What is the surface area of the solid formed by revolving $f\left(x\right)=\cos x$ on $[0,\frac{\pi }{3}]$ about the $x$-axis?

$\frac{2\pi }{3}(\sqrt{7}+3\ln \left(\frac{\sqrt{7}+\sqrt{2}}{3}\right)){\text{units}}^2$

$\frac{\pi }{4}(\sqrt{21}+4\ln \left(\frac{\sqrt{7}+\sqrt{3}}{2}\right)){\text{units}}^2$

$\frac{2\pi }{3}(2+3\ln \left(\frac{\sqrt{5}+\sqrt{2}}{3}\right)){\text{units}}^2$

$\frac{\pi }{4}(1+4\ln \left(\frac{\sqrt{5}+\sqrt{2}}{2}\right)){\text{units}}^2$