# How do you find the volume of the solid generated by revolving the region bounded by the curves y = 1/x, y = x^2, x = 0, and y = 2 rotated about the x-axis?

I have solved considering the solid is bounded to the right by the interception of the two functions. See below:

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To find the volume of the solid generated by revolving the region bounded by the curves (y = 1/x), (y = x^2), (x = 0), and (y = 2) rotated about the x-axis, we use the method of cylindrical shells.

First, we need to find the points of intersection between the curves (y = 1/x) and (y = x^2). Setting them equal to each other:

[ 1/x = x^2 ]

We solve for (x):

[ x^3 - 1 = 0 ]

This gives us (x = 1).

Next, we integrate with respect to (x) from (x = 0) to (x = 1), because that's the region enclosed between the curves:

[ V = \int_{0}^{1} 2\pi x(2 - x^2 - 1/x)^2 dx ]

Now, simplify and solve this integral to find the volume.

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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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