Answer
a. $\sqrt 3\pi-\frac{\pi^2}{6}$
b. $4\sqrt 3-\frac{2\pi}{3}$
Work Step by Step
a. Step 1. Draw a diagram as shown in the figure. The range of integration is $[-\frac{\pi}{3},\frac{\pi}{3}]$ and the diameter of the cross section disk is $d=2r=sec(x)-tan(x)$
Step 2. The volume can be found as $V=\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}(\pi r^2)dx=\frac{\pi}{4}\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}(sec(x)-tan(x))^2dx=\frac{\pi}{4}\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}(sec^2(x)-2sec(x)tan(x)+tan^2(x))dx=\frac{\pi}{4}\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}(2sec^2(x)-1-\frac{2sin(x)}{cos^2(x)})dx$
Step 3. Use the antiderivatives:
$\int sec^2(x)dx=tan(x)+C$ and $\int \frac{sin(x)}{cos^2(x)}dx=-\int \frac{d(cos(x))}{cos^2(x)}=\frac{1}{cos(x)}+C$, we have $V=\frac{\pi}{4}(2tan(x)-x-\frac{2}{cos(x)})|_{-\frac{\pi}{3}}^{\frac{\pi}{3}}=\frac{\pi}{4}[(2tan(\frac{\pi}{3})-\frac{\pi}{3}-\frac{2}{cos(\frac{\pi}{3})})-(2tan(\frac{-\pi}{3})+\frac{\pi}{3}-\frac{2}{cos(-\frac{\pi}{3})})]=\frac{\pi}{4}[4\sqrt 3-\frac{2\pi}{3}]=\sqrt 3\pi-\frac{\pi^2}{6}$
b. Step 1. Draw a diagram as shown in the figure. The range of integration is $[-\frac{\pi}{3},\frac{\pi}{3}]$ and the edge length of the cross section square is $a=sec(x)-tan(x)$
Step 2. The volume can be found as
$V=\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}(a^2)dx=\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}(sec(x)-tan(x))^2dx=\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}(sec^2(x)-2sec(x)tan(x)+tan^2(x))dx=\int_{-\frac{\pi}{3}}^{\frac{\pi}{3}}(2sec^2(x)-1-\frac{2sin(x)}{cos^2(x)})dx$
Step 3. Use the antiderivatives:
$\int sec^2(x)dx=tan(x)+C$ and $\int \frac{sin(x)}{cos^2(x)}dx=-\int \frac{d(cos(x))}{cos^2(x)}=\frac{1}{cos(x)}+C$, we have $V=(2tan(x)-x-\frac{2}{cos(x)})|_{-\frac{\pi}{3}}^{\frac{\pi}{3}}=(2tan(\frac{\pi}{3})-\frac{\pi}{3}-\frac{2}{cos(\frac{\pi}{3})})-(2tan(\frac{-\pi}{3})+\frac{\pi}{3}-\frac{2}{cos(-\frac{\pi}{3})})=4\sqrt 3-\frac{2\pi}{3}$
