Find the volume of the solid obtained by rotating the region bounded by the curves
y^2=x
 and x=2y
about the line y=0
.
Volume = 

To find the volume of the solid obtained by rotating the region bounded by the curves \( y^2 = x \) and \( x = 2y \) about the line \( y = 0 \) (which is the x-axis), we'll use the method of disks (or washers).

### Step 1: Find the points of intersection
We first need to determine the points of intersection between the curves \( y^2 = x \) and \( x = 2y \).

- For \( y^2 = x \):
  \[
  x = y^2
  \]
- For \( x = 2y \):
  \[
  y^2 = 2y
  \]

Solving the equation \( y^2 = 2y \):
\[
y^2 - 2y = 0
\]
\[
y(y - 2) = 0
\]

So, \( y = 0 \) or \( y = 2 \). 

For \( y = 0 \), \( x = 0 \) (intersection at \((0,0)\)).
For \( y = 2 \), \( x = 4 \) (intersection at \((4,2)\)).

### Step 2: Express \( x \) as a function of \( y \)
- The curve \( y^2 = x \) can be written as \( x = y^2 \).
- The line \( x = 2y \) can be rewritten as \( y = \frac{x}{2} \), which is already in terms of \( y \).

### Step 3: Set up the volume integral
The volume of the solid is given by:
\[
V = \pi \int_{y_1}^{y_2} \left[ R(y)^2 - r(y)^2 \right] dy
\]
where:
- \( R(y) \) is the outer radius (from the axis of rotation to the outer curve).
- \( r(y) \) is the inner radius (from the axis of rotation to the inner curve).
- \( y_1 \) and \( y_2 \) are the limits of integration.

Here, \( R(y) = 2y \) and \( r(y) = y^2 \).

Thus, the volume integral becomes:
\[
V = \pi \int_{0}^{2} \left[ (2y)^2 - (y^2)^2 \right] dy
\]

### Step 4: Compute the integral
\[
V = \pi \int_{0}^{2} \left[ 4y^2 - y^4 \right] dy
\]

Now, integrate term by term:
\[
V = \pi \left[ \int_{0}^{2} 4y^2 \, dy - \int_{0}^{2} y^4 \, dy \right]
\]

- The first integral:
  \[
  \int 4y^2 \, dy = \frac{4y^3}{3} \bigg|_0^2 = \frac{4(2)^3}{3} - \frac{4(0)^3}{3} = \frac{32}{3}
  \]

- The second integral:
  \[
  \int y^4 \, dy = \frac{y^5}{5} \bigg|_0^2 = \frac{(2)^5}{5} - \frac{(0)^5}{5} = \frac{32}{5}
  \]

### Step 5: Calculate the volume
\[
V = \pi \left[ \frac{32}{3} - \frac{32}{5} \right]
\]
To subtract these fractions, find a common denominator:
\[
\frac{32}{3} - \frac{32}{5} = \frac{32 \times 5}{15} - \frac{32 \times 3}{15} = \frac{160 - 96}{15} = \frac{64}{15}
\]

Thus, the volume is:
\[
V = \frac{64\pi}{15}
\]

### Final Answer:
\[
\text{Volume} = \frac{64\pi}{15}
\]

Would you like further details on any part of this solution or have any additional questions?

---

Here are 5 related questions for further exploration:
1. What is the volume if the region was rotated about the line \( y = 1 \) instead of \( y = 0 \)?
2. How would the method change if we used cylindrical shells instead of the disk method?
3. What is the area of the region bounded by \( y^2 = x \) and \( x = 2y \)?
4. How would the problem change if we rotated the region about the line \( x = 0 \) (the y-axis)?
5. What if the region was rotated about the line \( y = 3 \)? How would the volume change?

**Tip:** Always check the symmetry of the region and the axis of rotation; it can sometimes simplify the integral setup.

Learn how to find the volume of a solid obtained by rotating the region bounded by the curves y^2 = x and x = 2y about the line y = 0 (x-axis). Detailed step-by-step solution using the method of disks/washers and integration. Suitable for advanced high school students and calculus enthusiasts.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation