Find the exact area of the surface obtained by rotating the curve about the x-axis.
y = x3,    0 ≤ x ≤ 3

To find the exact area of the surface obtained by rotating the curve \( y = x^3 \) about the \( x \)-axis for \( 0 \leq x \leq 3 \), we use the formula for the surface area of a solid of revolution:

\[
A = 2\pi \int_{a}^{b} y \sqrt{1 + \left(\frac{dy}{dx}\right)^2} \, dx
\]

Here, \( y = x^3 \) and \( \frac{dy}{dx} = 3x^2 \). Substituting these into the formula:

1. Compute \( \left(\frac{dy}{dx}\right)^2 \):

\[
\left(\frac{dy}{dx}\right)^2 = (3x^2)^2 = 9x^4
\]

2. Substitute \( y \) and \( \left(\frac{dy}{dx}\right)^2 \) into the surface area formula:

\[
A = 2\pi \int_{0}^{3} x^3 \sqrt{1 + 9x^4} \, dx
\]

To evaluate this integral, we can use a trigonometric substitution or numerical methods. However, given the structure of the problem, this integral is best approached using substitution. Let’s set up the substitution:

Let \( u = 1 + 9x^4 \). Then, \( du = 36x^3 dx \) and \( dx = \frac{du}{36x^3} \).

When \( x = 0 \), \( u = 1 \).

When \( x = 3 \), \( u = 1 + 9(3)^4 = 1 + 9 \cdot 81 = 730 \).

So, we change the integral from \( x \) to \( u \):

\[
A = 2\pi \int_{1}^{730} x^3 \sqrt{u} \cdot \frac{du}{36x^3}
\]

The \( x^3 \) terms cancel out:

\[
A = 2\pi \int_{1}^{730} \frac{\sqrt{u}}{36} \, du
\]

\[
A = \frac{2\pi}{36} \int_{1}^{730} u^{1/2} \, du
\]

\[
A = \frac{\pi}{18} \int_{1}^{730} u^{1/2} \, du
\]

Now, integrate \( u^{1/2} \):

\[
\int u^{1/2} \, du = \frac{2}{3} u^{3/2}
\]

So,

\[
A = \frac{\pi}{18} \left[ \frac{2}{3} u^{3/2} \right]_{1}^{730}
\]

\[
A = \frac{\pi}{18} \cdot \frac{2}{3} \left[ u^{3/2} \right]_{1}^{730}
\]

\[
A = \frac{\pi}{27} \left[ 730^{3/2} - 1^{3/2} \right]
\]

\[
A = \frac{\pi}{27} \left[ 730^{3/2} - 1 \right]
\]

Calculate \( 730^{3/2} \):

\[
730^{3/2} = 730 \sqrt{730}
\]

So, the final area is:

\[
A = \frac{\pi}{27} \left[ 730 \sqrt{730} - 1 \right]
\]

Thus, the exact area of the surface obtained by rotating the curve \( y = x^3 \) about the \( x \)-axis from \( 0 \leq x \leq 3 \) is:

\[
A = \frac{\pi}{27} \left( 730 \sqrt{730} - 1 \right)
\]

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

Here are 5 related questions to expand your understanding:

1. How would the surface area change if the curve \( y = x^3 \) were rotated about the y-axis instead?
2. Can you find the surface area of the solid of revolution generated by rotating the curve \( y = x^2 \) about the x-axis from \( x = 0 \) to \( x = 2 \)?
3. What is the surface area of the solid formed by rotating the curve \( y = \sqrt{x} \) about the x-axis from \( x = 1 \) to \( x = 4 \)?
4. How do you apply the shell method to find the volume of a solid of revolution?
5. What are the key differences between the disk/washer method and the shell method in calculus?

**Tip:** When dealing with integrals involving square roots and polynomial expressions, consider substitution methods to simplify the integrand for easier evaluation.

Find the exact area of the surface obtained by rotating the curve about the x-axis.
y = x^3, 0 ≤ x ≤ 3

Learn how to find the exact surface area of revolution for the curve y = x^3 rotated about the x-axis from x = 0 to x = 3. This problem involves advanced calculus concepts, including surface area integrals, substitution methods, and applying the surface area of revolution formula.

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