# How do you find the volume of the solid obtained by rotating the region bounded by the curves #1/(1+x^2)#, #y=0#, #x=0#, and #x=2# rotated around the #x=2#?

Given:

Required Volume?

Solution Strategy: Use the "Volume by Cylindrical Shells Method"

The "shell method" give the volume calculated in the region bound by

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To find the volume of the solid obtained by rotating the region bounded by the curves ( y = \frac{1}{1+x^2} ), ( y = 0 ), ( x = 0 ), and ( x = 2 ) about the line ( x = 2 ), you can use the method of cylindrical shells.

The volume ( V ) of the solid generated by rotating a curve ( y = f(x) ) from ( x = a ) to ( x = b ) about the line ( x = c ) is given by:

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

In this case:

- The curve is ( y = \frac{1}{1+x^2} ).
- The bounds of integration are ( x = 0 ) and ( x = 2 ).
- The axis of rotation is ( x = 2 ).

Substitute these values into the formula:

[ V = 2\pi \int_{0}^{2} x \cdot \frac{1}{1+x^2} , dx ]

Now, compute the integral:

[ V = 2\pi \int_{0}^{2} \frac{x}{1+x^2} , dx ]

To solve this integral, perform a substitution:

Let ( u = 1 + x^2 ), then ( du = 2x , dx ).

When ( x = 0 ), ( u = 1 ) and when ( x = 2 ), ( u = 5 ).

The integral becomes:

[ V = 2\pi \int_{1}^{5} \frac{1}{u} , du ] [ V = 2\pi [\ln|u|]_{1}^{5} ] [ V = 2\pi [\ln(5) - \ln(1)] ] [ V = 2\pi \ln(5) ]

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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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