# How do you find the volume of the solid obtained by rotating the region bounded by the given curves about the specified line: y= x, #y = sqrt(x)#; about x = 2?

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To find the volume of the solid obtained by rotating the region bounded by the curves ( y = x ) and ( y = \sqrt{x} ) about the line ( x = 2 ), we can use the method of cylindrical shells.

The intersection points of the curves are found by setting ( y = x ) equal to ( y = \sqrt{x} ):

[ x = \sqrt{x} ]

Solving this equation, we find that ( x = 1 ) is the intersection point.

The radius of the cylindrical shell at any ( x ) value is the distance from the line of rotation ( x = 2 ) to the curve, which is ( 2 - x ).

The height of each cylindrical shell is the difference between the functions ( y = x ) and ( y = \sqrt{x} ), so it's ( x - \sqrt{x} ).

The integral to find the volume ( V ) is given by:

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

After evaluating this integral, you will get the volume of the solid of revolution.

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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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