# How do you find the area of the surface generated by revolving the curve #y=x^3/3# on the interval [0,3], about the x-axis?

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I'm unsure how to answer this, having a bit of trouble. This is in Calculus 2.

I'm unsure how to answer this, having a bit of trouble. This is in Calculus 2.

See the answer below:

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To find the area of the surface generated by revolving the curve (y = \frac{x^3}{3}) on the interval ([0,3]) about the x-axis, we use the formula for surface area of revolution:

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

where (y = \frac{x^3}{3}) and (a = 0), (b = 3).

First, we find ( \frac{dy}{dx} ): [ \frac{dy}{dx} = x^2 ]

Next, we plug the values into the formula: [A = 2\pi \int_{0}^{3} \frac{x^3}{3} \sqrt{1 + (x^2)^2} dx]

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

Now, we integrate: [A = 2\pi \int_{0}^{3} \frac{x^3}{3} \sqrt{1 + x^4} dx]

This integral may require techniques such as substitution or integration by parts to solve, but once evaluated, it will give the area of the surface of revolution.

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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