### Do you need further clarification or a breakdown of any step?

Here are 5 related questions:
1. How does the Washer method differ from the Shell method for finding volume?
2. How would the volume change if the curve were rotated around the x-axis instead of \( y = 1 \)?
3. What happens if the region is bounded by a different upper line, say \( y = 2 \)?
4. How would we handle the integral if the curve were more complex, such as \( y = \sin(x) \)?
5. How does the method change if the axis of rotation is \( y = -1 \)?

**Tip:** Always identify the radii (inner and outer) carefully when using the washer method; this ensures you subtract the correct areas.

Learn how to calculate the volume of a solid generated by rotating the region bounded by y = sqrt(x), y = 1, and x = 4 about y = 1 using the washer method. Detailed step-by-step solution with integrals and formulas provided.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation