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Find the area of the surface generated by revolving the curve [tex]$y=\frac{x^3}{9}$[/tex], [tex]$0 \leq x \leq 2$[/tex], about the [tex][tex]$x$[/tex][/tex]-axis.

Set up the integral that gives the area of the given surface.

Sagot :

GinnyAnswer
To solve this problem, we need to determine the area of the surface generated by revolving the given curve around the [tex]\( x \)[/tex]-axis.

The general formula for the surface area [tex]\( S \)[/tex] of a solid of revolution about the [tex]\( x \)[/tex]-axis is:

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

Given the curve [tex]\( y = \frac{x^3}{9} \)[/tex] and the interval [tex]\( 0 \leq x \leq 2 \)[/tex]:

1. Determine [tex]\( y \)[/tex] and [tex]\( \frac{dy}{dx} \)[/tex]:
[tex]\[ y = \frac{x^3}{9} \][/tex]
[tex]\[ \frac{dy}{dx} = \frac{d}{dx} \left( \frac{x^3}{9} \right) = \frac{3x^2}{9} = \frac{x^2}{3} \][/tex]

2. Compute [tex]\( 1 + \left(\frac{dy}{dx}\right)^2 \)[/tex]:
[tex]\[ \left(\frac{dy}{dx}\right)^2 = \left(\frac{x^2}{3}\right)^2 = \frac{x^4}{9} \][/tex]
[tex]\[ 1 + \left(\frac{dy}{dx}\right)^2 = 1 + \frac{x^4}{9} \][/tex]

3. Set up the integral:
[tex]\[ S = 2\pi \int_{0}^{2} \left( \frac{x^3}{9} \right) \sqrt{1 + \frac{x^4}{9}} \, dx \][/tex]

Now, we have to integrate this expression:

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

Simplify the constant factor:
[tex]\[ S = \frac{2\pi}{9} \int_{0}^{2} x^3 \sqrt{1 + \frac{x^4}{9}} \, dx \][/tex]

So, the integral set up for the area of the surface generated by revolving the curve [tex]\( y = \frac{x^3}{9} \)[/tex] around the [tex]\( x \)[/tex]-axis over the interval [tex]\([0, 2]\)[/tex] is:

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