Answer
$\pi$($\frac{\pi}{2}$-1)
Work Step by Step
When applying the method of disk to find V(I+II) and V(II), we find:
V= V(I+II) - V(II)= $\int$ $\pi$$y_{1}^{2}$dx - $\int$ $\pi$$y_{2}^{2}$dx
--We know, $y_{1}$=1 and $y_{2}$=tanx & x∈[0, $\frac{\pi}{4}$].
So V=$\int$ $\pi$$(1)^{2}$dx- $\int$ $\pi$$tan^{2}x$dx
We apply the identity $tan^{2}x$=$sec^{2}x$-1: V=$\int$ $\pi$$(1)^{2}$dx- $\int$ $\pi$($sec^{2}x-1)$dx
Then we split the ingetral and rearrange the expression: V=2$\int$ $\pi$dx -$\pi$$\int$$sec^{2}x$dx
Now $\int$$sec^{2}x$dx=tanx, so V=(2$\pi$x-$\pi$tanx)|($x_{i}$=0, $x_{f}$=$\frac{\pi}{4}$)=$\pi$($\frac{\pi}{2}$-1)
