$$ y=x^{3}, \quad 0 \leq x \leq 2 $$ 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 interv $$ 0 \leq x \leq 2 $$. Recall the following formula for the surface area of a function of $$ x $$ rotated about the $$ x $$-axis. Note t as the curve rotates in a circular manner about the $$ x $$-axis, the expression $$ 2 \pi y $$ is the circumference of radius 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, $$ \begin{aligned} S & \left.=\int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+\left(\sqrt{3 x^{2}}\right.} \sqrt{3 x^{2}}\right)^{2} d x \ & =\int_{0}^{2} 2 \pi\left(x^{3}\right) \sqrt{1+9} \sqrt[9]{ } x^{4} d x \end{aligned} $$ 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. $$ \begin{array}{l} d u=36 x^{3} \ x=0 \rightarrow u=1 \ x=2 \rightarrow u=\square 14 \end{array} $$ 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.**  
The surface area for a curve $$ y = f(x) $$ rotated about the $$ x $$-axis is given by  
$$
S = \int_{a}^{b} 2\pi y \sqrt{1+\left(\frac{dy}{dx}\right)^2}\, dx.
$$  
For the curve $$ y=x^3 $$ on $$ 0 \le x \le 2 $$, we first compute  
$$
\frac{dy}{dx} = 3x^2.
$$  
Substituting into the formula, we obtain  
$$
S = \int_{0}^{2} 2\pi x^3 \sqrt{1+\left(3x^2\right)^2}\, dx = \int_{0}^{2} 2\pi x^3 \sqrt{1+9x^4}\, dx.
$$

**Step 2.**  
We next make the substitution  
$$
u = 1 + 9x^4.
$$
Then, differentiating with respect to $$ x $$ gives  
$$
du = 36x^3\, dx \quad \Longrightarrow \quad x^3\, dx = \frac{du}{36}.
$$
The limits change as follows:  
- When $$ x = 0 $$: $$ u = 1 + 9(0)^4 = 1 $$.  
- When $$ x = 2 $$: $$ u = 1 + 9(2)^4 = 1 + 9\cdot 16 = 145 $$.

The integral becomes  
$$
S = 2\pi \int_{x=0}^{2} x^3 \sqrt{1+9x^4}\, dx = 2\pi \int_{u=1}^{145} \sqrt{u} \left(\frac{du}{36}\right) = \frac{2\pi}{36} \int_{1}^{145} \sqrt{u}\, du.
$$

Simplify the constant:  
$$
\frac{2\pi}{36} = \frac{\pi}{18}.
$$  
Thus,  
$$
S = \frac{\pi}{18} \int_{1}^{145} \sqrt{u}\, du.
$$

**Step 3.**  
Evaluate the integral  
$$
\int \sqrt{u}\, du = \int u^{1/2}\, du = \frac{2}{3}u^{3/2} + C.
$$  
Applying the limits of integration, we obtain  
$$
\int_{1}^{145} \sqrt{u}\, du = \frac{2}{3}\left[145^{3/2} - 1^{3/2}\right] = \frac{2}{3}\left(145^{3/2} - 1\right).
$$

Substitute back into the expression for $$ S $$:  
$$
S = \frac{\pi}{18} \cdot \frac{2}{3}\left(145^{3/2} - 1\right).
$$

Multiply the constants:  
$$
\frac{\pi}{18} \cdot \frac{2}{3} = \frac{2\pi}{54} = \frac{\pi}{27}.
$$

Thus, the final answer is  
$$
S = \frac{\pi}{27}\left(145^{3/2} - 1\right).
$$
The surface area $$ S $$ of the curve $$ y = x^3 $$ rotated about the $$ x $$-axis from $$ x = 0 $$ to $$ x = 2 $$ is $$ S = \frac{\pi}{27}\left(145^{3/2} - 1\right). $$

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