# How do you find volume by rotating area enclosed by #y=x^3# and #y=sqrt(x)# about x=1?

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To find the volume by rotating the area enclosed by (y = x^3) and (y = \sqrt{x}) about (x = 1), you would use the method of cylindrical shells. The formula for this method is:

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

where (f(x)) and (g(x)) are the functions bounding the region, and (a) and (b) are the x-values where the functions intersect.

First, find the intersection points of (y = x^3) and (y = \sqrt{x}) by setting them equal to each other and solving for (x):

[ x^3 = \sqrt{x} ] [ x^6 = x ] [ x^6 - x = 0 ] [ x(x^5 - 1) = 0 ] [ x = 0 \text{ or } x^5 = 1 ] [ x = 0 \text{ or } x = 1 ]

So, (x = 1) is the only intersection point.

The integral becomes:

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

Now integrate to find the volume.

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