91Ó°ÊÓ

Plot the curves \(y=\tan(x)\), \(y=0\), and \(x=\frac{\pi}{3}\) on the same graph to visualize the region we are revolving. The region is bounded by these curves and lies completely in the first quadrant.

To set up the integral using the disk method, we need to find the radius of the disks formed when the region is revolved about the x-axis. In this case, the height of the region at any point x is given by \(y=\tan(x)\), so the radius of the disks is simply \(y\). The area of a disk can be found using the formula \(A = \pi r^2\), where A is the area and r is the radius. So in this case, the area of a disk is \(A = \pi (\tan(x))^2\). Now, we need to integrate the area of the disks along the length of the region which is from \(0\) to \(\frac{\pi}{3}\): \[V = \int_{0}^{\frac{\pi}{3}} A \, dx = \int_{0}^{\frac{\pi}{3}} \pi (\tan(x))^2 \, dx\]

Integrate the function with respect to x: \[V = \pi \int_{0}^{\frac{\pi}{3}} (\tan(x))^2 \, dx = \pi \int_{0}^{\frac{\pi}{3}} \left(\frac{\sin^2(x)}{\cos^2(x)}\right) \, dx\] We can perform a substitution: let \(u=\cos(x)\), so \(-du = \sin(x) \, dx\), and our limits of integration become \(u(0)=1\) and \(u\left(\frac{\pi}{3}\right)=\frac{1}{2}\): \[V = \pi \int_{1}^{\frac{1}{2}}\left(-\frac{1-u^2}{u^2}\right)(-du) = \pi \int_{1}^{\frac{1}{2}}\left(\frac{1-u^2}{u^2}\right) \, du\] Now, integrate with respect to u: \[V = \pi \left[-\int_{1}^{\frac{1}{2}} \frac{1}{u^2} \, du + \int_{1}^{\frac{1}{2}} u^2 \, du \right] = \pi \left[\left.\frac{1}{u}\right|_{1}^{\frac{1}{2}} + \left.\frac{1}{3}u^3\right|_{1}^{\frac{1}{2}}\right]\]

Plug in the limits of integration and simplify: \[V = \pi\left[\left(\frac{1}{1}-\frac{1}{\frac{1}{2}}\right) + \frac{1}{3}\left(\frac{1}{8}-1\right)\right] = \pi\left[-1+\frac{1}{2}\left(\frac{1}{8}-1\right)\right]\] \[V = \pi\left[-\frac{1}{2}+\frac{1}{16}\right] = \pi\left(-\frac{8}{16}+\frac{1}{16}\right) = -\frac{7}{16}\pi\] However, since volume cannot be negative, we take the absolute value: \[V = \frac{7}{16}\pi\] The volume of the solid generated by revolving the region bounded by the given curves about the x-axis is \(\frac{7}{16}\pi\).