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Compute the volume of a right circular cone generated by f(x) = cx

Consider the ordinate set of the function f(x) = cx on the interval [0,b]. Revolving this ordinate set about the x-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, V = \frac{1}{3} a(B) h, where a(B) is the area of the base.

Proof. First, the area of a cross-section of this solid of revolution is \pi r^2 = \pi c^2 x^2. So, using the formula for the volume of a Cavalieri solid (Theorem 2.7 in Apostol), we have

    \begin{align*}    v(R) = \int_0^b \pi c^2 x^2 \, dx &= \pi c^2 \left.\frac{x^3}{3}\right|_0^b \\   &= \pi c^2 \frac{b^3}{3}. \end{align*}

Since the area of the base is \pi (cb)^2 (since cb is the height of f(x) = cx at x = b, and this is then rotated around the x-axis), we have,

    \[ V = \frac{1}{3} a(B) h = \pi c^2 \frac{b^3}{3}. \qquad \blacksquare \]