Question

Find the volume of the solid generated by revolving the region bounded by the graphs of the equations about the \(x\) -axis. $$ y=e^{x / 2}, \quad y=0, \quad x=0, \quad x=4 $$

Short answer

The volume of the solid generated by revolving the region bounded by the graphs of the given equations about the \(x\)-axis is \(\pi[e^4 - 1]\) cubic units.

Step-by-step solution

Identify the Bounded Region

The region bounded by the graphs of the equations \(y=e^{x / 2}\), \(y=0\), \(x=0\), and \(x=4\) should be identified. It is bounded below by the \(x\)-axis (\(y=0\)), on the left by the \(y\)-axis (\(x=0\)), on the right by the vertical line (\(x=4\)), and above by the graph of the equation \(y=e^{x / 2}\).

Implement the Disk Method

Use the Disk Method to solve this. It involves integrating the cross-sectional area of the solid. For disks/washers, the trivial cross section is a circle, with an area \(A = \pi r^2\). Thus, the differential volume element \(dV = A\: dx = \pi [r(x)]^2\: dx\). Here, \(r(x) = y = e^{x / 2}\), since the radius of a disk here will be the distance from the \(x\)-axis to a point on the curve \(y=e^{x / 2}\). So, \(dV = \pi [e^{x / 2}]^2\: dx\). Before substituting this into volume integral, simplify \(r(x)\) to work with.

Simplify and Setup the Volume Integral

\([r(x)]^2 = [e^{x / 2}]^2 = e^x\). Substitute this into the volume integral and set the limits of integration from \(x=0\) to \(x=4\). This gives: \(V = \int_{0}^{4} \pi e^x\: dx\).

Evaluate the Definite Integral

Evaluate the definite integral \(V = \int_{0}^{4} \pi e^x\: dx = \pi[e^x]_{0}^{4} = \pi[e^4 - e^0] = \pi[e^4 - 1]\).