# How do you find the volume of the solid obtained by rotating the region bounded by the given curves about the specified axis y=x^2, y=1, about y=2?

See below.

First we find the volume of the area A+B and then subtract the volume of the area B to give us the volume of the area A:

Volume A+B:

It can be see from the diagram that the radius cd of volume A+B is

Plugging in upper and lower bounds:

Volume of B:

This produces a cylinder of radius ab= 1 and length ( this is the length of the interval

Volume of A= volume(A+B)- volumeB =

Volume of A:

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To find the volume of the solid obtained by rotating the region bounded by the curves (y = x^2), (y = 1), and (y = 2) about the axis (y = 2), we can use the disk method.

The region is bounded by (y = x^2) and (y = 1), and it lies between (x = -1) and (x = 1).

The volume of the solid can be calculated as:

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

Simplifying this integral gives:

[V = \pi \int_{-1}^{1} (4 - 4x^2 + x^4) - 1 , dx]

[V = \pi \int_{-1}^{1} 3 - 4x^2 + x^4 , dx]

[V = \pi \left[3x - \frac{4}{3}x^3 + \frac{1}{5}x^5\right]_{-1}^{1}]

[V = \pi \left[(3 - \frac{4}{3} + \frac{1}{5}) - (-3 + \frac{4}{3} - \frac{1}{5})\right]]

[V = \pi \left[\frac{14}{15} + \frac{14}{15}\right]]

[V = \pi \cdot \frac{28}{15}]

So, the volume of the solid is (\frac{28\pi}{15}) cubic units.

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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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