Question Use the washer method to calculate the volume of the solid formed when the region bounded by $$ y=2 x^{2} $$ and $$ y=2 \sqrt{x} $$ revolved around the the line $$ y=3 $$.

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To calculate the volume of the solid formed by revolving the region bounded by $$ y = 2x^2 $$ and $$ y = 2\sqrt{x} $$ around the line $$ y = 3 $$ using the washer method, follow these steps:

### 1. **Find the Points of Intersection**
First, determine where the curves $$ y = 2x^2 $$ and $$ y = 2\sqrt{x} $$ intersect.

$$
2x^2 = 2\sqrt{x} \quad \Rightarrow \quad x^2 = \sqrt{x}
$$

Let $$ t = \sqrt{x} $$, then:

$$
t^2 = t^4 \quad \Rightarrow \quad t^4 - t = 0 \quad \Rightarrow \quad t(t^3 - 1) = 0
$$

This gives $$ t = 0 $$ or $$ t = 1 $$, so the points of intersection are $$ x = 0 $$ and $$ x = 1 $$.

### 2. **Determine the Radii for the Washer Method**
Since we are rotating around $$ y = 3 $$, the radii for the washers are:

- **Outer radius ($$ R(x) $$)**: Distance from $$ y = 3 $$ to the lower curve $$ y = 2x^2 $$.
  
  $$
  R(x) = 3 - 2x^2
  $$

- **Inner radius ($$ r(x) $$)**: Distance from $$ y = 3 $$ to the upper curve $$ y = 2\sqrt{x} $$.
  
  $$
  r(x) = 3 - 2\sqrt{x}
  $$

### 3. **Set Up the Integral Using the Washer Method**
The volume $$ V $$ is given by:

$$
V = \pi \int_{0}^{1} \left[ (R(x))^2 - (r(x))^2 \right] dx
$$

Substituting $$ R(x) $$ and $$ r(x) $$:

$$
V = \pi \int_{0}^{1} \left[ (3 - 2x^2)^2 - (3 - 2\sqrt{x})^2 \right] dx
$$

### 4. **Expand and Simplify the Integrand**
Expand both squared terms:

$$
(3 - 2x^2)^2 = 9 - 12x^2 + 4x^4
$$
$$
(3 - 2\sqrt{x})^2 = 9 - 12\sqrt{x} + 4x
$$

Subtract the second expansion from the first:

$$
(9 - 12x^2 + 4x^4) - (9 - 12\sqrt{x} + 4x) = 4x^4 - 12x^2 - 4x + 12\sqrt{x}
$$

### 5. **Integrate Term by Term**
Now, integrate each term separately:

$$
V = \pi \int_{0}^{1} \left(4x^4 - 12x^2 - 4x + 12\sqrt{x}\right) dx
$$

$$
= \pi \left[ \frac{4x^5}{5} - 4x^3 - 2x^2 + 8x^{3/2} \right]_0^1
$$

Evaluate from 0 to 1:

$$
= \pi \left( \frac{4}{5} - 4 - 2 + 8 \right)
$$
$$
= \pi \left( \frac{4}{5} + 2 \right)
$$
$$
= \pi \cdot \frac{14}{5}
$$

### 6. **Final Answer**

$$
V = \frac{14}{5} \pi
$$

**Thus, the volume of the solid is $$ \dfrac{14}{5}\pi $$ cubic units.**
The volume of the solid is $$ \frac{14}{5} \pi $$ cubic units.

Question Use the washer method to calculate the volume of the solid formed when the region bounded by \( y=2 x^{2} \) and \( y=2 \sqrt{x} \) revolved around the the line \( y=3 \).

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