Consider the ordinate set of the function on the interval . Revolving this ordinate set about the -axis generates a right circular cone. Using integration, compute the volume of this right circular cone. Show that the result agrees with our usual formula for the volume of a right circular cone, namely, , where is the area of the base.

* Proof. * First, the area of a cross-section of this solid of revolution is . So, using the formula for the volume of a Cavalieri solid (Theorem 2.7 in Apostol), we have

Since the area of the base is (since is the height of at , and this is then rotated around the -axis), we have,

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