# How do you find the volume of the solid obtained by rotating the region bound by the curve and #y=x^2+1# and #x#-axis in the interval #(2,3)#?

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To find the volume of the solid obtained by rotating the region bound by the curve ( y = x^2 + 1 ) and the x-axis in the interval (2, 3), we use the method of cylindrical shells.

The formula for the volume of the solid generated by revolving a curve ( y = f(x) ) around the x-axis in the interval ( [a, b] ) is given by:

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

In this case, the curve is ( y = x^2 + 1 ), and the interval is ( (2, 3) ). So, we have:

[ V = 2\pi \int_{2}^{3} x \cdot (x^2 + 1) , dx ]

We integrate this expression with respect to ( x ) over the given interval, and then we evaluate the definite integral to find the volume of the solid.

After integrating and evaluating, we will obtain the volume of the solid generated by rotating the region bound by the curve ( y = x^2 + 1 ) and the x-axis in the interval (2, 3) around the x-axis using cylindrical shells.

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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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