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**ThePerfectHacker** Draw yourself the graph and the curve. Now convince yourself the rotating $\displaystyle \sec^{-1}(x)$ around the line $\displaystyle y=\pi/2$ is the same as moving the curve down $\displaystyle \pi/2$ units and rotating about $\displaystyle y=0$. When we move the curve down the new curve is $\displaystyle \sec^{-1} x - \pi / 2$. Thus, find the surface area of revolution of this new curve using the formula that you learned from $\displaystyle x=2/\sqrt{3}$ to $\displaystyle x=2$.