How do you use the method of cylindrical shells to find the volume of the solid obtained by rotating the region bounded by #125 y = x^3# , y = 8 , x = 0 revolved about the xaxis?
See below
Here is the region:
A representative slice taken perpendicular to the axis of rotation has volume
# = pi[640  10^7/(7(5^6))]#
# = pi[640  (10 * 2^6)/7]#
# = pi[640  640/7]#
# = pi[(7*640)/7  640/7]#
# = pi (6 * 640)/7#
# = (3840pi)/7# We can also use cylindrical shells to get the same answer:
# V =int_(y=a)^(y=b)2pi \ y \ g(y) \ dy #
# \ \ =2pi int_0^8 \ y \ root(3)(125y) \ dy #
# \ \ =2pi int_0^8 5 \ y \ y^(1/3) \ dy #
# \ \ =10pi int_0^8 y^(4/3) \ dy #
# \ \ =10pi [y^(7/3)/(7/3)]_0^8 #
# \ \ =(30pi)/7 [y^(7/3)]_0^8 #
# \ \ =(30pi)/7 (1280) #
# \ \ =(3840pi)/7 #
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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 = x^3 ), ( y = 8 ), and ( x = 0 ) about the xaxis, follow these steps:

Determine the limits of integration. In this case, the limits of integration for x are from 0 to the xcoordinate where ( y = x^3 ) intersects ( y = 8 ). Solving ( x^3 = 8 ) gives ( x = 2 ).

Determine the radius of each cylindrical shell. The radius of each shell is the distance from the axis of rotation (the xaxis) to the curve ( y = x^3 ) at a particular value of x. So, the radius ( r ) is ( x ).

Determine the height of each cylindrical shell. The height of each shell is the difference between the upper and lower functions at a particular xvalue. Here, the upper function is ( y = 8 ), and the lower function is ( y = x^3 ). So, the height ( h ) is ( 8  x^3 ).

Write the expression for the volume of each cylindrical shell. The volume ( V ) of a cylindrical shell is given by ( V = 2\pi rh ), where ( r ) is the radius and ( h ) is the height.

Integrate the volume expression with respect to ( x ) from 0 to 2.
[ V = \int_{0}^{2} 2\pi x(8  x^3) , dx ]
 Evaluate the integral to find the volume.
[ V = 2\pi \int_{0}^{2} (8x  x^4) , dx ]
[ = 2\pi \left[ 4x^2  \frac{x^5}{5} \right]_{0}^{2} ]
[ = 2\pi \left( 4(2)^2  \frac{(2)^5}{5} \right)  2\pi \left( 4(0)^2  \frac{(0)^5}{5} \right) ]
[ = 2\pi \left( 16  \frac{32}{5} \right)  0 ]
[ = 2\pi \left( \frac{80}{5}  \frac{32}{5} \right) ]
[ = 2\pi \left( \frac{48}{5} \right) ]
[ = \frac{96\pi}{5} ]
Therefore, the volume of the solid obtained by rotating the given region about the xaxis using the method of cylindrical shells is ( \frac{96\pi}{5} ) cubic units.
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