How do you use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by #y = 2 x^4#, y = 0, x = 1 revolved about the x=2?
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To use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by ( y = 2x^4 ), ( y = 0 ), ( x = 1 ) revolved about ( x = 2 ), follow these steps:

Determine the limits of integration. In this case, the region is bounded by ( y = 2x^4 ), ( y = 0 ), and ( x = 1 ). The rotation is about the vertical line ( x = 2 ), so the limits of integration will be from ( x = 1 ) to ( x = 2 ).

Set up the integral for the volume using the formula for cylindrical shells:
[ V = 2\pi \int_{a}^{b} x \cdot h(x) , dx ]
where ( h(x) ) is the height of the shell and ( a ) and ( b ) are the lower and upper limits of integration, respectively.

Express the height of the shell, ( h(x) ), as the difference between the outer radius and the inner radius. The outer radius is the distance from the axis of rotation ( ( x = 2 ) in this case) to the curve ( y = 2x^4 ), which is ( 2  x ). The inner radius is the distance from the axis of rotation to the line ( x = 1 ), which is ( 2  1 = 1 ).

Calculate the integral:
[ V = 2\pi \int_{1}^{2} x \cdot (2  x  1) , dx ]
 Integrate the expression:
[ V = 2\pi \int_{1}^{2} (2x  x^2  x) , dx ]
[ V = 2\pi \int_{1}^{2} (2x  x^2  x) , dx ]
[ V = 2\pi \left[ \frac{x^2}{2}  \frac{x^3}{3}  \frac{x^2}{2} \right]_{1}^{2} ]
[ V = 2\pi \left[ \left(2  \frac{8}{3}  2\right)  \left(\frac{1}{2}  \frac{1}{3}  \frac{1}{2}\right) \right] ]
[ V = 2\pi \left[ \frac{4}{3}  \left(\frac{1}{6}\right) \right] ]
[ V = 2\pi \left( \frac{7}{6} \right) ]
[ V = \frac{7\pi}{3} ]
Therefore, the volume of the solid obtained by rotating the region about ( x = 2 ) using the method of cylindrical shells is ( \frac{7\pi}{3} ).
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