Answer
$\pi (\dfrac{ 7 \pi }{9}-\sqrt 3 )$
Work Step by Step
The Washer method to compute the volume of revolution: When the function $f(x)$ is continuous and $f(x) \geq g(x) \geq 0$ on the interval $[m,n]$, then the volume of a solid obtained by rotating the region under $y=f(x)$ over an interval $[m,n]$ about the x-axis is given by:
$V=\pi \int_{m}^{n} (R^2_{outside}-R^2_{inside}) \ dy$
where, $R_{outside}=1+\dfrac{3x}{\pi} $ and $ R_{inside}=\sec x$
Now, $V=\pi \int_0^{\pi/3} [\pi [(1+\dfrac{3x}{\pi})^2 -\sec^2 x] \ dx \\ = \pi \int_0^{\pi/3} [1+\dfrac{9x^2}{\pi}+\dfrac{6x}{\pi}-\sec^2 x ] \ dx \\=\pi [x+\dfrac{3x^3}{\pi^2}+\dfrac{3x^2}{\pi}-\tan x ]_0^{\pi/3} \\=\pi (\dfrac{ 7 \pi }{9}-\sqrt 3 )$
