To solve the problem of finding the volume obtained by revolving the region bounded by the curves $y = x^2$ and $y = 2x$ about the x-axis, we will use the method of disks or washers.

Step 1: Find the points of intersection

We need to determine the points where $y = x^2$ and $y = 2x$ intersect: $x^2 = 2x$ Rearrange the equation: $x^2 - 2x = 0$ Factoring out $x$: $x(x - 2) = 0$ So, the points of intersection are $x = 0$ and $x = 2$.

Step 2: Set up the integral using the washer method

The volume of the solid formed by revolving the region around the x-axis is given by: $V = \pi \int_{a}^{b} \left( R^2 - r^2 \right) dx$ where:

• $R(x)$ is the outer radius (distance from the x-axis to the outer curve).
• $r(x)$ is the inner radius (distance from the x-axis to the inner curve).

For this problem:

• $R(x) = 2x$ (since $y = 2x$ is farther from the x-axis).
• $r(x) = x^2$ (since $y = x^2$ is closer to the x-axis).

The bounds are $a = 0$ and $b = 2$.

Step 3: Write the integral expression

$V = \pi \int_{0}^{2} \left( (2x)^2 - (x^2)^2 \right) dx$ Simplify the integrand: $V = \pi \int_{0}^{2} \left( 4x^2 - x^4 \right) dx$

Step 4: Integrate

Calculate the integral: $V = \pi \left[ \int_{0}^{2} 4x^2 dx - \int_{0}^{2} x^4 dx \right]$ Integrate each term separately: $\int 4x^2 dx = \frac{4x^3}{3}$ $\int x^4 dx = \frac{x^5}{5}$ Substituting these into the expression for $V$: $V = \pi \left[ \frac{4x^3}{3} \bigg|_{0}^{2} - \frac{x^5}{5} \bigg|_{0}^{2} \right]$

Step 5: Evaluate the bounds

Calculate at the upper bound $x = 2$ and lower bound $x = 0$: $\frac{4(2)^3}{3} = \frac{4 \cdot 8}{3} = \frac{32}{3}$ $\frac{(2)^5}{5} = \frac{32}{5}$

Substituting these values: $V = \pi \left( \frac{32}{3} - \frac{32}{5} \right)$

Step 6: Simplify the expression

Find a common denominator (15): $\frac{32}{3} = \frac{160}{15}, \quad \frac{32}{5} = \frac{96}{15}$ $V = \pi \left( \frac{160}{15} - \frac{96}{15} \right) = \pi \cdot \frac{64}{15}$

Final Answer

$V = \frac{64\pi}{15}$

The volume of the solid formed by revolving the region bounded by $y = x^2$ and $y = 2x$ about the x-axis is $\frac{64\pi}{15}$ cubic units.

Would you like more details or have any questions about this solution?

Related Questions:

1. How would the volume change if the region was revolved around the y-axis instead?
2. What if the curves were different, say $y = x^3$ and $y = 3x$? How would that affect the volume calculation?
3. Can the shell method be used to solve this problem? If yes, how would the integral be set up?
4. What does the geometric interpretation of this volume look like?
5. How can we verify the points of intersection between two different functions analytically?

Tip:

When revolving regions around an axis, always sketch the graphs and identify the outer and inner radii carefully to set up the integral correctly.