Step 1 We are asked to find the surface area of the curve defined by $$ y=x^{3} $$ rotated about the $$ x $$-axis over the interval $$ 0 \leq x \leq 2 $$. Recall the following formula for the surface area of a function of $$ x $$ rotated about the $$ x $$-axis. Note that as the curve rotates in a circular manner about the $$ x $$-axis, the expression $$ 2 \pi y $$ is the circumference of radius $$ y $$ and the radical measures the arc length that is the width of a band. $$ S=\int_{a}^{b} 2 \pi y \sqrt{1+\left(\frac{d y}{d x}\right)^{2}} d x $$ We begin by substituting $$ y=x^{3} $$ and its derivative in the surface area formula and simplifying. $$ \begin{aligned} S & \left.=\int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+\left(\sqrt{3 x^{2}}\right.} \sqrt{3 x^{2}}\right)^{2} d x \ & =\int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+9}, 9^{x^{4}} d x \end{aligned} $$ Step 2 We have found the following integral for the surface area. $$ S=\int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+9 x^{4}} d x $$ To evaluate the integral we will first make the substitution $$ u=1+9 x^{4} $$. We also will need the following to complete the substitution. $$ \begin{array}{l} d u=36 x^{3} \ x=0 \rightarrow u=1 \ x=2 \rightarrow u=\square 146 x^{d x} \end{array} $$ Step 3 We can now make the substitution $$ u=1+(9 x)^{4} $$ and evaluate the definite integral with respect to $$ u $$. $$ \begin{aligned} \int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+(9 x)^{4}} d x & =\int_{1}^{145} 2 \pi \sqrt{u}\left(\frac{d u}{36}\right) \ & =\frac{2 \pi}{36} \int_{1}^{145} \sqrt{u} d u \end{aligned} $$ $$ =\frac{2 \pi}{36}\left[\frac{2}{3} u^{\frac{2}{3}}\right]_{1}^{145} $$

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**Step 1. Write the Surface Area Integral**

For a curve $$ y=f(x) $$ rotated about the $$ x $$-axis, the surface area is given by  
$$
S=\int_{a}^{b} 2\pi y \sqrt{1+\left(\frac{dy}{dx}\right)^{2}}\, dx.
$$  
Here, $$ y=x^3 $$ so that  
$$
\frac{dy}{dx}=3x^2.
$$  
Then,  
$$
\left(\frac{dy}{dx}\right)^2 = 9x^4.
$$  
Substituting into the surface area formula, we have  
$$
S=\int_{0}^{2} 2\pi\,(x^3) \sqrt{1+9x^4}\, dx.
$$

---

**Step 2. Use the Substitution**

Let  
$$
u=1+9x^4.
$$  
Differentiate with respect to $$ x $$:  
$$
\frac{du}{dx} = 36x^3 \quad \Longrightarrow \quad du=36x^3\, dx.
$$  
Thus,  
$$
x^3\, dx=\frac{du}{36}.
$$

The limits of integration change as follows:  
- When $$ x=0 $$: $$ u=1+9(0)^4=1 $$.  
- When $$ x=2 $$: $$ u=1+9(2)^4=1+9\cdot 16=145 $$.

Replacing in the integral, we get  
$$
S=\int_{x=0}^{2} 2\pi\, x^3\, \sqrt{1+9x^4}\, dx 
=\int_{u=1}^{145} 2\pi\, \sqrt{u}\cdot \frac{du}{36}.
$$  
This simplifies to  
$$
S=\frac{2\pi}{36}\int_{1}^{145} \sqrt{u}\, du
=\frac{\pi}{18}\int_{1}^{145} u^{\frac{1}{2}}\, du.
$$

---

**Step 3. Integrate**

Now, we compute the integral  
$$
\int u^{\frac{1}{2}}\, du.
$$  
Recall that  
$$
\int u^{\frac{1}{2}}\, du=\frac{2}{3}u^{\frac{3}{2}}+C.
$$  
Thus, the definite integral becomes  
$$
\int_{1}^{145} u^{\frac{1}{2}}\, du = \frac{2}{3}\left[u^{\frac{3}{2}}\right]_{1}^{145}
=\frac{2}{3}\Bigl(145^{\frac{3}{2}}-1\Bigr).
$$

Substituting back, we have  
$$
S=\frac{\pi}{18}\cdot \frac{2}{3}\Bigl(145^{\frac{3}{2}}-1\Bigr).
$$

Simplify the constant coefficient:  
$$
\frac{\pi}{18}\cdot \frac{2}{3} = \frac{2\pi}{54}=\frac{\pi}{27}.
$$

---

**Final Answer**

The surface area of the rotated curve is  
$$
S=\frac{\pi}{27}\Bigl(145^{\frac{3}{2}}-1\Bigr).
$$
The surface area of the rotated curve $$ y = x^3 $$ over $$ 0 \leq x \leq 2 $$ is $$ S = \frac{\pi}{27} \left(145^{\frac{3}{2}} - 1\right). $$

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