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

Gr = cos(θ)i + rsin(θ)j - 1/√( 1-r2)k
Gθ= -rsin(θ)i + rcos(θ)j
Gr X Gθ = rcos(θ)/√( 1-r2)i + rsin(θ)/√( 1-r2)j + rk

Now, surface area = |Gr X Gθ|drdθ = r(√( 2-r2)/√( 1-r2))drdθ

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