# How do you find the volume of a solid that is enclosed by #y=x+1#, #y=x^3+1#, x=0 and y=0 revolved about the x axis?

The volume

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To find the volume of the solid generated by revolving the region bounded by the curves (y = x + 1), (y = x^3 + 1), (x = 0), and (y = 0) about the x-axis, you can use the method of cylindrical shells. The volume (V) can be calculated using the integral:

[ V = \int_{a}^{b} 2\pi x \cdot (f(x) - g(x)) , dx ]

Where (a) and (b) are the x-values of the points of intersection of the curves, and (f(x)) and (g(x)) represent the functions that define the upper and lower boundaries of the region, respectively.

First, we find the intersection points of the curves: [ x + 1 = x^3 + 1 ] [ x^3 - x = 0 ] [ x(x^2 - 1) = 0 ] [ x(x + 1)(x - 1) = 0 ]

So, the intersection points are (x = -1), (x = 0), and (x = 1).

Next, we integrate from 0 to 1, as the region of interest lies between those x-values: [ V = \int_{0}^{1} 2\pi x \cdot ((x^3 + 1) - (x + 1)) , dx ]

[ V = \int_{0}^{1} 2\pi x \cdot (x^3 - x) , dx ]

[ V = \int_{0}^{1} 2\pi (x^4 - x^2) , dx ]

[ V = 2\pi \left[\frac{x^5}{5} - \frac{x^3}{3}\right]_{0}^{1} ]

[ V = 2\pi \left(\frac{1}{5} - \frac{1}{3}\right) ]

[ V = \frac{2\pi}{15} ]

So, the volume of the solid is (\frac{2\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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