Find the volume of the solid obtained by revolving the region enclosed by the curves y=x^2 and y=2x about the x-axis

The problem asks to find the volume of the solid obtained by revolving the region enclosed by the curves \( y = x^2 \) and \( y = 2x \) about the x-axis. We can use the **disk method** (since we revolve around the x-axis) to compute this volume.

### Steps:

1. **Find the points of intersection:**
   To find the bounds of integration, we need the points where the curves \( y = x^2 \) and \( y = 2x \) intersect.
   \[
   x^2 = 2x
   \]
   Rearranging this:
   \[
   x^2 - 2x = 0
   \]
   Factor the equation:
   \[
   x(x - 2) = 0
   \]
   So, \( x = 0 \) and \( x = 2 \).

2. **Set up the volume integral:**
   The volume is found by revolving the area between the two curves around the x-axis. The volume is given by the difference of two solids formed by revolving each curve. Using the disk method:

   \[
   V = \pi \int_{a}^{b} \left( R_{\text{outer}}(x)^2 - R_{\text{inner}}(x)^2 \right) dx
   \]
   where:
   - \( R_{\text{outer}}(x) = 2x \) (the upper curve),
   - \( R_{\text{inner}}(x) = x^2 \) (the lower curve),
   - and the bounds are \( a = 0 \) and \( b = 2 \).

   The volume formula becomes:
   \[
   V = \pi \int_{0}^{2} \left( (2x)^2 - (x^2)^2 \right) dx
   \]
   Simplifying:
   \[
   V = \pi \int_{0}^{2} \left( 4x^2 - x^4 \right) dx
   \]

3. **Integrate:**
   Now, compute the integral:
   \[
   \int_{0}^{2} 4x^2 dx = \left[ \frac{4x^3}{3} \right]_{0}^{2} = \frac{4(2)^3}{3} - \frac{4(0)^3}{3} = \frac{32}{3}
   \]
   \[
   \int_{0}^{2} x^4 dx = \left[ \frac{x^5}{5} \right]_{0}^{2} = \frac{(2)^5}{5} - \frac{(0)^5}{5} = \frac{32}{5}
   \]

4. **Calculate the volume:**
   Now subtract the integrals and multiply by \( \pi \):
   \[
   V = \pi \left( \frac{32}{3} - \frac{32}{5} \right)
   \]
   To subtract, first find a common denominator:
   \[
   \frac{32}{3} - \frac{32}{5} = \frac{160}{15} - \frac{96}{15} = \frac{64}{15}
   \]
   Therefore:
   \[
   V = \pi \cdot \frac{64}{15} = \frac{64\pi}{15}
   \]

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

Would you like further details or have any questions? 

Here are 5 related questions to expand on this topic:
1. How does the volume change if the curves are revolved around the y-axis?
2. What is the volume if we revolve the region between \( y = x^2 \) and \( y = 2x \) around the line \( y = -1 \)?
3. Can you apply the shell method to compute the same volume and compare the results?
4. What if the curves were \( y = x^3 \) and \( y = 3x \), how would the method differ?
5. How would you find the surface area of the solid formed by revolving the same region?

**Tip:** Always carefully determine the points of intersection when finding limits of integration in volume problems.

Find the volume of the solid obtained by revolving the region enclosed by the curves y=x^2 and y=2x about the x-axis.

Learn how to calculate the volume of the solid obtained by revolving the region enclosed by y=x^2 and y=2x around the x-axis. This solution uses the disk method and includes step-by-step integration. Suitable for students in Grades 11-12 and college-level calculus.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation