Problem 39 - Essential Calculus

Question

If the portion of the line \(y=\frac{1}{2} x\) lying in the first quadrant is revolved about the \(x\) -axis, a cone is generated. Find the volume of the cone extending from \(x=0\) to \(x=6\).

Step-by-Step Solution

Answer

The volume of the cone generated by the given line is \(18\pi\) cubic units.

Step 1: Determine the Radius

Recall one of the parameters of a cone is its radius, r. Along the line \(y=\frac{1}{2} x\), the y-coordinate at any point can represent the radius of the generated cone. Thus, calculate the radius at x = 6 which would be \(r = \frac{1}{2} * 6 = 3\).

Step 2: Determine the Height

The other parameter required to calculate volume of a cone is its height, h. Since we are revolving around the x-axis, the line from \(x=0\) to \(x=6\) acts as the height of the cone. Thus, the height is simply 6.

Step 3: Compute the Volume

With both parameters determined, the volume V of the cone can be obtained by using the formula: \(V = \frac{1}{3}\pi r^{2}h\). Substituting the values calculated in steps 1 and 2 (r=3, h=6) into the formula, compute the volume: \(V = \frac{1}{3}\pi * 3^{2} * 6\).

Step 4: Simplify the Result

After computing, simplify the resulting volume. Multiply and divide as directed, remembering that \(3^{2}\) (3 squared) is 9 and \(9 * 6\) is 54. Thus, the volume is \(V = \frac{1}{3}\pi * 54\). Reduce the frction and the final volume of the cone is \(V = 18\pi\).