# 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 = 1 + x^2#, #y = 0#, #x = 0#, #x = 2# rotated about the line #x=4#?

graph{(y-1-x^2)(y)( sqrt(2-x) )(sqrt(x)) / (sqrt(2-x))/(sqrt(x))<=0 [0, 6, -1.51, 6.39]}

Recall the general form of the volume of a solid of revolution using the shells method when you are revolving about a vertical line:

Thus:

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To use the shell method, you need to integrate the circumference of the shell (a cylindrical shell) multiplied by its height. The circumference of the shell is ( 2\pi r ), and the height is the difference in the y-values of the curve at the two x-values that define the shell.

For the given problem, the integral to find the volume using the shell method is:

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

This integral represents the sum of the volumes of infinitesimally thin cylindrical shells between x=0 and x=2.

Solving this integral gives the volume of the solid generated by revolving the region about the line x=4.

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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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