Question

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

Short answer

The volume of the solid generated by revolving the region bounded by the curves about the \(x\)-axis is \( \pi^2/4 \)

Step-by-step solution

Establish the Bounded Region

Firstly, we need to plot the curves \(y = \sin(x/2)\) and \(y = \cos(x/2)\), along with the lines \(x=0\) and \(x = \pi/2\). When we sketch these, we see that the bounded region lies above the x -axis, meaning that we can use the disc method for rotating about the x-axis.

Set Up the Integral for the Disk Method

The disk method formula for volume is \(V = \pi\int_{a}^{b}(g(x))^2 dx\), where [a,b] is the interval [0 , \(\pi/2\)] on the x-axis, and g(x) is the radius of the disk, which is here given by the upper function minus the lower function, i.e., \(g(x) = \cos(x/2) - \sin(x/2)\). Hence, set up this integral: \(V = \pi\int_{0}^{\pi/2}(\cos^2(x/2) - 2\cos(x/2)\sin(x/2) + \sin^2(x/2)) dx\).

Simplify and Compute the Integral

The integrand simplifies to \(V = \pi\int_{0}^{\pi/2}(1 - \sin(x)) dx\), because the sum of squares of sine and cosine is one and the double-angle identity for sine can be expressed as \(\sin(2u) = 2\sin(u)*\cos(u)\). Then, compute the integral \(V = \pi [ x - \frac{1}{2}(-\cos(x)) ]_{0}^{\pi/2}\).

Evaluate the Integral

Finally, evaluate the integral at the bounds \(x=0\) and \(x= \pi/2\). This yields \(V = \pi * \pi/4 = \pi^2/4\)