Find the area of surface generated by revolving the curve $$x= \int_{0}^{y} \tan t \ dt, \ 0 \leq y \leq\frac {\pi}{3}$$ about the $y$-axis
My try
$$\frac {dx}{dy}=\tan y \to \left(\frac {dx}{dy}\right)^2=(\tan y)^2$$
$$ S= 2\pi \int_{0}^{\pi/3} \int_{0}^{y} \tan t ~dt ~\sqrt {1+(\tan y)^2} ~dy$$
$$ S= 2\pi \int_{0}^{\pi/3} \int_{0}^{y} \tan t ~dt ~\sec y ~dy $$
I am stuck here , What should I do now?