Question - Let S be the "football" surface formed by rotating the curve x=cos(z), y=0, -π/2 ≤ z ≤ π/2 around the z-axis. Find a parametrization for S and compute its surface area.

My attempt - Considering r to be the radius of ring formed at any value of z, x=rcos(θ), y=rsin(θ) , r=cos(z) I parametrize S as vector G = rcos(θ)i + rsin(θ)j + cos^{-1}(r)k

G_{r }= cos(θ)i + rsin(θ)j - 1/√( 1-r^{2})k

G_{θ}= -rsin(θ)i + rcos(θ)j

G_{r }X G_{θ }= rcos(θ)/√( 1-r^{2})i + rsin(θ)/√( 1-r^{2})j + rk

Now, surface area =∫∫|G_{r }X G_{θ}|drdθ =∫∫r(√( 2-r^{2})/√( 1-r^{2}))drdθ

However, I don't understand what'll be the limit of r in this expression. Please help.