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0 Please help me to prove that helicoid whose parametric equation is given by $$x=ucos v, y= usin v, z=pu$$ is developable, where $p$ is a constant and $u,v$ are the curvelinear coordinates of the surface. Thank you. differential-geometry surfaces share | cite | improve this question asked Dec 4 '18 at 17:05 chandan mondal 175 8 add a comment  |  0 ...

0 0 Find using Residue Theorem $$int_{-1}^1 dfrac{dx}{(sqrt {1-x^2})(1+x^2)}$$ My try : I took the contour $C=C_1+C_2 $ where $C_1$ is the upper half of the circle with center at $0$ and and radius $1$ . and $C_2$ is the line joining $-1$ to $1$ . I now consider the function in $Bbb C$ to be $$int_C dfrac{dz}{(sqrt {1-z^2})(1+z^2)}$$ Over $C_2$ I get that $$int_{C_2} dfrac{dz}{(sqrt {1-z^2})(1+z^2)}$$ = $$int_{-1}^1 dfrac{dx}{(sqrt {1-x^2})(1+x^2)}$$ But I dont know if I am doing it right or not because every singularity of the function lies on the boundary of $C$ . Also I cant figure out how to calculate $$int_{C_1} dfrac{dz}{(sqrt {1-z^2})(1+z^2)}$$ Any help from someone here? Thanks for reading my post complex-analysis residue-calculus ...