How do you use the shell method to set up and evaluate the integral that gives the volume of the solid generated by revolving the plane region #y=2x#, #2<=x<=4# rotated about the xaxis?
Always graph the constraints first, then determine the correct integration formula.
Here is a graph of the constraints:
Now, without using calculus, you should visualize rotating this line around the xaxis will generate a cone with radius 2 and height 2. The volume of a cone is:
So, in advance of using calculus, we know the answer is
Now, using calculus and the shell method :
Because we rotate around the xaxis, integrate with respect to y:
hope that helped
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To use the shell method to set up and evaluate the integral for finding the volume of the solid generated by revolving the region about the xaxis, follow these steps:

Determine the limits of integration for the xvalues. In this case, the limits are from x = 2 to x = 4.

Determine the radius of each shell. For the shell method, the radius is the distance from the axis of rotation (in this case, the xaxis) to the outer edge of the region. Since we're revolving the region about the xaxis, the radius is the yvalue of the function (2  x).

Determine the height of each shell. The height of each shell represents the thickness of the region being revolved. In this case, the height is the differential element dx, representing an infinitesimally small change in x.

Write the integral to find the volume using the shell method: [ V = \int_{a}^{b} 2\pi r(x) h(x) , dx ]
where:
 ( a ) and ( b ) are the lower and upper limits of integration, in this case, 2 and 4, respectively.
 ( r(x) ) is the radius function, which is ( 2  x ).
 ( h(x) ) is the height function, which is ( dx ) (since it represents an infinitesimally small change in x).

Substitute the expressions for ( r(x) ) and ( h(x) ) into the integral: [ V = \int_{2}^{4} 2\pi (2  x) \cdot dx ]

Integrate with respect to x: [ V = \int_{2}^{4} 4\pi  2\pi x , dx ] [ V = \left[4\pi x  \pi x^2\right]_{2}^{4} ]

Evaluate the integral: [ V = \left[4\pi(4)  \pi(4)^2\right]  \left[4\pi(2)  \pi(2)^2\right] ] [ V = (16\pi  16\pi)  (8\pi  4\pi) ] [ V = 0  4\pi ] [ V = 4\pi ]
Thus, the volume of the solid generated by revolving the region about the xaxis is ( 4\pi ) cubic units. However, since volume cannot be negative, the correct answer is ( 4\pi ) cubic units.
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To use the shell method to set up and evaluate the integral for finding the volume of the solid generated by revolving the plane region ( y = 2  x ), where ( 2 \leq x \leq 4 ), about the xaxis, follow these steps:

First, visualize the region by sketching the graph of ( y = 2  x ) and identifying the bounds of integration, which are ( x = 2 ) and ( x = 4 ).

Determine the radius of each cylindrical shell. Since we're revolving the region about the xaxis, the radius of each shell will be the distance from the axis of rotation (xaxis) to the curve at each xvalue within the interval [2, 4]. This radius can be expressed as ( r = y = 2  x ).

Determine the height of each cylindrical shell. The height of each shell will be the difference between the upper and lower bounds of the region, which is ( h = f(x)  g(x) ), where ( f(x) ) is the upper curve (here, ( f(x) = 2 )) and ( g(x) ) is the lower curve (here, ( g(x) = 2  x )).

Write the integral to find the volume using the shell method. The volume ( V ) is given by the integral:
[ V = \int_{a}^{b} 2\pi rh , dx ]
where ( a ) and ( b ) are the lower and upper bounds of integration, ( r ) is the radius of the shell (in this case, ( r = 2  x )), and ( h ) is the height of the shell (in this case, ( h = 2  (2  x) = x )).
 Substitute the values of ( r ) and ( h ) into the integral:
[ V = \int_{2}^{4} 2\pi (2  x)(x) , dx ]
 Simplify the integrand and evaluate the integral:
[ V = 2\pi \int_{2}^{4} (2x  x^2) , dx ] [ V = 2\pi \left[ x^2  \frac{x^3}{3} \right]_{2}^{4} ] [ V = 2\pi \left[ (4^2  \frac{4^3}{3})  (2^2  \frac{2^3}{3}) \right] ] [ V = 2\pi \left[ (16  \frac{64}{3})  (4  \frac{8}{3}) \right] ] [ V = 2\pi \left[ \frac{48}{3}  \frac{56}{3} \right] ] [ V = 2\pi \left[ \frac{8}{3} \right] ] [ V = \frac{16}{3} \pi ]
So, the volume of the solid generated by revolving the region about the xaxis using the shell method is ( \frac{16}{3} \pi ) cubic units. Note that the negative sign indicates that the orientation of the solid is such that it is below the xaxis.
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