# How do you find the area of the surface generated by rotating the curve about the y-axis #x=t^2, y=1/3t^3, 0<=t<=3#?

I have skipped the mechanical integration steps. you can start with a sub:

Which is trivial but protracted

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To find the area of the surface generated by rotating the curve about the y-axis ( x = t^2 ), ( y = \frac{1}{3}t^3 ) for ( 0 \leq t \leq 3 ), you can use the formula for surface area of revolution:

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

First, express ( x ) in terms of ( y ) to get ( x = (3y)^{2/3} ). Then, compute ( \frac{dy}{dx} ) and substitute the expressions into the formula to calculate the integral. This will give you the area of the surface generated by rotating the curve about the y-axis.

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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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