Find the area of the surface generated when the given curve is revolved about the given axis.

yequals=StartFraction x cubed Over 20 EndFractionx320​, 

for 

0 less than or equals x less than or equals 2 StartRoot 5 EndRoot0≤x≤25​; 

about the​ x-axis

We need to find the surface area generated when the curve \( y = \frac{x^3}{20} \) is revolved around the x-axis for \( 0 \leq x \leq 2\sqrt{5} \).

### Formula for Surface Area of Revolution:

The formula for the surface area \( A \) of a curve \( y = f(x) \) revolved around the x-axis between \( x = a \) and \( x = b \) is:

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

### Step 1: Compute the derivative \( \frac{dy}{dx} \)

Given \( y = \frac{x^3}{20} \), let's first compute the derivative \( \frac{dy}{dx} \):

\[
\frac{dy}{dx} = \frac{d}{dx} \left( \frac{x^3}{20} \right) = \frac{3x^2}{20}
\]

### Step 2: Plug into the surface area formula

We can now substitute \( y = \frac{x^3}{20} \) and \( \frac{dy}{dx} = \frac{3x^2}{20} \) into the surface area formula.

The surface area becomes:

\[
A = 2\pi \int_0^{2\sqrt{5}} \frac{x^3}{20} \sqrt{1 + \left( \frac{3x^2}{20} \right)^2} \, dx
\]

\[
A = 2\pi \int_0^{2\sqrt{5}} \frac{x^3}{20} \sqrt{1 + \frac{9x^4}{400}} \, dx
\]

This integral can be solved either by substitution or using numerical methods to get an exact or approximate value for the surface area.

### Step 3: Set up the integral for numerical solution

We now have the following integral to compute:

\[
A = 2\pi \int_0^{2\sqrt{5}} \frac{x^3}{20} \sqrt{1 + \frac{9x^4}{400}} \, dx
\]

Let me compute this integral for you.The surface area of the surface generated when the curve \( y = \frac{x^3}{20} \) is revolved about the x-axis for \( 0 \leq x \leq 2\sqrt{5} \) is:

\[
A = \frac{20\pi \left( -1 + 10\sqrt{10} \right)}{27}
\]

Would you like more details on any part of this solution or have any further questions?

Here are 5 related questions you might find useful:
1. How would the formula change if the curve were revolved around the y-axis?
2. How does the surface area change if the upper limit of \( x \) is different?
3. Can this method be used for non-polynomial curves?
4. What is the physical meaning of surface area of revolution in real-world applications?
5. How would the calculation change if the curve were parametric?

**Tip:** For complex integrals involving square roots, numerical methods can often provide quicker approximate solutions.

Find the area of the surface generated when the given curve y = x^3 / 20 is revolved about the x-axis for 0 ≤ x ≤ 2√5.

Find the surface area generated when the curve y = x^3 / 20 is revolved about the x-axis for 0 ≤ x ≤ 2√5. This problem involves calculus concepts such as surface area of revolution, definite integrals, and differentiation. Suitable for advanced high school students studying calculus.

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